Practice Test

Probability

Question 1 of 40

2026 Practice Test B Math, question 13

Question

A basket contains 3 apples, 4 oranges, 5 bananas, and 8 pears.

If one piece of fruit is randomly pulled from the basket, what is the probability that it will be a pear?

Select the place on the number line to plot the point.

On the test you click a point on the number line; here, choose it from the menu.

Point

Show the answer

Correct answer: the point 0.4

The total number of fruits is:

$3(\text{apples}) + 4(\text{oranges}) + 5(\text{bananas}) + 8(\text{pears}) = 20\text{ fruits}$

The probability of picking a pear is:

$\frac{8\ (\text{pears})}{20\ (\text{total})} = \frac{8}{20} \text{ or } 0.4$

Question analysis

A basket holds apples, oranges, bananas and pears, and students find the probability that a piece of fruit pulled at random is a pear, then plot it as a point on a number line from 0 to 1. It tests simple probability as a part of a whole. With no choices to pick from, the fraction has to be turned into tenths to find its place.

NYS standards: 6.SP.8a uniform probability models
6.SP.6 probability from 0 to 1
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of a simple event from counts is grade 6 work in New York. This one adds four whole numbers, forms one fraction and plots it among marked tenths, a shorter and more routine task than grade 7 probability.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Adding the four counts to 20 and writing 8 over 20 is routine, so most students succeed. The losses come from the number line: its decimal labels make it easy to read the 8 pears as 0.8, and a part-to-part fraction such as 8/12 falls between marks and has to be rounded, which should prompt a second look. Simplifying 8/20 to 2/5 and then to 4/10 makes the mark plain.

Something you must already know minor Working it out
The question does not say that a probability is the favorable count over the whole count, here the 8 pears over all 20 pieces of fruit rather than over the other 12. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The number line is labeled only in tenths from 0 to 1, so 8/20 has to be rewritten as a decimal before the right mark can be chosen. (Enright, Morley & Sheehan, 2002)

Strategy: Add up all the fruit for the total, write the number of pears over that total, rewrite the fraction in tenths, and select that mark on the line.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

0.8 Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Takes the 8 pears straight to the mark labeled 0.8, never dividing by the 20 pieces of fruit in the basket.
0.7 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Compares the 8 pears with the 12 other pieces of fruit, 8/12 or about 0.67, a part-to-part ratio, and plots the nearest tenth.
0.6 Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Puts the 12 apples, oranges and bananas over the 20 pieces of fruit, plotting the chance of not getting a pear instead of getting one.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of results you want}}{\text{total number of results}}$; P is the probability, the chance that something happens. It goes from 0 (never) to 1 (always). The top counts the results you want, here pears. The bottom counts every possible result, here every piece of fruit. It works when each result is equally likely, like grabbing fruit without looking.
Equal fractions: $\frac{a}{b} = \frac{a \div n}{b \div n}$; a is the top of a fraction, and b is the bottom. n is any number except 0. Dividing the top and the bottom by the same number gives an equal fraction. Use it to get 10 on the bottom, which is easy to write as a decimal.

Sort out the question
The basket has 3 apples, 4 oranges, 5 bananas and 8 pears. You pull out one piece without looking. You need the probability that it is a pear. Then you plot that number as a point. The number line runs from 0 to 1, with a mark at every tenth.
Count all the fruit
The bottom of the fraction is every piece of fruit you could pull out. Add all four kinds:
$3 + 4 + 5 + 8 = 20$
There are 20 pieces of fruit in all.
Write the probability as a fraction
Use the probability formula. There are 8 pears out of 20 pieces of fruit:
$P = \frac{8}{20}$
Watch out: the bottom is all the fruit, not just the other fruit. $\frac{8}{12}$ compares the pears with the 12 fruits that are not pears. It is about 0.67, and it leads you to the mark at 0.7. And 12 over 20 is the chance of not getting a pear.
Change it to tenths
The number line is marked in tenths. So rewrite the fraction with 10 on the bottom. Divide the top and the bottom by 2:
$\frac{8}{20} = \frac{8 \div 2}{20 \div 2} = \frac{4}{10} = 0.4$
Four tenths is written 0.4.
Watch out: the 8 pears do not make 0.8. That skips dividing by the 20 pieces of fruit.
Plot the point
Start at 0 and count four marks to the right: 0.1, 0.2, 0.3, 0.4. On the test, you click the mark labeled 0.4. On this site, you choose 0.4 from the menu.
So the answer is the point at 0.4.

Check your answer

The chance of not getting a pear is $\frac{12}{20} = 0.6$ . A pear and not a pear cover every result, so the two must add to 1: $0.4 + 0.6 = 1$ . Also, pears are less than half of the fruit, so the point should be left of 0.5. It is.

More questions like this

The original DOE question

Question 13 as shown on the DOE’s digital test, page 85 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test B Math, page 85, question 13.

Question 2 of 40

2026 Practice Test C Math, question 12

Question

A spinner has 5 congruent sections labeled as shown. One side of a fair coin is heads, and the other side is tails.

$A spinner shaped like a regular pentagon, divided into 5 congruent triangular sections that meet at the center, where an arrow points straight up. Going clockwise from the upper left, the sections are labeled 1, 2, 1, 4 and 3. Beside the spinner is a coin (a dime) showing its heads side, captioned Heads.$

The spinner is spun one time, and the coin is flipped. What is the probability that the spinner lands on 1 and the coin lands on tails?

Select the place on the number line that represents the probability.

On the test you click a point on the number line; here, choose it from the menu.

Point

Show the answer

Correct answer: the point 0.2

The spinner has five equal-sized sections. Two of the sections are labeled with a 1. So, the probability of landing on a 1 is two out of five, or 2/5.

A coin has two sides, one heads and one tails. So, the probability of landing on tails is one out of two, or 1/2.

The compound probability of the spinner landing on a 1 and the coin landing on tails can be calculated by multiplying 2/5 and 1/2 as shown:

$\frac{2}{5} \cdot \frac{1}{2} = \frac{2}{10} = 0.2$

Question analysis

A spinner with five equal sections is spun once and a fair coin is flipped. Students find the probability that the spinner lands on 1 and the coin shows tails, then select that point on a number line from 0 to 1. Counting the right sections and combining the two events correctly are the tricky parts.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.6 probability from 0 to 1
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the chance of two independent events together is grade 7 compound probability, here with one spinner and one coin. Counting sections from the picture and placing the result on a number line add two small steps.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 59 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The probability itself takes one multiplication of two simple fractions; the losses come before and after it. Before, the spinner has to be searched for every section labeled 1, and the coin's chance brought in as well. After, 1/5 has to be rewritten as 2/10 to find its point, since only 0, 1/2 and 1 are labeled. Checking that the result is smaller than both separate chances catches an addition.

Reading a figure, table or graph moderate Understanding the question
The two sections labeled 1, upper left and right, are not side by side, and the number line labels only 0, 1/2 and 1, so its nine points must be read as tenths. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Something you must already know minor Working it out
Nothing says that the chance of two independent events both happening is the product of their chances, 2/5 × 1/2, rather than their sum. (Anderson & Krathwohl, 2001)

Strategy: Count the sections labeled 1 to get the spinner's chance, take the coin's chance of tails, multiply the two because both must happen, and write the result in tenths.

Traps

Stops one step early: A choice is a number from partway through the solution.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Misreads the figure: Reads a value, label or tick spacing in a figure, table or graph wrongly.

The wrong answers, classified

0.4 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the spinner's chance of landing on 1, two sections out of five, and plots 2/5 without multiplying by the coin's chance of tails.
0.9 Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds the two chances, 2/5 + 1/2 = 9/10, instead of multiplying them, though both events must happen together.
0.1 Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Counts only one section labeled 1, so multiplies 1/5 by 1/2 and plots 1/10, missing the second 1 on the spinner.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of results you want}}{\text{total number of results}}$; P is the probability, the chance that something happens. It goes from 0 (never) to 1 (always). The top counts the results you want. The bottom counts all possible results. It works when every result is equally likely, like equal sections of a spinner.
Probability that two things both happen: $P(A \text{ and } B) = P(A) \times P(B)$; A and B are two events that do not affect each other, like a spin and a coin flip. P(A) and P(B) are their chances. Multiply them to get the chance that both happen. Needing both makes the chance smaller, never bigger.

Sort out the question
The spinner has 5 congruent sections. Congruent means the same size and shape, so each is equally likely. The coin has a heads side and a tails side. You need the chance that the spinner lands on 1 and the coin lands on tails.
Find the spinner’s chance
Two sections show 1: the upper left one and the one on the right. So 2 of the 5 equal sections give a 1:
$P(1) = \frac{2}{5}$
Watch out: the two 1s are not next to each other. Counting only one gives 1/5 here, and 0.1 at the end.
Find the coin’s chance
A fair coin is equally likely to land on either side. Tails is 1 of its 2 sides:
$P(\text{tails}) = \frac{1}{2}$
Multiply the two chances
The spin and the flip do not affect each other, and both must happen. So multiply:
$\frac{2}{5} \times \frac{1}{2} = \frac{2}{10} = 0.2$
Watch out: stopping at 2/5 gives 0.4. Adding gives $\frac{2}{5} + \frac{1}{2} = \frac{9}{10}$ , or 0.9. Both are wrong.
Select the point
On the test, the number line labels only 0, 1/2 and 1. Between 0 and 1 are 9 targets, one at each tenth. 0.2 is the second target to the right of 0. On this site, choose 0.2 from the Point menu.
So the answer is the point at 0.2.

Check your answer

List every result. The spinner has 5 sections and the coin has 2 sides, so there are $5 \times 2 = 10$ results, all equally likely. Just 2 of them are a 1 with tails, one for each section labeled 1. That is 2 out of 10, or 0.2.

More questions like this

The original DOE question

Question 12 as shown on the DOE’s digital test, page 85 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 85, question 12.

Question 3 of 40

2026 Practice Test C Math, question 33

Question

A bag contains 40 tiles that are either red, green, or blue. A tile is selected at random, the color is recorded, and the tile is put back in the bag. This process is carried out 120 times, and 18 of those times a red tile is selected. Based on the information, what is the most likely number of red tiles in the bag?

Show the answer

Correct answer: A

First, determine what fraction of the sample were red tiles, given 18 of the 120 tiles drawn were red.

$\frac{18 \div 6}{120 \div 6} = \frac{3}{20}$

To predict how many of the actual 40 tiles in the bag are red, multiply 40 by the fraction of red tiles in the sample.

$\frac{3}{20}(40) = \frac{120}{20} = 6$

Question analysis

A bag holds 40 red, green and blue tiles, and a tile is drawn and put back 120 times, with red coming up 18 times. Students predict the most likely number of red tiles in the bag. It tests using experimental results to estimate what is in the bag, and the draws outnumber the tiles, which can confuse the scaling.

NYS standards: 6.SP.7 long-run relative frequency
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Predicting from repeated trials is grade 6 probability, and grade 7 students make similar estimates from samples. The twist is scaling down, from 120 draws to a bag of 40 tiles, rather than up to a larger group.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 61 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

There are more draws than tiles, which is what makes this unusual: the 120 draws must be scaled down to a bag of 40, not up to a bigger group. Red came up 18 times because each tile went back in the bag, so 18 is not a count of red tiles. Since there are three draws for every tile, dividing the 18 red draws by 3 gives the same estimate as taking 3/20 of 40.

Turning words into math minor Understanding the question
The 18 red draws come from 120 draws with the tile put back each time, so they give a share, 3/20, to take of the 40 tiles, not a count of red tiles. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Students must know that the share of red in repeated draws estimates the share of red tiles in the bag; the stem never says to multiply 40 by that share. (Anderson & Krathwohl, 2001)

Strategy: Write the red draws as a fraction of all 120 draws, simplify it, and take that same fraction of the 40 tiles in the bag.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds that red came up in 18 of 120 draws, or 15%, and gives 15 as the count, stopping before taking that share of the 40 tiles.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives 18, the number of times red was drawn, as the number of red tiles, though each tile went back and could be drawn many times.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working from 18 red draws in 120 gives 30 red tiles, three quarters of the bag, when red came up in fewer than one draw in six.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Predicting from repeated draws: $\text{red tiles} \approx \frac{\text{red draws}}{\text{all draws}} \times \text{tiles in the bag}$; The fraction is the share of draws that came up red. That share is your best guess for the share of tiles that are red. Multiply it by the number of tiles in the bag to predict the red tiles. The sign ≈ means about equal.

Sort out the question
The bag has 40 tiles, each red, green or blue. A tile is drawn, its color is written down, and it goes back in the bag. This happens 120 times, and red comes up 18 times.
You need the most likely number of red tiles in the bag.
See why 18 is not the answer
Each tile goes back in the bag after it is drawn. So the same red tile can be drawn again and again. That means 18 counts draws, not tiles.
Watch out: choice C, 18, mixes up the number of red draws with the number of red tiles.
Find the share of red draws
Red came up 18 times out of 120. Simplify by dividing the top and bottom by 6:
$\frac{18}{120} = \frac{3}{20}$
As a decimal, that is 0.15, or 15%.
Watch out: 15 is a percent here, not a number of tiles. Stopping now gives choice B.
Take that share of the bag
Use the same share of the 40 tiles in the bag:
$\frac{3}{20} \times 40 = \frac{120}{20} = 6$
So about 6 of the 40 tiles are red.
Give the answer
The best prediction is 6 red tiles. The other 34 tiles are probably green or blue.
So the answer is A, 6.

Check your answer

There were 120 draws for 40 tiles. That is 3 draws for every tile. So divide the red draws by 3: $18 \div 3 = 6$ . Same answer. Choice D, 30 red tiles, would be three quarters of the bag. Then red would come up far more than 18 times.

More questions like this

The original DOE question

Question 33 as shown on the DOE’s digital test, page 106 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 106, question 33.

Question 4 of 40

2026 Practice Test C Math, question 34

Question

A teacher will randomly select 3 students to be on a school committee. He will choose either a student from Classroom A or a student from Classroom B with each selection. This tree diagram shows the selection possibilities.

$A tree diagram with three columns of boxes. From a starting point, two arrows lead to boxes A and B. From each of those, two arrows lead to boxes A and B, four boxes in all. From each of those four, two arrows lead to boxes A and B, so the last column has eight boxes: A, B, A, B, A, B, A, B.$

What is the probability that the teacher will select 2 students from Classroom A and 1 student from Classroom B (in any order) for the committee?

A $\frac{1}{8}$
B $\frac{3}{8}$
C $\frac{1}{2}$
D $\frac{3}{4}$

Show the answer

Correct answer: B

The tree diagram shows that each of the 3 randomly selected students is from Classroom A or Classroom B. So the first chosen is either A or B, represented by the left column of the diagram.

The second student is either A or B, so if the first is A, the first two chosen will be either AA or AB. If the first is B, then the first two chosen will be either BA or BB, represented by the middle column.

The third student is either A or B, so if the first 2 were AA, the group of 3 would be either AAA or AAB. These outcomes are represented by the top two boxes of the right column. This pattern is followed so that 8 distinct outcomes can be visualized in order:

AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB

Since the order does not matter, there are 3 groups made up of 2 students from Classroom A and 1 student from Classroom B: AAB, ABA, and BAA.

Out of 8 possible outcomes, 3 outcomes have 2 students from Classroom A and 1 student from Classroom B. Therefore, the probability is 3 out of 8, or $\frac{3}{8}$ .

Question analysis

A teacher picks 3 students for a committee, each from Classroom A or Classroom B, and a tree diagram shows every possible sequence of picks. Students find the probability of getting 2 students from A and 1 from B in any order. Reading the tree as a list of outcomes, and counting each order, is the key.

NYS standards: 7.SP.8b sample spaces for compound events
7.SP.8a compound event as fraction of outcomes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a compound probability from a tree diagram is a grade 7 standard, and here the tree is already drawn. Listing its paths and counting the ones that fit the event, in any order, is typical grade 7 classwork.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 65 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The tree does the hard part of the setup, but it shows each outcome only as a path through three columns, not as a list. The parenthesis “in any order” is the key phrase: missing it leaves a single path, while reading the event loosely, as at least two from Classroom A, adds A, A, A. Writing out all eight paths and marking those with exactly two A’s settles the count.

Reading a figure, table or graph moderate Understanding the question
The tree prints only A or B in each box, so the eight three-pick outcomes, such as A, B, A, have to be read by tracing each path across all three columns. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Much to keep track of minor Working it out
All eight paths must be tracked while counting those with exactly two A’s and one B, and “in any order” means three of them count, not just A, A, B. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Follow each path through the tree to write out every three-letter outcome, mark the ones with exactly two A's and one B, and divide by the total number of outcomes.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Counts only the path A, A, B, missing the stem’s “in any order”, so it finds 1 of the 8 outcomes instead of 3.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Counts every outcome with at least two picks from Classroom A, adding A, A, A to the three right paths, so 4 of 8, or 1/2.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Counts every outcome that mixes the two classrooms, including those with two from B, so 6 of the 8 outcomes, or 3/4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance something happens. The top counts the outcomes that fit the event. The bottom counts every outcome that can happen. This works when every outcome is equally likely, like each path through this tree.

Sort out the question
The teacher picks 3 students, one at a time. Each pick is from Classroom A or Classroom B.
You need the chance of getting 2 students from A and 1 from B. The words “in any order” matter. They mean the B can be the first, second or third pick.
List every path in the tree
Each path from left to right is one outcome, or possible result. The top path is A, then A, then A. Read all of them:
AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB
Each pick has 2 choices, so there are $2 \times 2 \times 2 = 8$ outcomes in all.
Mark the ones that fit
Look for exactly two A’s and one B. Three outcomes fit: AAB, ABA and BAA.
Watch out: counting only AAB misses “in any order” and gives $\frac{1}{8}$ , choice A. AAA has three A’s, not two. Counting it too gives $\frac{4}{8} = \frac{1}{2}$ , choice C. Counting every mixed outcome gives $\frac{6}{8} = \frac{3}{4}$ , choice D, but three of those have two B’s.
Find the probability
Use the probability formula. 3 outcomes fit, out of 8:
$P = \frac{3}{8}$
So the answer is B, $\frac{3}{8}$ .

Check your answer

Count every outcome by type. Two A’s and one B: 3 outcomes. Two B’s and one A: 3 outcomes (ABB, BAB, BBA). All A’s: 1. All B’s: 1. $3 + 3 + 1 + 1 = 8$ , so each outcome was counted once.

More questions like this

The original DOE question

Question 34 as shown on the DOE’s digital test, page 107 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 107, question 34.

Question 5 of 40

2026 Practice Test C Math, question 36

Question

A deck of 52 playing cards contains 13 hearts. Sara has 12 cards from this deck in her hand, 3 of which are hearts. The other cards remain in the deck. What is the probability that a card drawn at random from the remainder of the deck will be a heart?

A $\frac{1}{10}$
B $\frac{1}{4}$
C $\frac{3}{13}$
D $\frac{1}{3}$

Show the answer

Correct answer: B

First, determine the number of cards in the remaining deck:

$52 - 12 = 40$

Next, determine the number of hearts left in the remaining deck:

$13 - 3 = 10$

Last, determine the probability of drawing a heart from the remaining deck:

$\frac{10\text{ hearts remaining}}{40\text{ total cards remaining}} = \frac{10}{40} = \frac{1}{4}$

Question analysis

A standard deck of 52 playing cards has 13 hearts, and Sara holds 12 cards, 3 of them hearts. Students find the probability that a card drawn at random from the rest of the deck is a heart. Her hand changes both the number of hearts and the number of cards left, so both counts must be updated.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Finding a simple probability is grade 6 content. This one adds a step many grade 7 problems skip: the cards in Sara's hand must be taken out of both the hearts and the whole deck first.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of minor
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 51 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The careful route takes two subtractions and a fraction: 10 hearts among 40 cards. The numbers are kind, because Sara’s hand is a quarter hearts, just like the full deck, so a student who ignores her hand and divides 13 by 52 also reaches 1/4. The wrong choices come from describing her hand instead of the deck, or from comparing hearts with the cards that are not hearts.

Much to keep track of minor Working it out
Four counts are in play, 52 cards, 13 hearts, Sara’s 12 cards and her 3 hearts, and both the hearts and the total must be reduced before the fraction is formed. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
“The other cards remain in the deck” and “the remainder of the deck” point to the 40 cards Sara is not holding, not to the full 52 or to her hand. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Work out how many cards are left in the deck and how many of them are hearts, then write the hearts left over the cards left and simplify.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the 10 hearts left in the deck to the 10 in 1/10, never dividing them by the 40 cards that remain.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Puts Sara’s 3 hearts over the deck’s 13, or her 12 cards over 52, both 3/13, which describes her hand rather than the cards left.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Compares the 10 hearts left with the 30 other cards left, a part-to-part ratio of 1 to 3, instead of with all 40 cards.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance something happens. The top counts the cards that are hearts. The bottom counts every card you could draw. Each card is equally likely to be drawn, so the fraction gives the chance.

Sort out the question
A full deck has 52 cards, and 13 of them are hearts. Sara holds 12 cards, and 3 of those are hearts. The rest of the cards stay in the deck.
You need the chance that a card drawn from the rest of the deck is a heart. Her hand changes both counts.
Count the cards left
Take Sara’s 12 cards out of the deck:
$52 - 12 = 40$
So 40 cards are left in the deck.
Count the hearts left
Sara holds 3 of the 13 hearts. Take those out:
$13 - 3 = 10$
So 10 hearts are left in the deck.
Find the probability
Use the probability formula. Put the hearts left over all the cards left, then simplify:
$P = \frac{10}{40} = \frac{1}{4}$
Watch out: comparing the 10 hearts with the 30 cards that are not hearts gives $\frac{1}{3}$ , choice D. The bottom must be all 40 cards. Choice C, $\frac{3}{13}$ , describes Sara’s hand, not the deck. Choice A, $\frac{1}{10}$ , uses the 10 hearts but forgets the 40 cards.
Give the answer
The chance of drawing a heart from the rest of the deck is 1 in 4.
So the answer is B, $\frac{1}{4}$ .

Check your answer

10 is one quarter of 40, because $4 \times 10 = 40$ . Here is another check. Sara’s hand is 3 hearts out of 12, which is also one quarter. Taking a hand that is one quarter hearts out of a deck that is one quarter hearts leaves one quarter hearts.

More questions like this

The original DOE question

Question 36 as shown on the DOE’s digital test, page 109 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 109, question 36.

Question 6 of 40

2026 Practice Test C Math, question 37

Question

The table shows the probability of randomly picking each flavor of candy from a bowl.

PICKING CANDY FROM
A BOWL

Flavor	Probability
Butterscotch	0.57
Lemon	0.095
Maple	0.125
Strawberry	0.21

Which flavor is most likely to be randomly picked?

Abutterscotch
Blemon
Cmaple
Dstrawberry

Show the answer

Correct answer: A

The flavor most likely to be picked at random is butterscotch, because its probability is the greatest of the probabilities in the list. More than half the candies in the bowl are butterscotch.

Question analysis

A table lists the probability of picking butterscotch, lemon, maple or strawberry candy at random from a bowl, each as a decimal. Students decide which flavor is most likely to be picked. It tests the meaning of probability and comparing decimals, which have different numbers of digits here and must be compared by place value.

NYS standards: 6.SP.6 probability from 0 to 1
6.NS.7 ordering and absolute value
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Reading probabilities as likelihoods is grade 6 content, and this needs only a comparison of four decimals. The one demand is comparing decimals with different numbers of digits, a skill from earlier grades.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Error-prone arithmetic minor
Solution: 1 step
Reading level: Question and choices: about grade 9, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One idea answers the question: the most likely flavor has the greatest probability. The only obstacle is comparing decimals of different lengths, where 0.125 and 0.095 look bigger than 0.57 to students who count digits. Writing each as thousandths, 570, 95, 125 and 210, or noticing that 0.57 alone is more than one half, makes the order plain.

Error-prone arithmetic minor Working it out
The probabilities have two and three decimal places, 0.57 beside 0.095 and 0.125, so comparing them by length, or by their digits as whole numbers, ranks them wrongly. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Line the four decimals up by place value, adding zeros so each has three decimal places, and compare them digit by digit starting from the tenths place.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

B Misreads the figure (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Reveals a slip Expected pull: weak
Reads lemon’s 0.095 as 0.95, overlooking the 0 in the tenths place, so it looks almost certain and larger than 0.57.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the decimal with more digits as the larger, or compares 125 with 57 as whole numbers, so 0.125 seems bigger than 0.57.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common misreading makes 0.21 the largest, since it is neither the longest decimal nor the one with the biggest digits, so it mostly draws guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{candies of that flavor}}{\text{all candies in the bowl}}$; P is the probability of picking a flavor. It is a number from 0 to 1. The more candies of a flavor there are, the bigger its probability. So the flavor with the greatest probability is the most likely to be picked.

Sort out the question
The table gives the probability of picking each of four flavors. You need the flavor that is most likely to be picked.
The greater the probability, the more likely the pick. So you need the greatest of the four decimals.
Line up the decimals
The decimals have different numbers of digits. Add zeros at the end so each has three decimal places. This does not change their values:
Flavor Probability
Butterscotch 0.570
Lemon 0.095
Maple 0.125
Strawberry 0.210
Compare the tenths digits
Start with the first digit after the decimal point, the tenths place. The tenths digits are 5, 0, 1 and 2. The biggest is 5, so 0.570 is the greatest.
Watch out: 0.125 has more digits than 0.57, but it is smaller. Compare place by place, not by length. That mistake leads to choice C. Also, lemon is 0.095, not 0.95, so choice B is out.
Give the answer
Butterscotch has the greatest probability, 0.57. So it is the flavor most likely to be picked.
So the answer is A, butterscotch.

Flavor	Probability
Butterscotch	0.570
Lemon	0.095
Maple	0.125
Strawberry	0.210

Check your answer

0.57 is more than 0.5, which is one half. So more than half the candies are butterscotch. The other three flavors add up to $0.095 + 0.125 + 0.21 = 0.43$ , less than half. So butterscotch must be the most likely.

More questions like this

The original DOE question

Question 37 as shown on the DOE’s digital test, page 110 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test C Math, page 110, question 37.

Question 7 of 40

2026 Practice Test D Math, question 5

Question

In a school raffle, 1 out of every 25 students wins a prize. Each prize includes a randomly selected trading card from a set of 100 different cards, labeled C1 through C100, with each card equally likely to be chosen. If one student is selected at random, what is the probability that the student both wins a prize and receives Card C17?

Write an expression that represents the probability that a student wins the prize and receives Card C17.

Enter your answer in the space provided. Enter only your answer.

On the test you build the answer on an on-screen keypad; here, type it. Use ^ for a power, sqrt( ) or √ for a root, pi for π, and <= or >= for ≤ and ≥.

Show the answer

Correct answer: 1/25 · 1/100

To calculate the joint probability, first determine the probability of each independent event.

The probability that a randomly selected student wins the prize is $\frac{1}{25}$ .

The probability that a randomly selected student will receive Card C17 is $\frac{1}{100}$ , because Card C17 is 1 card out of 100 labeled cards.

The probability that a student wins the prize and receives Card C17 is the product of those two probabilities, $\frac{1}{25} \cdot \frac{1}{100}$ .

Question analysis

In a school raffle, 1 out of every 25 students wins a prize, and each prize includes one of 100 equally likely trading cards. Students type an expression for the probability that a randomly chosen student both wins and gets one particular card. It tests the probability of two independent events, and combining the two chances correctly is the main hurdle.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.8a uniform probability models
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 students find compound probabilities, usually by listing outcomes for coins or number cubes. These two stages have 25 and 100 outcomes, far too many pairs to list, so a rule must be used, and the answer is typed as an expression.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 87 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each chance is stated almost directly, 1 out of 25 and 1 out of 100, so the question tests one rule: both events must happen, and the chances multiply. The word “and” pulls students toward adding, which gives a result larger than either chance, a sign that something is wrong. Others write only the card's chance and forget that the student must first win a prize.

Something you must already know moderate Working it out
The question never says that the chance of two independent events both happening is the product of their chances, so students must bring that rule to the 1/25 and the 1/100 themselves. (Anderson & Krathwohl, 2001)
An answer in an unexpected form minor Choosing the answer
The answer is typed as an expression with no choices to compare, so a sum of the two chances, which looks just as finished, gets no warning on screen. (Enright, Morley & Sheehan, 2002)

Strategy: Write each event's chance as a fraction, then apply the rule for two independent events that must both happen, and check that the result is smaller than either chance alone.

Traps

Follows a key word: Picks the operation from a single word rather than the situation: "removed" so subtract, "of" so multiply, "more" so add, when the situation needs something else.
Stops one step early: A choice is a number from partway through the solution.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

1/25 + 1/100 Follows a key word (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Reads “both wins a prize and receives Card C17” as a cue to add, so sums the chances, though needing both events must be less likely than either.
1/100 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Writes only the chance of receiving Card C17 and leaves out that the student must first be one of the 1 in 25 who win a prize.
1/125 A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: weak
Combines “1 out of 25” and “1 out of 100” by adding the totals, 25 + 100, a faulty rule that gives neither the sum nor the product.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance that something happens, from 0 to 1. The top counts the outcomes you want. The bottom counts all the outcomes that could happen. Use it when every outcome is equally likely, like one card picked from 100.
Probability that two independent events both happen: $P(A \text{ and } B) = P(A) \times P(B)$; A and B are two events. Independent means one does not change the chance of the other. P(A) is the chance of A, and P(B) is the chance of B. Multiply them to get the chance that both happen. The result is smaller than either chance, since both things must happen.

Sort out the question
Two things must both happen. First, the student wins a prize. Second, the card in that prize is Card C17. You need an expression for that chance. An expression is a math phrase made of numbers and signs, with no equal sign. You do not have to work it out to one number.
Find the chance of winning
The question says “1 out of every 25 students wins a prize.” One student is the part that wins, and 25 is the whole group. So:
$P(\text{win}) = \frac{1}{25}$
Find the chance of Card C17
Each prize has one card, picked at random from 100 cards. Every card is equally likely. Card C17 is 1 of those 100 cards:
$P(\text{C17}) = \frac{1}{100}$
Multiply the two chances
Winning a prize does not change which card is inside it. So the two events are independent. To find the chance that both happen, multiply:
$P(\text{win and C17}) = \frac{1}{25} \cdot \frac{1}{100}$
The dot · means times. Worked out, this is 1/2500.
Watch out: the word and does not mean add here. 1/25 + 1/100 is bigger than either chance, but needing both things must be less likely. Writing only 1/100 forgets that the student must win a prize first.
Enter the expression
On the test, press the fraction key, a small box over a small box. Type 1 on top and 25 on the bottom. Move the cursor out of the fraction with the → key. Press the dot key for times. Then use the fraction key again for 1 over 100. On this site, type 1/25 * 1/100 or 1/25 · 1/100. The value 1/2500 also counts.
So you would enter 1/25 · 1/100.

Check your answer

Picture 2,500 students. 1 in 25 wins, so about 100 students win a prize. Of those 100, about 1 gets Card C17. That is 1 student out of 2,500, which matches 1/25 · 1/100 = 1/2500. It is smaller than 1/25 and 1/100, as it should be.

More questions like this

The original DOE question

Question 5 as shown on the DOE’s digital test, page 78 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 78, question 5.

Question 8 of 40

2026 Practice Test D Math, question 7

Question

A middle school library has a shelf with 40 books, each labeled with a different number from 201 to 240. One book is selected at random. What is the probability, as a percentage, of selecting book number 223?

Enter your answer in the space provided.

Fill-in question: work out the answer and type it as a number.

Show the answer

Correct answer: 2.5%

The probability that book number 223 will be selected at random from 40 books is 1 out of 40. To find the probability as a percentage, convert the fraction to a decimal.

The chance of selecting one specific book is $\frac{1}{40} = 0.025$ .

Convert the decimal to a percentage: $0.025 \times 100 = 2.5\%$

Question analysis

A library shelf holds 40 books numbered 201 to 240, and one is picked at random. Students find the probability of picking one particular book and type it as a percentage. It tests simple probability and converting a fraction to a percent; the book numbers play no part, and the decimal point needs care.

NYS standards: 6.SP.8a uniform probability models
6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of one equally likely outcome is grade 6 content. The only further step is rewriting the fraction as a percent, which needs care with place value, so it is lighter than grade 7 compound probability.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: An answer in an unexpected form moderate
Error-prone arithmetic minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The book numbers, 201 to 240, and the chosen number 223 play no part, since every book is equally likely and the shelf holds 40. The chance is 1 in 40, and the real demand is the form: changing it to a percent means a decimal with a zero in the tenths place, then a shift of two places. Students who know that 1/4 is 25% can check that 1/40 must be a tenth of that.

An answer in an unexpected form moderate Choosing the answer
Only the bold word percentage sets the form, since the empty box shows no % sign, so the probability 1/40 must be turned into 2.5, not entered as the fraction or as 0.025. (Enright, Morley & Sheehan, 2002)
Error-prone arithmetic minor Working it out
Dividing 1 by 40 gives 0.025, with a zero right after the decimal point, and losing that zero turns the percent into 25. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the books, write the chance of picking one particular book as a fraction, then turn the fraction into a decimal and multiply by 100 to get the percent.

Traps

Stops one step early: A choice is a number from partway through the solution.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Information you don't need: The question gives numbers or facts that play no part in the answer.

The wrong answers, classified

0.025 Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds the probability as a decimal, 1 ÷ 40 = 0.025, and enters it without multiplying by 100, though the box asks for a percentage.
25 An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: moderate
Loses the zero after the decimal point when dividing, taking 1 ÷ 40 as 0.25, as 1 ÷ 4 would be, and so enters 25.
1/40 Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Enters the probability as the fraction 1/40, the form it was found in, overlooking the bold word percentage in the question.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of one outcome: $P = \frac{\text{outcomes you want}}{\text{all possible outcomes}}$; P is the probability, the chance of an event. The top counts the outcomes you want. The bottom counts all the equally likely outcomes. Use it when each outcome has the same chance, like picking one book at random.
Changing a fraction to a percent: $\text{percent} = \frac{a}{b} \times 100$; a/b is the fraction. Percent means out of 100. Multiplying by 100 tells you how many out of 100 the fraction is. For example, 1/2 × 100 = 50, so 1/2 is 50%.

Sort out the question
The shelf has 40 books, numbered 201 to 240. One book is picked at random, so every book has the same chance. You need the chance of picking book 223, written as a percentage. Book 223 is on the shelf, since 223 is between 201 and 240.
Write the probability as a fraction
Only 1 book is number 223, and there are 40 books in all:
$P = \frac{1}{40}$
The numbers on the labels do not matter beyond this. They just give each book its own name.
Change the fraction to a decimal
Divide 1 by 40. It helps to start from 1/4 = 0.25. Since 40 is ten times 4, 1/40 is one tenth of 1/4:
$1 \div 40 = 0.025$
Watch out: there is a 0 right after the decimal point. Losing it gives 0.25 and then 25, which is wrong.
Change the decimal to a percent
Multiply by 100 to get a percent. That moves the decimal point two places to the right:
$0.025 \times 100 = 2.5$
So the chance is 2.5%.
Watch out: 0.025 is the probability as a decimal, and 1/40 is the same chance as a fraction. The question asks for a percentage, so neither one is the answer here.
Enter the number
The box shows no % sign. Type 2.5. Adding the sign, as 2.5%, also counts.
So you would enter 2.5.

Check your answer

2.5% means 2.5 out of every 100. Out of 40 books, that is 40 × 0.025 = 1 book, the one numbered 223. Also, one book out of 40 should be a small chance, and 2.5% is small.

More questions like this

The original DOE question

Question 7 as shown on the DOE’s digital test, page 80 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 80, question 7.

Question 9 of 40

2026 Practice Test D Math, question 29

Question

Karl has one red spinner and one blue spinner. Each spinner is divided into 4 equal sections, numbered 1 through 4. He spins each spinner once and writes down the number that each lands on. What is the probability that the two numbers, when multiplied together, will have 4 as a product?

A $\frac{1}{16}$
B $\frac{3}{16}$
C $\frac{4}{7}$
D $\frac{7}{16}$

Show the answer

Correct answer: B

There are four possible outcomes for each spin of the spinner:

1, 2, 3, 4

The possible outcomes of two spins are:

$\begin{array}{cccc} 11 & 12 & 13 & \boxed{14} \\ 21 & \boxed{22} & 23 & 24 \\ 31 & 32 & 33 & 34 \\ \boxed{41} & 42 & 43 & 44 \end{array}$

Of the 16 possible outcomes, only 3 pairings have a product of 4: 14, 22, and 41.

Therefore, the probability that the two numbers resulting from two consecutive spins have a product of 4 is $\frac{3}{16}$ .

Question analysis

Karl spins a red spinner and a blue spinner, each with four equal sections numbered 1 to 4, and multiplies the two results. This compound probability question asks for the chance that the product is 4. Students build the sample space of outcomes, and listing every pair rather than the ones that come to mind is the safe route.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding a compound probability from a list or table of equally likely outcomes is standard grade 7 work. Two four-section spinners give a small sample space that is quick to write out, and the event needs one careful scan of the products.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Much to keep track of minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 52 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question is decided by the count of favorable outcomes, not by any calculation. Students who think of 4 only as 2 × 2 find one outcome, and those who think in factor pairs find two, which is not a listed choice and should send them back to the table. A 4-by-4 grid of products, red down the side and blue across, shows all three cells at once.

Something you must already know moderate Working it out
Nothing in the stem says that red 1 with blue 4 and red 4 with blue 1 are separate outcomes, or that two four-part spinners give 16 equally likely pairs. (Anderson & Krathwohl, 2001)
Much to keep track of minor Working it out
All 16 pairs must be scanned for a product of 4, and the two pairs that use 1 and 4 are easy to miss beside the obvious 2 times 2. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Make a 4-by-4 table with the red spinner's numbers down the side and the blue spinner's across the top, write each product in its cell, and count the cells that match.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Finds only the double, 2 and 2, and stops, missing the two outcomes in which a 1 on one spinner meets a 4 on the other.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No count of the 16 outcomes gives a denominator of 7, so 4/7 has no working behind it and mostly attracts guesses.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Counts the 7 outcomes in which either spinner shows a 4, reading “4 as a product” as a pair that simply contains a 4.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of outcomes you want}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. An outcome is one possible result. This works when every outcome is equally likely, like landing on equal sections of a spinner.
Counting outcomes of two actions: $\text{total outcomes} = a \times b$; a is the number of results for the first action, and b is the number for the second. Each result of the first can go with each result of the second, so you multiply.

Sort out the question
Karl spins two spinners, one red and one blue. Each has 4 equal sections, numbered 1 to 4. He multiplies the two numbers. You need the probability that the product, the answer to a multiplication, is 4.
Count all the outcomes
The red spinner has 4 results, and so does the blue one. Each red result can go with each blue result:
$4 \times 4 = 16$
So there are 16 outcomes, all equally likely. Red 1 with blue 4 is not the same as red 4 with blue 1. They come from different spinners.
Make a table of products
Write each product in a table. Red goes down the side and blue goes across the top.
Red × Blue 1 2 3 4
1 1 2 3 4
2 2 4 6 8
3 3 6 9 12
4 4 8 12 16
Three cells show 4: red 1 with blue 4, red 2 with blue 2, and red 4 with blue 1.
Watch out: thinking of 4 only as 2 × 2 finds one outcome. That gives 1/16, choice A.
Find the probability
Use the probability rule. Three outcomes give 4, out of 16 in all:
$P = \frac{3}{16}$
Watch out: 7 outcomes have a 4 on at least one spinner. But that counts pairs that show a 4, not pairs whose product is 4. That mistake gives 7/16, choice D.
So the answer is B, $\frac{3}{16}$ .

Red × Blue	1	2	3	4
1	1	2	3	4
2	2	4	6	8
3	3	6	9	12
4	4	8	12	16

Check your answer

List the ways to multiply two numbers from 1 to 4 and get 4: $1 \times 4$ , $2 \times 2$ and $4 \times 1$ . That is 3 ways. The table has 16 cells, and 13 of them are not 4. So a small chance like 3/16 makes sense.

More questions like this

The original DOE question

Question 29 as shown on the DOE’s digital test, page 102 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 102, question 29.

Question 10 of 40

2026 Practice Test D Math, question 34

Question

Last week, Emily rode the bus to school on 3 of the 5 mornings and rode the bus home on 4 of the 5 afternoons. Based on last week’s events, what is the probability that Emily will ride the bus in the morning and the afternoon next Monday?

A $\frac{1}{5}$
B $\frac{7}{25}$
C $\frac{12}{25}$
D $\frac{7}{10}$

Show the answer

Correct answer: C

Using the joint probability formula, find the product of the two independent events. The probability that Emily will ride the bus to school on any given morning is $\frac{3}{5}$ . The probability that Emily will ride the bus home on any given afternoon is $\frac{4}{5}$ . Therefore, the probability that Emily will ride the bus in the morning and the afternoon is $\frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$ .

Question analysis

Last week Emily rode the bus to school on 3 of 5 mornings and home on 4 of 5 afternoons. This probability question asks how likely she is to ride the bus both ways next Monday, based on those frequencies. Students turn each count into a probability and combine the two, deciding how chances join when both events must happen.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.SP.8b probability models from observed frequencies
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Compound probability is grade 7 work, often done with a table of outcomes. Each chance here comes from one week's frequencies, and joining the morning and afternoon events in one step is about the usual grade 7 demand.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 48 words.
Corrected from the DOE handbook: An inaccurate detail in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the two chances are written, the question is how they combine. Pooling the week into 7 bus rides out of 10 trips, or adding fractions in some way, gives listed choices that look reasonable, since each lies between 0 and 1. Picturing the 25 pairings of a morning with an afternoon, 3 × 4 = 12 of which have both rides, shows why the chances multiply.

Something you must already know moderate Working it out
The rule that two independent events both happen with the product of their chances, 3/5 times 4/5, is not given, and “the morning and the afternoon” is the only signal for it. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The counts “3 of the 5 mornings” and “4 of the 5 afternoons” have to become two separate probabilities based on last week, not one pool of 10 trips. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Turn each count into a probability, then multiply them, since both rides must happen; picturing a table of 5 mornings by 5 afternoons shows why multiplying works.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Subtracts the morning chance from the afternoon chance, 4/5 − 3/5 = 1/5, an operation with no meaning for two rides that must both happen.
B A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Adds the numerators, 3 + 4, but multiplies the denominators, 5 × 5, mixing the rule for adding fractions with the rule for multiplying them.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Pools the week’s trips into 7 bus rides out of 10, a rate for a single trip, as if riding both ways next Monday were just as likely.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “The probability that Emily will ride the bus to school on any given afternoon is 4/5.” The question says she rode the bus home on 4 of the 5 afternoons; her rides to school are the morning ones. The site now says “ride the bus home on any given afternoon”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability from past results: $P = \frac{\text{times it happened}}{\text{times it could have happened}}$; P is the probability, the chance something happens. You use what happened before to predict the chance it happens again. Emily rode in 3 of 5 mornings, so the chance for one morning is 3/5.
Both of two independent events: $P(A \text{ and } B) = P(A) \times P(B)$; A and B are two events. Independent means one does not change the chance of the other. To find the chance that both happen, multiply their chances. The result is smaller than either one, since both must happen.

Sort out the question
Last week Emily:
- rode the bus to school on 3 of the 5 mornings;
- rode the bus home on 4 of the 5 afternoons.
You need the chance that she rides the bus in the morning and in the afternoon next Monday. The word and means both rides must happen.
Write each chance
Use last week to find each chance:
- Bus to school on a morning: $\frac{3}{5}$
- Bus home on an afternoon: $\frac{4}{5}$
Keep them apart. The morning rides go to school, and the afternoon rides go home.
Multiply the two chances
Both rides must happen, so multiply:
$\frac{3}{5} \times \frac{4}{5} = \frac{12}{25}$
Why multiply? Pair each of the 5 mornings with each of the 5 afternoons. That makes 25 pairs. In $3 \times 4 = 12$ of them, she rides both times. So the chance is 12 out of 25.
Avoid the tempting wrong choices
Watch out: pooling the week gives 7 rides out of 10 trips, or 7/10, choice D. That is the chance for one trip, not for two trips on one day.
Adding the tops but multiplying the bottoms gives 7/25, choice B. Subtracting the chances gives 1/5, choice A.
So the answer is C, $\frac{12}{25}$ .

Check your answer

The chance of both rides must be less than the chance of each ride alone. As decimals, 12/25 = 0.48. That is less than 3/5 = 0.6 and 4/5 = 0.8. Choice D, 7/10 = 0.7, is more than 3/5, so it cannot be right.

More questions like this

The original DOE question

Question 11 of 40

2026 Practice Test D Math, question 39

Question

The table shows the probabilities of a hockey team scoring different numbers of goals in a game.

HOCKEY GOAL PROBABILITIES

Number of Goals	Probability
0	0.16
1	0.21
2	0.29
3	0.21
4	0.09
5	0.04

What is the probability that the team will score three or more goals in the next game?

A0.13
B0.21
C0.34
D0.50

Show the answer

Correct answer: C

To find the probability that any one of several mutually exclusive events occurs, use the addition rule, and add the probabilities of each event:

$\mathrm{P}(\mathrm{A} \text{ or } \mathrm{B}) = \mathrm{P}(\mathrm{A}) + \mathrm{P}(\mathrm{B})$

Using the data in the table, the probability that the team will score 3 or more goals in the next game is represented by

$\mathrm{P}(3) + \mathrm{P}(4) + \mathrm{P}(5) =$

$0.21 + 0.09 + 0.04 = 0.34.$

Question analysis

A table gives the probabilities of a hockey team scoring 0 to 5 goals in a game, and students find the chance that it scores three or more goals in the next one. This probability question tests adding the probabilities of separate outcomes. Reading exactly which rows the phrase covers is the main point of care.

NYS standards: 6.SP.8 probability models for simple events
6.NS.3 multi-digit decimal operations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of an event from a given probability model is grade 6 work in New York. Here the model is a table, and the task is choosing the right rows and adding decimals, shorter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Adding three decimals is easy, so the question turns on which rows the words cover. Including the 3-goal row, and not stopping at it, are the two decisions, and each misreading gives a listed choice. Checking the complement, 1 − (0.16 + 0.21 + 0.29), confirms both the sum and the choice of rows.

Turning words into math minor Understanding the question
The phrase “three or more goals” has to become the three rows for 3, 4 and 5 goals, starting at the 3-goal row itself. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The table lists six probabilities, two of them both 0.21, so the rows for 3, 4 and 5 goals must be picked out by their goal counts, not by their values. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Decide which rows the words "three or more" cover, add their probabilities, and check by subtracting the total of the other rows from 1.

Traps

Counting the ends: Inclusive or exclusive ends, fence posts, consecutive numbers: an answer that is one too many or too few, or an endpoint drawn open where it belongs closed (or the reverse).
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Counting the ends (Brown & Burton, 1978)
Reveals a slip Expected pull: moderate
Adds only the rows for 4 and 5 goals, 0.09 + 0.04 = 0.13, reading three or more as more than three and leaving out the end row.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads only the row for exactly 3 goals, 0.21, overlooking the words “or more” and the two rows below it.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Counts 3 of the 6 rows, for 3, 4 and 5 goals, as half the outcomes, ignoring that the table gives them different probabilities.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Adding probabilities: $P(A \text{ or } B) = P(A) + P(B)$; A and B are two events that cannot both happen at once, like scoring exactly 3 goals and exactly 4 goals. The chance that one or the other happens is the sum of their chances.
The chance it does not happen: $P(\text{not } A) = 1 - P(A)$; The chances of all possible outcomes add up to 1. So the chance an event does not happen is 1 minus the chance it does. Use it to check an answer from the other side.

Sort out the question
The table gives the chance of each number of goals, from 0 to 5. You need the chance the team scores three or more goals in the next game.
Pick the right rows
Three or more means 3 goals, 4 goals or 5 goals. The row for 3 counts too. Read those three rows:
- 3 goals: 0.21
- 4 goals: 0.09
- 5 goals: 0.04
Watch out: leaving out the 3-goal row gives 0.09 + 0.04 = 0.13, choice A. Using only the 3-goal row gives 0.21, choice B.
Add the chances
A game can end with only one number of goals. So add the three chances:
$0.21 + 0.09 + 0.04 = 0.34$
Watch out: three of the six rows is half the rows, but that does not make the chance 0.50, choice D. The rows have different chances.
So the answer is C, 0.34.

Check your answer

Find the chance of fewer than 3 goals, the other rows: $0.16 + 0.21 + 0.29 = 0.66$ . Then the chance of 3 or more is $1 - 0.66 = 0.34$ . Both ways give 0.34.

More questions like this

The original DOE question

Question 39 as shown on the DOE’s digital test, page 112 of its printable version. — As printed in the DOE’s printable version of 2026 Practice Test D Math, page 112, question 39.

Question 12 of 40

2024 Test A Math, question 15

Question

A program on a computer randomly generates a sequence of whole numbers from 1 to 9, inclusive. If the computer generates a sequence of 300 numbers, what is the best prediction of the number of odd numbers in the sequence?

A120
B133
C150
D167

Show the answer

Correct answer: D

There are five odd numbers in the range 1 through 9: 1, 3, 5, 7, and 9.

So the proportion of odd numbers in the range 1 through 9 is $\frac{5}{9}$ .

The expected number of odd values in the sequence from the computer program is

$300 \bullet \frac{5}{9} = 166.\overline{66}.$

So the best prediction for the number of odd numbers in the sequence is 167.

Question analysis

A probability question: a computer randomly generates 300 whole numbers from 1 to 9, and students predict how many will be odd. It asks students to find the probability of an odd number and use it to predict a count. Counting carefully which of the nine numbers are odd, and rounding the prediction sensibly, are the points to watch.

NYS standards: 6.SP.7 long-run relative frequency
6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Using a probability to predict how often an event happens is grade 6 and 7 work. This takes two short steps plus a rounding decision, about the demand of a typical grade 7 probability task.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Something you must already know minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 11, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The method is short, a probability times 300, so the question is decided by the probability. Odd and even feel like an even split, and half of 300 is a listed choice, but the numbers 1 to 9 hold five odd and four even. Listing the nine numbers before computing removes the trap, and the result, about 167, then needs only rounding.

Turning words into math minor Understanding the question
The phrase "whole numbers from 1 to 9, inclusive" means nine equally likely values, five of them odd, so the chance of odd is 5/9, not the one half that odd and even suggest. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The stem never says that an expected count is the probability times the number of trials, here 5/9 of the 300 numbers generated. (Anderson & Krathwohl, 2001)
Error-prone arithmetic minor Working it out
The product 300 × 5/9 is 166.67, not a whole number, so the prediction has to be rounded to the nearest whole number. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the odd numbers from 1 to 9, write the probability as that count over 9, and multiply by 300, rounding to the nearest whole number.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No count of the nine numbers gives 120, which is 40% of 300; reaching it takes two errors at once, so it mostly attracts guesses.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Counts the four even numbers instead of the five odd ones and predicts 4/9 of 300, about 133.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Assumes odd and even numbers are equally likely and takes half of 300, ignoring that 1 to 9 holds one more odd number than even.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{wanted results}}{\text{possible results}}$; P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of the results that you want.
Expected count: $E = P \times n$; E is the expected count, your best guess of how many times something will happen. P is its probability, and n is the number of tries. If something happens 5 times out of 9, it should happen about 5 ninths of the time in many tries.

Sort out the question
The computer picks whole numbers from 1 to 9. Inclusive means 1 and 9 count too. Randomly means each number has the same chance. It picks 300 numbers. You need the best guess of how many of them are odd.
Count the odd numbers
An odd number does not split evenly by 2. From 1 to 9, the odd numbers are 1, 3, 5, 7 and 9. That is 5 of the 9 numbers. The even ones, 2, 4, 6 and 8, are only 4. Watch out: odd and even are not equally likely here, since there is one more odd number. Taking half of 300 gives 150, choice C.
Find the chance of odd
$P = \frac{\text{wanted results}}{\text{possible results}}$
$P = \frac{5}{9}$
Each number the computer picks has a 5 in 9 chance of being odd. Watch out: do not use the 4 even numbers. That gives about 133, choice B.
Multiply by the 300 numbers
$E = P \times n$
$E = \frac{5}{9} \times 300 = \frac{1500}{9} \approx 166.7$
The sign ≈ means about. You can’t have part of a number, so round to the nearest whole number. 166.7 is closer to 167 than to 166. So the answer is D, 167.

Check your answer

The odd numbers are a bit more than half of the nine. So the answer should be a bit more than half of 300, which is 150. Only 167 is a bit more than 150. Also, 167 odd plus about 133 even makes 300.

More questions like this

The original DOE question

Question 72 as printed on page 79 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 79, question 72. Open page 79 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 13 of 40

2024 Test A Math, question 17

Question

The probability of drawing a red candy at random from a bag of 25 candies is $\frac{2}{5}$ . After 5 red candies are removed from the bag, what is the probability of randomly drawing a red candy from the bag?

A0
B $\frac{1}{10}$
C $\frac{1}{5}$
D $\frac{1}{4}$

Show the answer

Correct answer: D

Start with a proportion to determine the total number of red candies in the bag:

$\frac{2}{5} = \frac{r}{25}$

$\frac{2}{5}(25) = r$

$10 = r$

There are currently 10 red candies out of 25 candies in the bag. After 5 red candies are removed, there would be 5 red candies left in the bag and 20 total candies.

The probability of randomly drawing a red candy would then be $\frac{5}{20} = \frac{1}{4}$ .

Question analysis

A probability question about a bag of 25 candies in which the chance of drawing red is 2/5; then 5 red candies are taken out. Students find the new probability of drawing a red candy. They must work out how many red candies there were, then rebuild the fraction from the counts left in the bag.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability as a fraction of outcomes is grade 6 and 7 content. This adds a step back from the probability to a count and an update after removal, which is typical grade 7 multi-step work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 7, 40 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question gives a probability, not a count, so the first job is to rebuild the bag: 10 red candies among 25. Removing candies changes two numbers at once, and the strongest wrong choice comes from updating the red count while leaving the total at 25. A student who writes both new counts before forming the fraction avoids it.

Turning words into math moderate Understanding the question
The probability 2/5 has to be turned into a count, 10 red of 25, and "after 5 red candies are removed" lowers both the red count and the bag's total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Three steps follow in order: 2/5 of 25 = 10, then 5 red among 20 candies left, then 5/20 reduced to lowest terms. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use 2/5 of 25 to find the number of red candies, subtract 5 from both the red count and the total, and write the new fraction in lowest terms.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Takes 2/5 of 25 as 25 ÷ 5 = 5, forgetting to multiply by 2, so removing 5 red candies seems to leave none.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working gives 1/10, which would need 2 red candies among the 20 left; it mostly draws students who are guessing.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Subtracts the 5 removed candies from the 10 red ones but keeps 25 as the total, finding 5/25 = 1/5, though the bag now holds 20.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{red candies}}{\text{all candies}}$; P is the probability, the chance of drawing a red candy. Red candies is how many red ones are in the bag. All candies is how many candies are in the bag in all. Each candy is equally likely, so the chance is the fraction of the bag that is red.

Sort out the question
The bag has 25 candies. The chance of drawing red is $\frac{2}{5}$ . Then 5 red candies are taken out. You need the new chance of drawing red. Taking out candies changes both the red count and the total.
Find the number of red candies
A chance of $\frac{2}{5}$ means 2 out of every 5 candies are red. Call the red count r.
$P = \frac{\text{red candies}}{\text{all candies}}$
$\frac{2}{5} = \frac{r}{25}$
$r = \frac{2}{5} \times 25 = 2 \times 5 = 10$
There are 10 red candies. Watch out: do not stop at $25 \div 5 = 5$ . You must also multiply by 2. Thinking there are 5 red candies leads to 0, choice A.
Take out the 5 red candies
$\text{red: } 10 - 5 = 5$
$\text{all: } 25 - 5 = 20$
Now 5 of the 20 candies are red. Watch out: the total goes down too. Keeping 25 gives $\frac{5}{25} = \frac{1}{5}$ , choice C.
Find the new probability
$P = \frac{5}{20} = \frac{1}{4}$
To simplify, divide the top and bottom by 5. So the answer is D, $\frac{1}{4}$ .

Check your answer

At first, $\frac{2}{5}$ is 40% of the bag. Taking out only red candies should lower the chance. $\frac{1}{4}$ is 25%, which is lower but not zero. That makes sense, because 5 red candies are still in the bag.

More questions like this

The original DOE question

Question 74 as printed on page 80 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 80, question 74. Open page 80 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 14 of 40

2024 Test A Math, question 54

Question

The probability of an event occurring is 0.05. What is the chance that the event will occur?

Alikely
Bunlikely
Cimpossible
Dneither likely nor unlikely

Show the answer

Correct answer: B

The probability of the event is 0.05, which equals 5%, which is less than 50%.

Option A:

An event that is likely has a probability over 50%.

Option B:

An event that is unlikely has a probability less than 50% but greater than 0%.

Option C:

An event that is impossible has a 0% probability.

Option D:

An event that is neither likely nor unlikely has a 50% probability.

Question analysis

This probability question gives an event a probability of 0.05 and asks how likely the event is to occur, described in words. Students place the decimal on the scale from 0, impossible, to 1, certain. Reading the place value of the decimal correctly is the only real hurdle.

NYS standards: 6.SP.6 probability from 0 to 1
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Describing in words how likely an event is from its probability between 0 and 1 is a grade 6 standard. This is one quick judgment with no computation, well below grade 7 probability work with compound events.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly understand
What makes it hard: Something you must already know minor
Solution: 1 step
Reading level: Question and choices: about grade 10, 23 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

One judgment answers the question: 0.05 is a small probability, but it is not zero. The two tempting errors are treating a small chance as no chance at all and misreading 0.05 as 0.5, an even chance. Writing the probability as 5% or as 1 in 20 makes its size plain.

Something you must already know minor Working it out
The probability scale, 0 for impossible, 1/2 for an even chance and 1 for certain, is assumed, and 0.05 has to be placed near its low end. (Anderson & Krathwohl, 2001)

Strategy: Picture the probability scale from 0 to 1, with 0 for impossible, 1/2 for an even chance and 1 for certain, and see where 0.05 falls.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common reading makes 0.05 likely, which would need a probability above one half, so it mostly draws guesses.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats a probability close to 0 as exactly 0, as if an event with a 1 in 20 chance could never happen.
D An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: weak
Reads 0.05 as 0.5, a place-value slip, which would mean an even chance of happening or not.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability scale: $0 \leq P \leq 1$; P is the probability of an event, a number that tells how likely it is to happen. It is always from 0 to 1. 0 means impossible, and 1 means certain. One half, or 0.5, means an even chance. The closer P is to 0, the less likely the event.

Sort out the question
Probability is a number that tells how likely an event is. An event is something that might happen. Here the probability is 0.05. You need the word that best describes that chance.
Change 0.05 to a percent
$0.05 = \frac{5}{100}$
So 0.05 is 5 out of 100, or 5%. That is the same as 1 chance in 20. Watch out: 0.05 is not 0.5. A probability of 0.5 would be an even chance, choice D.
Place it on the scale
$0 \leq P \leq 1$
Probability Chance
0 impossible
more than 0, less than 0.5 unlikely
0.5 neither likely nor unlikely
more than 0.5, less than 1 likely
1 certain
0.05 is more than 0 but much less than 0.5. So the event is unlikely.
Rule out the other choices
Watch out: a small chance is still a chance. 0.05 is more than 0, so the event can happen. It is not impossible, choice C. Likely, choice A, would need more than 0.5.
So the answer is B, unlikely.

Probability	Chance
0	impossible
more than 0, less than 0.5	unlikely
0.5	neither likely nor unlikely
more than 0.5, less than 1	likely
1	certain

Check your answer

Picture 20 tries. With a 5% chance, the event would happen about once, since 5% of 20 is 1. It would not happen about 19 times. That is rare but possible, so unlikely fits.

More questions like this

The original DOE question

Question 111 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 90, question 111. Open page 90 of the handbook (PDF)

Question 15 of 40

2024 Test B Math, question 17

Question

A bowl contains 12 green candies, 4 yellow candies, and some red candies. The probability of choosing a green candy at random is twice as great as the probability of choosing a red candy at random. What is the probability of choosing a yellow candy at random?

A $\frac{2}{11}$
B $\frac{2}{9}$
C $\frac{1}{4}$
D $\frac{3}{11}$

Show the answer

Correct answer: A

The probability of choosing a green candy is twice as great as the probability of choosing a red candy. There are 12 green candies. Let r = the number of red candies.

$12=2r$

$r=6$

So there are 6 red candies. 12 green candies + 4 yellow candies + 6 red candies = 22 total candies.

There are 4 yellow candies out of a total of 22 candies. So the probability of choosing a yellow candy is

$\frac{4}{22}=\frac{2}{11}$ .

Question analysis

A probability question about a bowl of 12 green candies, 4 yellow candies and an unknown number of red ones. A green candy is twice as likely to be picked as a red one, and students find the chance of picking a yellow candy. The missing red count has to be worked out before any probability can be found.

NYS standards: 6.SP.8a uniform probability models
6.EE.7 one-step equations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Above grade 7 Simple probability is grade 6 content in New York, and grade 7 problems usually give every count. Working backward from a comparison of two probabilities to a missing count, then forward to a third probability, is more than typical grade 7 work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Combines ideas moderate
Solution: 3 steps
Reading level: Question and choices: about grade 7, 49 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question withholds the red count, and everything depends on recovering it from the comparison of probabilities. Students who cannot make that step tend to leave red out altogether, while those who find 6 red candies can still slip at the end by using the wrong total or the wrong color. Writing all three counts and the total of 22 before forming any fraction prevents both endings.

Turning words into math moderate Understanding the question
“Twice as great as the probability of choosing a red candy” must become 12 = 2 × the red count, which works only because both probabilities share one total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Combines ideas moderate Working it out
An unknown count has to be found from a ratio before any probability can be written, joining proportional reasoning with probability as a part over the total. (Embretson, 1998; Daniel & Embretson, 2010)

Strategy: Turn twice as likely into a statement about numbers of candies to find the red count, add all three colors for the total, then put the yellow count over that total and simplify.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Finds the 6 red candies but compares the 4 yellow with the 18 green and red ones, 4/18, a part-to-part ratio instead of part to whole.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Leaves out the red candies, whose number is not given, and divides the 4 yellow by the 16 green and yellow candies.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Finds the 6 red candies and the total of 22 correctly but gives the chance of red, 6/22 = 3/11, instead of yellow.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of the kind you want}}{\text{total number}}$; P is the probability, the chance that something happens. The top is how many candies are the kind you want. The bottom is how many candies there are in all. Each candy is just as likely to be picked, so this fraction gives the chance.

Sort out what you know
The bowl has 12 green candies, 4 yellow candies and some red candies. You are not told how many are red. Let r be the number of red candies.
The chance of picking green is twice the chance of picking red. You need the probability (the chance) of picking yellow.
Compare green and red
Write each chance with the probability formula.
$P = \frac{\text{number of the kind you want}}{\text{total number}}$
Green and red have the same bottom number, the total number of candies. So if the chance of green is twice the chance of red, there are twice as many green candies as red ones.
$12 = 2 \times r$
Find the red candies and the total
$r = 12 \div 2 = 6$
There are 6 red candies. Now add up all the candies.
$12 + 4 + 6 = 22$
There are 22 candies in all. Watch out: do not leave out the red candies just because their number was not given. Using 16 as the total gives 1/4, choice C.
Find the chance of yellow
There are 4 yellow candies out of 22.
$P = \frac{4}{22} = \frac{2}{11}$
Divide the top and the bottom by 2 to simplify. Watch out: the question asks for yellow, not red. 6/22 = 3/11 is the chance of red, choice D. The bottom must be all 22 candies. Using 18 gives 2/9, choice B.
So the answer is A, $\frac{2}{11}$ .

Check your answer

Check the clue. The chance of green is $\frac{12}{22}$ , and the chance of red is $\frac{6}{22}$ . Twelve is twice 6, so green is twice as likely. Also, the three chances add up to 1: $\frac{12}{22} + \frac{4}{22} + \frac{6}{22} = \frac{22}{22} = 1$ .

More questions like this

The original DOE question

Question 74 as printed on page 203 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 203, question 74. Open page 203 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 16 of 40

2024 Test B Math, question 29

Question

Jar Q contains 12 balls, and Jar R contains 8 balls. The probability of drawing a yellow ball at random from Jar Q is $\frac{1}{3}$ . The probability of drawing a yellow ball at random from Jar R is $\frac{3}{4}$ . All the balls are poured into an empty jar, Jar S. What is the probability of drawing a yellow ball at random from Jar S?

A $\frac{1}{4}$
B $\frac{1}{3}$
C $\frac{1}{2}$
D $\frac{4}{7}$

Show the answer

Correct answer: C

Jar Q contains 12 balls, and the probability of drawing a yellow ball at random is $\frac{1}{3}$ .

$\frac{1}{3}\times\frac{4}{4}=\frac{4}{12}$

There are 4 yellow balls in Jar Q.

Jar R contains 8 balls, and the probability of drawing a yellow ball at random is $\frac{3}{4}$ .

$\frac{3}{4}\times\frac{2}{2}=\frac{6}{8}$

There are 6 yellow balls in Jar R.

Jar S contains all the balls from Jar Q and Jar R.

$12+8=20$

There are 20 balls in Jar S.

$4+6=10$

There are 10 yellow balls in Jar S.

$\frac{10}{20}\div\frac{10}{10}=\frac{1}{2}$

The probability of drawing a yellow ball at random from Jar S is $\frac{1}{2}$ .

Question analysis

Two jars hold different numbers of balls and give different chances of drawing a yellow ball, and then everything is poured into a third jar. Students find the probability of drawing yellow from the combined jar. The tricky part is that the jars are different sizes, so their probabilities cannot be merged without first counting the yellow balls.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Grade 7 probability tasks usually give the counts and ask for a fraction. Here each jar's probability must be turned back into a count of yellow balls before the jars are pooled, a reverse step that typical classwork rarely asks for.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
A very close wrong answer minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 66 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light; the hurdle is seeing that chances from jars of different sizes cannot be combined as fractions. Every wrong choice works on 1/3 and 3/4 directly, by multiplying them, by keeping one, or by adding tops and bottoms, while the right path runs through counts of balls. Asking how many yellow balls each jar holds before anything is poured avoids all three.

Turning words into math moderate Understanding the question
Each jar's probability, 1/3 of 12 balls and 3/4 of 8 balls, has to be turned back into a count of yellow balls, 4 and 6, before the two jars can be combined. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The straight-across 4/7, about 0.57, sits just above the right 1/2, and both lie between the jars' chances of 1/3 and 3/4, so a rough estimate does not rule it out. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Turn each jar's probability into a number of yellow balls by multiplying it by that jar's total, then divide all the yellow balls by all the balls.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Multiplies the jars' chances, 1/3 × 3/4 = 1/4, the rule for two draws that must both be yellow, though only one ball is drawn from the mixture.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Repeats Jar Q's 1/3 from the stem, as if the bigger jar's chance carried over to the mixture unchanged.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: strong
Adds the fractions straight across, (1 + 3)/(3 + 4) = 4/7, treating 1/3 and 3/4 as if they were counts of balls rather than chances.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P(\text{yellow}) = \frac{\text{yellow balls}}{\text{all balls}}$; P(yellow) is the probability, or chance, of drawing a yellow ball. It is a number from 0 to 1. The top counts the yellow balls. The bottom counts all the balls. It works because every ball is equally likely to be drawn.

Sort out what you know
Jar Q holds 12 balls. The probability, or chance, of drawing yellow from it is $\frac{1}{3}$ .
Jar R holds 8 balls. The chance of drawing yellow from it is $\frac{3}{4}$ .
All the balls are poured into Jar S. You need the chance of drawing yellow from Jar S.
Count the yellow balls in Jar Q
The jars are different sizes, so you can’t mix their chances directly. Count the yellow balls instead. The chance tells what fraction of the balls are yellow. So multiply it by the number of balls.
$\text{yellow balls} = \frac{1}{3} \times 12 = 4$
Jar Q has 4 yellow balls.
Count the yellow balls in Jar R
Do the same for Jar R. Three fourths of its 8 balls are yellow.
$\text{yellow balls} = \frac{3}{4} \times 8 = 6$
Jar R has 6 yellow balls.
Count the balls in Jar S
Jar S gets every ball from both jars. Add all the balls. Then add the yellow balls.
$12 + 8 = 20$
$4 + 6 = 10$
Jar S has 20 balls, and 10 of them are yellow.
Find the chance for Jar S
Use the probability formula with the Jar S counts.
$P(\text{yellow}) = \frac{10}{20} = \frac{1}{2}$
Watch out: Adding the tops and the bottoms, 1 + 3 over 3 + 4, gives $\frac{4}{7}$ , choice D. Those numbers are parts of fractions, not balls. Multiplying $\frac{1}{3} \times \frac{3}{4}$ gives $\frac{1}{4}$ , choice A. That is the chance of two yellow draws, one from each jar.
So the answer is C, $\frac{1}{2}$ .

Check your answer

Half of the 20 balls is 10, and 4 + 6 = 10 yellow balls. That matches. Also, the mixed jar’s chance should land between the two jars’ chances, $\frac{1}{3}$ and $\frac{3}{4}$ . One half does.

More questions like this

The original DOE question

Question 86 as printed on page 207 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 207, question 86. Open page 207 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 17 of 40

2024 Test B Math, question 41

Question

A bag contains 2 green, 4 blue, and 3 yellow disks. Alana selects 4 disks from the bag at random, one at a time, without replacing them. What is the probability that she selects all 4 blue disks?

A $\frac{8}{2{,}187}$
B $\frac{1}{126}$
C $\frac{256}{6{,}561}$
D $\frac{16}{9}$

Show the answer

Correct answer: B

If Alana picks 4 blue disks in a row, the number of blue disks and the total number of disks will both decrease by 1 after each selection.

For the first selection, there are 4 blue disks out of 9 total disks, so the probability of selecting the first blue disk is $\frac{4}{9}$ .

The probabilities of selecting blue for the next three selections are $\frac{3}{8}$ , $\frac{2}{7}$ , and $\frac{1}{6}$ .

$\frac{4}{9}\times\frac{3}{8}\times\frac{2}{7}\times\frac{1}{6}=\frac{24}{3{,}024}=\frac{1}{126}$

Question analysis

A bag holds green, blue and yellow disks, and Alana draws four of them one at a time without putting any back. Students find the probability that all four disks are blue. It tests the probability of dependent events, and the difficulty is keeping track of how the bag changes after each draw.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability is usually found with lists or tree diagrams for two stages. Four dependent draws, each changing the counts, are too many to list and need the multiplication rule, which grade 7 does not formally teach.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Error-prone arithmetic moderate
Solution: 3 steps
Reading level: Question and choices: about grade 5, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Students who know the multiplication rule for successive draws still have two adjustments to make on every draw, and each wrong choice skips at least one of them. Leaving the bag at 9 disks, or putting each disk back, both produce fractions over 6,561 that look as exact as the right one. Writing the four fractions in a row before multiplying, and cancelling as you go, keeps the arithmetic small.

Something you must already know moderate Working it out
The rule for successive draws without replacement is assumed: each draw's chance is multiplied in, with both the blue count and the total lowered by one. (Anderson & Krathwohl, 2001)
Error-prone arithmetic moderate Working it out
The product 4/9 × 3/8 × 2/7 × 1/6 = 24/3,024 has to be reduced to 1/126, and two wrong choices hide behind 9 to the fourth power, 6,561. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Write the chance of blue for each of the four draws in turn, lowering both the blue count and the total by one each time, and multiply.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: moderate
Lowers the blue count on each draw but keeps all 9 disks in the bag, 4 × 3 × 2 × 1 over 9 × 9 × 9 × 9, which is 24/6,561 = 8/2,187.
C Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Uses 4/9 for every draw, (4/9) to the fourth power = 256/6,561, as if each disk were put back before the next draw.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: weak
Adds the first draw's 4/9 four times, or multiplies it by 4, instead of multiplying the draws' chances, and misses that a probability cannot exceed 1.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(A \text{ then } B) = P(A) \times P(B \text{ after } A)$; A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

Sort out the question
The bag holds 2 green, 4 blue and 3 yellow disks. That is 9 disks in all. Alana takes 4 disks, one at a time.
“Without replacing them” means a disk she takes does not go back in the bag. You need the chance that all 4 disks are blue.
Find the chance for each pick
For all 4 to be blue, each pick must be blue. After each blue pick, one blue disk is gone. One disk is gone from the total, too. So the top and the bottom each go down by 1.
Pick Blue disks left Disks left Chance of blue
1st 4 9 $\frac{4}{9}$
2nd 3 8 $\frac{3}{8}$
3rd 2 7 $\frac{2}{7}$
4th 1 6 $\frac{1}{6}$
Multiply the four chances
All four picks must go right, one after another. Use the rule for events in a row and multiply.
$\frac{4}{9} \times \frac{3}{8} \times \frac{2}{7} \times \frac{1}{6} = \frac{4 \times 3 \times 2 \times 1}{9 \times 8 \times 7 \times 6} = \frac{24}{3{,}024}$
Simplify the fraction
Divide the top and the bottom by 24. That keeps the value the same.
$\frac{24 \div 24}{3{,}024 \div 24} = \frac{1}{126}$
Watch out: Using $\frac{4}{9}$ for every pick gives $\frac{256}{6{,}561}$ , choice C. That is right only if each disk goes back in. Keeping 9 on the bottom each time gives $\frac{8}{2{,}187}$ , choice A. Choice D is more than 1, and no chance can be more than 1.
So the answer is B, $\frac{1}{126}$ .

Pick	Blue disks left	Disks left	Chance of blue
1st	4	9	$\frac{4}{9}$
2nd	3	8	$\frac{3}{8}$
3rd	2	7	$\frac{2}{7}$
4th	1	6	$\frac{1}{6}$

Check your answer

Cancel before you multiply. On top, $4 \times 2 = 8$ cancels the 8 on the bottom. That leaves $\frac{3}{9 \times 7 \times 6} = \frac{3}{378}$ . Divide the top and bottom by 3 to get $\frac{1}{126}$ again.

More questions like this

The original DOE question

Question 98 as printed on page 211 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 211, question 98. Open page 211 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 18 of 40

2024 Test B Math, question 45

Question

A box contains a total of 30 paper clips with colors in the ratio $\text{red}:\text{white}:\text{blue}=2:5:3$ . Suppose that one paper clip of each color is removed from the box and not replaced. What is the probability that the next paper clip chosen will be blue?

A $\frac{5}{27}$
B $\frac{2}{7}$
C $\frac{8}{27}$
D $\frac{3}{10}$

Show the answer

Correct answer: C

The ratio is $\text{red}:\text{white}:\text{blue}=2:5:3$ . Find the number of each color in a box of 30 paper clips.

$2x+3x+5x=30$

$10x=30$

$x=3$

$\text{Red: }2x=2(3)=6$

$\text{White: }5x=5(3)=15$

$\text{Blue: }3x=3(3)=9$

If one paper clip of each color is removed (3 total), there will be 8 blue paper clips out of 27 in the box. The probability that the next paper clip chosen will be blue is $\frac{8}{27}$ .

Question analysis

A box of 30 paper clips has red, white and blue clips in the ratio 2 to 5 to 3, and one clip of each color is taken out. Students find the probability that the next clip drawn is blue. It tests ratios and simple probability; the ratio gives shares rather than counts, and the removals change both the blue count and the total.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Splitting a total by a three-part ratio and finding a simple probability are both familiar skills. Combining them, then adjusting the counts after the removals, adds steps but stays within typical grade 7 multi-step work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Many steps moderate
A very close wrong answer moderate
Turning words into math minor
Solution: 4 steps
Reading level: Question and choices: about grade 5, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question chains a ratio problem into a probability problem, with a small change in between. Students who skip the removal get 3/10, a choice that is almost the right value and looks natural because it matches the ratio. The cleanest route lists the actual counts, 6, 15 and 9, takes one from each, and writes blue's 8 over the remaining 27.

Many steps moderate Working it out
The ratio 2:5:3 must become counts of 6, 15 and 9 clips, one clip of each color must come out, and only then can blue's 8 be put over 27. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
3/10 = 0.3 and 8/27, about 0.296, are almost equal, so an estimate cannot separate them; only subtracting the three removed clips does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
The removal of one clip of each color, not replaced, happens before the draw, so it changes both the blue count and the total in the box. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Divide 30 by the sum of the ratio parts to find each color's count, take one clip away from each color, then put the blue count over the new total.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Works out the counts correctly but puts red's remaining 5 clips over the 27 left, answering for the wrong color.
B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the ratio terms as actual counts and removes one from each, so blue is 3 − 1 = 2 of 10 − 3 = 7 clips.
D Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Gives blue's share of the full box, 9 of 30 or 3 of 10 ratio parts, ignoring the three clips removed before the draw.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Size of one ratio part: $\text{one part} = \frac{\text{total}}{\text{sum of the ratio numbers}}$; The ratio numbers tell how many parts each color gets. Their sum is the number of parts in all. total is the number of clips. Dividing gives the size of one part. Each color is then its ratio number times one part.
Probability: $P(\text{blue}) = \frac{\text{blue clips}}{\text{all clips}}$; P(blue) is the chance of picking a blue clip. The top counts the blue clips in the box. The bottom counts all the clips in the box. It works because each clip is equally likely to be picked.

Sort out the question
The box has 30 clips. The ratio red : white : blue = 2 : 5 : 3 means that for every 2 red clips, there are 5 white clips and 3 blue clips.
Then one clip of each color is taken out and not put back. You need the chance that the next clip picked is blue.
Find the size of one part
The ratio has $2 + 5 + 3 = 10$ parts in all. The 30 clips are shared among those 10 parts.
$\text{one part} = \frac{30}{10} = 3$
Each part is 3 clips.
Count the clips of each color
Multiply each ratio number by the size of one part.
$\text{red} = 2 \times 3 = 6$
$\text{white} = 5 \times 3 = 15$
$\text{blue} = 3 \times 3 = 9$
Check: 6 + 15 + 9 = 30 clips. That matches the total.
Take out one of each color
One red, one white and one blue clip come out. Each color drops by 1, and the total drops by 3.
Color Before After
Red 6 5
White 15 14
Blue 9 8
All 30 27
Find the chance of blue
$P(\text{blue}) = \frac{8}{27}$
Watch out: $\frac{9}{30} = \frac{3}{10}$ , choice D, ignores the 3 clips taken out. $\frac{5}{27}$ , choice A, is the chance of red. Taking 1 from each ratio number, as if the box held only 10 clips, gives $\frac{2}{7}$ , choice B.
So the answer is C, $\frac{8}{27}$ .

Color	Before	After
Red	6	5
White	15	14
Blue	9	8
All	30	27

Check your answer

Add the clips left: 5 + 14 + 8 = 27. That matches. Also, $\frac{8}{27}$ is about 0.296, a bit less than blue’s share of the full box, 0.3. That makes sense, since one blue clip came out.

More questions like this

The original DOE question

Question 102 as printed on page 212 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 102. Open page 212 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 19 of 40

2024 Test B Math, question 50

Question

Chris has 1 red shirt, 2 white shirts, and 3 black shirts. Chris picks a shirt at random. Without putting the first shirt back, he picks a second shirt at random. What is the probability that the first shirt is black and the second shirt is red?

A $\frac{1}{12}$
B $\frac{1}{10}$
C $\frac{1}{5}$
D $\frac{7}{10}$

Show the answer

Correct answer: B

Use the joint probability formula: P (black shirt first) • P (red shirt second, given black shirt first).

Initially there are 3 black shirts out of 6 total shirts, so the probability of picking a black shirt first is $\frac{3}{6}$ .

Once a black shirt is picked first, there is 1 red shirt out of 5 total shirts remaining.

So the probability of picking a red shirt second, given that a black shirt was picked first, is $\frac{1}{5}$ .

Then P (black shirt first) • P (red shirt second, given black shirt first)
$=\frac{3}{6}\bullet\frac{1}{5}=\frac{3}{30}=\frac{1}{10}$ .

Question analysis

Chris has red, white and black shirts and picks two at random, one after the other, without putting the first back. Students find the probability that the first shirt is black and the second is red. It tests the probability of two dependent events, where the second pick has one fewer shirt to choose from.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability is usually found with a list or tree of equally likely outcomes. Only two picks are involved, but adjusting the second pick for the shirt already taken goes beyond typical grade 7 work.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
A very close wrong answer moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two dependent picks with small numbers make this a textbook question, but the choices punish each half-step. Students who leave the first shirt in the drawer get a fraction just below the right one, students who find only the second pick's chance stop at a listed value, and students who add the two chances get a probability far too large. Writing both fractions, 3/6 and 1/5, and multiplying is the whole method.

Something you must already know moderate Working it out
The rule for two dependent picks, the first pick's chance times the second's given the first, is assumed, along with the drop to 5 shirts for the second pick. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
1/10 and 1/12 differ only in the second pick's denominator, 5 shirts or 6, so estimating cannot separate them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
"Without putting the first shirt back" sets the second pick's total at 5 shirts, and the order, black first then red, fixes which fraction comes first. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the chance of black on the first pick, then the chance of red on the second pick with one shirt fewer, and multiply the two.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Stops one step early: A choice is a number from partway through the solution.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Keeps all 6 shirts for the second pick, 3/6 × 1/6 = 1/12, as if the black shirt had been put back.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Gives only the chance that the second shirt is red once a black one is gone, 1/5, leaving out the first pick's 3/6.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Adds the two chances, 1/2 + 1/5 = 7/10, instead of multiplying them for two events that must both happen.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(A \text{ then } B) = P(A) \times P(B \text{ after } A)$; A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

Sort out the question
Chris has 1 red, 2 white and 3 black shirts. That is 6 shirts. He picks one and keeps it out. Then he picks a second shirt.
You need the chance that the first shirt is black and the second is red. Both must happen, in that order.
Find the chance of black first
3 of the 6 shirts are black. Each shirt is equally likely to be picked.
$P(\text{black first}) = \frac{3}{6}$
That is one half.
Find the chance of red second
A black shirt is now out and not put back. That leaves 5 shirts: 1 red, 2 white and 2 black.
$P(\text{red second}) = \frac{1}{5}$
Watch out: The bottom is 5, not 6. Using 6 gives $\frac{1}{12}$ , choice A, as if the shirt went back.
Multiply the two chances
Use the rule for events in a row. The second pick must go right too, inside the half of the time the first one did.
$P(\text{black, then red}) = \frac{3}{6} \times \frac{1}{5} = \frac{3}{30} = \frac{1}{10}$
Watch out: Adding the chances gives $\frac{7}{10}$ , choice D. Stopping at $\frac{1}{5}$ gives choice C.
So the answer is B, $\frac{1}{10}$ .

Check your answer

Count every ordered pair of picks. There are 6 choices for the first and 5 for the second, so $6 \times 5 = 30$ pairs. Pairs with black then red: 3 black shirts times 1 red shirt is 3. So the chance is $\frac{3}{30} = \frac{1}{10}$ .

More questions like this

The original DOE question

Question 107 as printed on page 213 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 213, question 107. Open page 213 of the handbook (PDF)

Question 20 of 40

2024 Test B Math, question 52

Question

$Two open boxes, labeled A and B. Box A holds two balls, numbered 2 and 6. Box B holds three balls, numbered 3, 4 and 5.$

Five numbered balls are placed in two boxes as shown in the figure above. Davina picks one ball at random from Box B and places it in Box A. Then Yusuf picks one ball from Box A without looking. What is the probability that Yusuf will pick a ball with an odd number?

A $\frac{2}{9}$
B $\frac{1}{3}$
C $\frac{2}{5}$
D $\frac{2}{3}$

Show the answer

Correct answer: A

The probability of Davina picking an odd-numbered ball from Box B is $\frac{2}{3}$ . If Davina places an odd-numbered ball in Box A, then the probability of Yusuf picking an odd-numbered ball from Box A is $\frac{1}{3}$ . The probability of both of these events happening is the product of the two probabilities.

$\frac{2}{3}\times\frac{1}{3}=\frac{2}{9}$

Question analysis

Two boxes hold five numbered balls, and Davina moves one ball at random from Box B into Box A before Yusuf draws one from Box A. Students find the probability that Yusuf's ball shows an odd number. It tests two-stage probability, and the tricky part is that the second draw depends on which ball was moved.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability lists outcomes for independent stages such as coins or spinners. Here the first random move changes what is in the second box, so students must follow two linked stages, a dependence typical grade 7 work does not include.
Estimated difficulty: Among the hardest on the SHSAT: a non-obvious approach, heavy reading, or a well-hidden trap.
Thinking it demands: Depth of knowledge 3, strategic thinking · mainly analyze
What makes it hard: No obvious method major
Something you must already know minor
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 4, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two chance events are hidden in what reads like a question about a single draw. Every listed wrong answer is a reasonable single-stage answer, either one of the two factors on its own or a count of odd balls across both boxes. Listing the nine equally likely pairs, Davina's ball and then Yusuf's, shows that only two of them end with an odd ball and makes the structure visible without any formula.

No obvious method major Working it out
The stem asks one question about Yusuf's draw, but the answer depends on Davina's earlier random pick; the key is that Box A starts with only even balls, so Yusuf can draw odd only by drawing the ball she moved. (Schoenfeld, 1985)
Something you must already know minor Working it out
Multiplying the chances of the two stages, 2/3 for an odd ball moving and 1/3 for Yusuf then drawing it, is a rule students must bring themselves. (Anderson & Krathwohl, 2001)
Reading a figure, table or graph minor Understanding the question
The numbers appear only in the figure, Box A holding 2 and 6 and Box B holding 3, 4 and 5, so the odd balls have to be spotted there. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: List every equally likely pair of a ball moved by Davina and a ball picked by Yusuf, and count the pairs that end with an odd ball.

Traps

Stops one step early: A choice is a number from partway through the solution.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Finds Yusuf's chance of drawing the odd ball once one is in Box A, 1 out of 3, leaving out the chance that Davina moves an odd ball at all.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Counts the odd balls in the figure, 3 and 5, over all five balls, 2/5, ignoring that Yusuf draws only from Box A.
D Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Gives the chance that Davina moves an odd ball, 2 out of 3, the first stage only, without multiplying by Yusuf's 1/3 chance of drawing it.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of wanted outcomes}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(A \text{ then } B) = P(A) \times P(B \text{ after } A)$; A and B are two events that happen one after the other. P(A) is the chance of the first. P(B after A) is the chance of the second, counting only what is left after the first. You multiply because the second must also happen inside the share of times the first did.

Read the figure
Box A holds balls numbered 2 and 6. Box B holds balls numbered 3, 4 and 5.
An odd number does not divide evenly by 2. So the only odd balls are 3 and 5, and both start in Box B. Box A starts with no odd balls at all.
Spot the two chance events
Two things happen by chance, one after the other:
1. Davina picks one of the 3 balls in Box B and moves it to Box A.
2. Then Yusuf picks one of the 3 balls now in Box A.
Yusuf gets an odd ball only if both go right. Davina must move an odd ball. Then Yusuf must pick that very ball.
The question asks only about Yusuf. But his chance depends on Davina’s pick, which comes first.
Find the chance for Davina’s pick
Box B has 3 balls, and 2 of them, the 3 and the 5, are odd.
$P(\text{odd ball moved}) = \frac{2}{3}$
If she moves the 4 instead, Box A holds only even balls. Then Yusuf can’t get an odd ball at all.
Find the chance for Yusuf’s pick
Say Davina moved an odd ball. Box A now holds 3 balls: the 2, the 6 and that one odd ball.
$P(\text{Yusuf picks it}) = \frac{1}{3}$
Watch out: $\frac{1}{3}$ alone is choice B. It forgets that Davina might move the 4.
Multiply the two chances
Both events must happen, so use the rule for events in a row.
$P(\text{odd}) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$
Watch out: $\frac{2}{3}$ alone, choice D, stops after Davina’s pick. $\frac{2}{5}$ , choice C, counts 2 odd balls out of all 5. But Yusuf picks only from Box A.
So the answer is A, $\frac{2}{9}$ .

Check your answer

List all 9 equally likely outcomes. Davina can move 3, 4 or 5. Then Yusuf picks one of 3 balls.

Ball moved	Box A then	Odd picks
3	2, 6, 3	1
4	2, 6, 4	0
5	2, 6, 5	1

That is 2 odd picks out of 9, or $\frac{2}{9}$ .

More questions like this

The original DOE question

Question 109 as printed on page 214 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 214, question 109. Open page 214 of the handbook (PDF)

Question 21 of 40

2023 Test A Math, question 23

Question

$\{0.1, 0.01, 0.001, 0.0001, 0.00001\}$

If a person chooses a number at random from the set above, what is the probability that the number is less than 0.005?

A $\frac{1}{5}$
B $\frac{2}{5}$
C $\frac{3}{5}$
D $\frac{2}{3}$

Show the answer

Correct answer: C

Three of the five numbers listed (0.001, 0.0001, and 0.00001) are less than 0.005. So, the probability of a person selecting a number less than 0.005 is $\frac{3}{5}$ .

Question analysis

A set holds five decimals, from one tenth down to one hundred-thousandth, and one number is chosen at random. Students find the probability that the number is less than 0.005. Comparing decimals with many zeros is the real task here, along with putting the count over the right total.

NYS standards: 6.SP.8a uniform probability models
6.NS.7a inequality as relative position
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Probability with equally likely outcomes is grade 6 content in New York, and this needs one comparison per number and one fraction. The only real demand is comparing decimals with many zeros correctly.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Error-prone arithmetic moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 27 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Each number in the set is a tenth of the one before, so only the three smallest fall below 0.005, which lies between 0.01 and 0.001. Students lose the point either by counting the numbers above 0.005 instead of below it or by writing the count over the wrong total. Lining the decimals up with the same number of places makes every comparison easy.

Error-prone arithmetic moderate Working it out
0.005 has to be compared with decimals from 0.1 down to 0.00001; written to five places, 0.00500 against 0.01000 and 0.00100, the order becomes a comparison of whole numbers. (Brown & Burton, 1978; Junker & Sijtsma, 2001)
Something you must already know minor Working it out
The probability is the count of numbers below 0.005 over all 5 numbers in the set; the stem leaves that part-to-whole rule to the student. (Anderson & Krathwohl, 2001)

Strategy: Write each number with five decimal places, compare it with 0.00500, count those that are smaller, and divide by the five numbers in the set.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Gives 1/5, the chance of picking any one particular number from the set, without counting how many of the five are below 0.005.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Counts the 2 numbers greater than 0.005, 0.1 and 0.01, instead of those less than it, finding the chance of the opposite event.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Writes a part-to-part ratio, the 2 numbers above 0.005 over the 3 below it, instead of dividing by all 5 numbers in the set.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{wanted results}}{\text{possible results}}$; P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.
Adding zeros to a decimal: $0.1 = 0.10 = 0.100$; Zeros added to the right end of a decimal don’t change its value. One tenth is the same as ten hundredths. Giving every number the same number of decimal places lets you compare them like whole numbers.

Sort out the question
The set has five numbers. A person picks one at random, so each number has the same chance. You need the probability, or chance, that the number picked is less than 0.005.
Give every number five decimal places
Decimals are easier to compare with the same number of places. Add zeros to the end of each.
Number With five places
0.1 0.10000
0.01 0.01000
0.001 0.00100
0.0001 0.00010
0.00001 0.00001
0.005 0.00500
Count the numbers below 0.005
Compare each number with 0.00500 like whole numbers: 10000, 1000, 100, 10 and 1 against 500. Two numbers are bigger: 0.1 and 0.01. Three are smaller: 0.001, 0.0001 and 0.00001. Watch out: 0.01 has fewer digits than 0.005, but it is bigger. Counting the 2 bigger numbers gives $\frac{2}{5}$ , choice B.
Write the probability
$P = \frac{\text{wanted results}}{\text{possible results}} = \frac{3}{5}$
Watch out: divide by all 5 numbers in the set. Comparing the 2 bigger numbers with the 3 smaller ones gives $\frac{2}{3}$ , choice D. So the answer is C, $\frac{3}{5}$ .

Number	With five places
0.1	0.10000
0.01	0.01000
0.001	0.00100
0.0001	0.00010
0.00001	0.00001
0.005	0.00500

Check your answer

The numbers below 0.005 and the numbers above it make up the whole set: 3 + 2 = 5. So the two chances, $\frac{3}{5}$ and $\frac{2}{5}$ , add up to 1, as they should.

More questions like this

The original DOE question

Question 80 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 80. Open page 81 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 22 of 40

2023 Test A Math, question 24

Question

A student uses Spinner R and Spinner T to generate a list of two-digit numbers.

$Two spinners, each with an arrow pointing out from its center. Spinner R is a circle divided into 4 equal sections, labeled 1 and 2 across the top and 3 and 4 across the bottom. Spinner T is a circle divided into 6 equal sections, labeled, clockwise from the top left, 1, 1, 2, 3, 4 and 4.$

Spinner R determines the digit in the tens place, and Spinner T determines the digit in the ones place. What is the probability that the two-digit number determined by spinning each spinner one time is a prime number?

A $\frac{3}{8}$
B $\frac{5}{12}$
C $\frac{3}{5}$
D $\frac{5}{8}$

Show the answer

Correct answer: A

Spinner R has 4 equal sections and Spinner T has 6 equal sections, so the number of possible two-digit numbers that can be created is found by multiplying 4 and 6, which is 24. List the possible outcomes:

11	11	12	13	14	14
21	21	22	23	24	24
31	31	32	33	34	34
41	41	42	43	44	44

There are 9 prime numbers out of 24 possible outcomes: 11, 11, 13, 23, 31, 31, 41, 41, and 43.

So the probability that the two-digit number created is a prime number is $\frac{9}{24}$ , which is equivalent to $\frac{3}{8}$ .

Question analysis

Spinner R, with four equal sections, gives the tens digit and Spinner T, with six equal sections and some labels repeated, gives the ones digit of a two-digit number. Students find the probability that the number is prime. Building the full table of outcomes, counting repeated sections separately, and checking whether each number is prime all take care.

NYS standards: 7.SP.8b sample spaces for compound events
7.SP.8a compound event as fraction of outcomes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Listing a sample space for two spinners is grade 7 work, usually with distinct labels and a simple event. Here one spinner repeats two numbers, the space has 24 outcomes, and each two-digit result must be tested for being prime.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Reading a figure, table or graph moderate
Much to keep track of moderate
Something you must already know minor
Solution: 4 steps
Reading level: Question and choices: about grade 7, 57 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question combines a compound sample space with a test for primes, and its figure adds a twist: two labels repeat on Spinner T. Counting the 24 section pairs rather than the 16 different numbers is the careful route, though here both happen to give the same fraction. The points are lost in the prime test, where 33 is easy to count as prime, and in writing the count as a share of all 24 outcomes.

Reading a figure, table or graph moderate Understanding the question
Spinner T has six equal sections but only four labels, with 1 and 4 each appearing twice, so the 24 equally likely outcomes include repeats such as 11 and 11. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Much to keep track of moderate Working it out
Twenty-four two-digit numbers have to be built and each tested for primality while the repeated sections are kept track of; a 4-by-6 table is the reliable way to hold them. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
Which two-digit numbers are prime is not given; students must test 21, 33, 43 and the rest themselves, and 21 = 3 × 7 and 33 = 3 × 11 look prime at a glance. (Anderson & Krathwohl, 2001)

Strategy: Make a 4-by-6 table with Spinner R's digits as tens and Spinner T's six sections as ones, mark each prime, and count every section, including repeated labels.

Traps

An arithmetic slip: The right method with a plausible arithmetic error: a digit, a place value or a times-table fact.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B An arithmetic slip (Junker & Sijtsma, 2001)
Reveals a slip Expected pull: strong
Counts 33 as prime, missing that 33 = 3 × 11, so 10 of the 24 outcomes look prime and 10/24 reduces to 5/12.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Finds the 9 prime outcomes but compares them with the 15 that are not prime, a part-to-part ratio, 9/15 = 3/5, instead of dividing by all 24.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Counts the 15 outcomes that are not prime and gives 15/24 = 5/8, the probability of the opposite event.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting outcomes: $\text{outcomes} = a \times b$; a is the number of equal sections on the first spinner, and b is the number on the second. Each section of the first can go with each section of the second. So the number of pairs is a times b.
Probability: $P = \frac{\text{wanted results}}{\text{possible results}}$; P is the probability, the chance that something happens. Possible results are all the results that are equally likely. Wanted results are the ones you are looking for. The chance is the fraction of all the results that you want.

Sort out the question
Spinner R gives the tens digit. Spinner T gives the ones digit. Together they make a two-digit number, like 23. You need the chance that the number is prime. A prime number has exactly two factors, 1 and itself. For example, 13 is prime, but 21 = 3 × 7 is not.
Count all the outcomes
Spinner R has 4 equal sections. Spinner T has 6 equal sections. Look closely: T has two 1s and two 4s. Each section counts as its own outcome.
$\text{outcomes} = 4 \times 6 = 24$
There are 24 outcomes, all equally likely.
Make a table of the numbers
Each row is a tens digit from R. Each column is a section of T. Prime numbers are in bold.
R 1 1 2 3 4 4
1 11 11 12 13 14 14
2 21 21 22 23 24 24
3 31 31 32 33 34 34
4 41 41 42 43 44 44
Count the prime numbers
Numbers ending in 2 or 4 are even, so 2 divides them. They are not prime. Check the rest. 11, 13, 23, 31, 41 and 43 are prime. But 21 = 3 × 7 and 33 = 3 × 11 are not. Count every section, so 11, 31 and 41 each count twice. That makes 3 + 1 + 2 + 3 = 9 primes. Watch out: counting 33 as prime gives $\frac{10}{24} = \frac{5}{12}$ , choice B.
Find the probability
$P = \frac{9}{24} = \frac{9 \div 3}{24 \div 3} = \frac{3}{8}$
Watch out: divide by all 24 outcomes. Comparing the 9 primes with the 15 others gives $\frac{9}{15} = \frac{3}{5}$ , choice C. So the answer is A, $\frac{3}{8}$ .

R	1	1	2	3	4	4
1	11	11	12	13	14	14
2	21	21	22	23	24	24
3	31	31	32	33	34	34
4	41	41	42	43	44	44

Check your answer

There are $24 - 9 = 15$ numbers that are not prime. Their chance is $\frac{15}{24} = \frac{5}{8}$ . And $\frac{3}{8} + \frac{5}{8} = 1$ , as it should be. Notice that $\frac{5}{8}$ , choice D, is the chance of not prime.

More questions like this

The original DOE question

Question 81 as printed on page 81 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 81, question 81. Open page 81 of the handbook (PDF)

Question 23 of 40

2023 Test A Math, question 27

Question

A basket contains red balls, green balls, and white balls. There are 12 red balls in the basket. The probability of randomly choosing a red ball is 1 in 3. If the probability of randomly choosing a green ball is 1 in 4, how many green balls are in the basket?

Show the answer

Correct answer: C

First, use the information about the red balls to calculate the total number of balls in the basket:

$\frac{1}{3} = \frac{12}{x}$	Cross multiply.
$x = 36$

There are a total of 36 balls in the basket. The probability of selecting a green ball is 1 in 4. Use that information to calculate how many green balls are in the basket:

$\frac{g}{36} = \frac{1}{4}$	Cross multiply.
$4g = 36$	Apply the multiplicative inverse property; divide both sides of the equation by 4.
$g = 9$

There are 9 green balls in the basket.

Question analysis

A basket holds red, green and white balls, including 12 red ones, and the chances of drawing red and green are 1 in 3 and 1 in 4. Students find how many green balls are in the basket. Getting the total from the red count is the step students skip when they apply a probability to the wrong number.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability is grade 6 content, but here it runs backward: a count and a probability give the total, which then gives another count. Two linked steps like this are typical of grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Many steps minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 46 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The basket's total is never stated, and that missing number is the whole problem: it has to be recovered from the red balls before the green probability means anything. Applying the green chance straight to the 12 red balls is the quick wrong move. Checking that 12 red and 9 green fit in 36 balls with 15 white left over confirms the setup.

Turning words into math minor Understanding the question
The sentence "The probability of randomly choosing a red ball is 1 in 3" has to become 12 = one third of the total, which makes the basket hold 36 balls. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Many steps minor Working it out
Two linked steps: the red balls give the total, 12 × 3 = 36, and only then does the green probability give a count, 36 ÷ 4 = 9. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Use the red balls to find the total, since 12 is one third of it, then take one fourth of that total for the green balls.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Takes one fourth of the 12 red balls, 12 ÷ 4 = 3, applying the green probability to the only count the stem gives instead of to the total.
B Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No use of the two probabilities gives 8, which is one third of the 24 balls that are not red, so it mostly catches guesses.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: moderate
Scales the 12 red balls by 4/3 instead of 3/4, taking the 1-in-4 chance to mean more balls than the 1-in-3 chance: 12 × 4/3 = 16.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{balls of one color}}{\text{all the balls}}$; P is the probability, the chance of picking that color. Each ball is equally likely to be picked. So the chance is the fraction of all the balls that have that color. A chance of 1 in 3 means one third of the balls.

Sort out the question
The basket has red, green and white balls. There are 12 red balls. The chance of picking red is 1 in 3. The chance of picking green is 1 in 4. You need the number of green balls. You don’t know the total number of balls yet, so find it first.
Find the total number of balls
1 in 3 means one third of all the balls are red. So 12 is one third of the total. Call the total number of balls b.
$\frac{12}{b} = \frac{1}{3}$
$b = 12 \times 3 = 36$
There are 36 balls in all.
Find the number of green balls
1 in 4 means one fourth of all 36 balls are green. Call the number of green balls g.
$\frac{g}{36} = \frac{1}{4}$
$g = 36 \div 4 = 9$
Watch out: take one fourth of all 36 balls, not of the 12 red ones. $12 \div 4 = 3$ gives choice A. Multiplying 12 by $\frac{4}{3}$ gives 16, choice D. So the answer is C, 9.

Check your answer

A 1 in 4 chance is smaller than a 1 in 3 chance. So there should be fewer green balls than red ones, and 9 is less than 12. Check: $\frac{9}{36} = \frac{1}{4}$ . The rest, $36 - 12 - 9 = 15$ , are white.

More questions like this

The original DOE question

Question 84 as printed on page 82 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 82, question 84. Open page 82 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 24 of 40

2023 Test A Math, question 30

Question

Lamel has a jar containing 6 red chips, 10 blue chips, and 4 yellow chips. If he removes one chip at random, what is the probability that it will not be red?

A $\frac{4}{5}$
B $\frac{7}{10}$
C $\frac{3}{10}$
D $\frac{1}{5}$

Show the answer

Correct answer: B

There are currently 20 chips in the jar ( $6 + 10 + 4$ ). Of those chips, 14 are not red. So the probability of choosing a chip that is not red is $\frac{14}{20} = \frac{7}{10}$ .

Question analysis

In this probability word problem, Lamel's jar holds red, blue and yellow chips, and students find the chance that one chip drawn at random is not red. It tests counting the favorable outcomes against the total and writing the probability as a fraction. Missing the bold word not, or counting the wrong colors, leads to tempting wrong fractions.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Probability of a single event with equally likely outcomes is grade 6 content in New York. Students add three counts and form one fraction, and the only extra care is the bold word not, so it sits below grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
An answer in an unexpected form minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 33 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-event probability with small whole numbers, so the bold not carries the question. A student who skims past it computes the chance of red, a tidy fraction that is waiting among the choices. The other wrong fractions come from removing the wrong color or keeping only one of the two colors that are not red. Adding blue and yellow first, then dividing by all 20 chips, avoids each of them.

Turning words into math minor Understanding the question
The bold not turns the question around: the chips that count are the 10 blue and 4 yellow ones, 14 in all, or equivalently 1 minus the red share of 6/20. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The count of 14 out of 20 matches no choice until it is reduced to 7/10, so a correct fraction left unsimplified can look missing from the list. (Enright, Morley & Sheehan, 2002)

Strategy: Add all the chips for the total, then count the chips that are blue or yellow, or subtract the chance of red from 1, and simplify the fraction.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Leaves out the 4 yellow chips instead of the 6 red ones, finding the chance of not yellow, 16/20 = 4/5.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Overlooks the bold not and gives the chance that the chip is red, 6 out of 20, which reduces to 3/10.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Counts only the 4 yellow chips as not red and forgets the 10 blue ones, so it finds 4/20 = 1/5, the chance of yellow alone.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. It works when each outcome is equally likely, like pulling one chip without looking.

Sort out the question
Lamel’s jar holds 6 red chips, 10 blue chips and 4 yellow chips. He takes one chip without looking. You need the probability that it is not red. That means the chip is blue or yellow.
Count all the chips
$6 + 10 + 4 = 20$
There are 20 chips in all. That is the bottom of the fraction, because any of the 20 could come out.
Count the chips that are not red
The blue and yellow chips are the ones that are not red.
$10 + 4 = 14$
So 14 chips are not red. Watch out: the question asks for not red. Counting the 6 red chips gives $\frac{6}{20} = \frac{3}{10}$ , choice C. Counting only the 4 yellow chips gives $\frac{1}{5}$ , choice D.
Find and simplify the probability
$P = \frac{\text{outcomes you want}}{\text{all outcomes}}$
$P = \frac{14}{20} = \frac{14 \div 2}{20 \div 2} = \frac{7}{10}$
Dividing the top and bottom by 2 puts the fraction in lowest terms. So the answer is B, $\frac{7}{10}$ .

Check your answer

A chip is either red or not red, so the two chances must add up to 1. The chance of red is $\frac{6}{20} = \frac{3}{10}$ . Then $\frac{3}{10} + \frac{7}{10} = 1$ . It works.

More questions like this

The original DOE question

Question 87 as printed on page 83 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 83, question 87. Open page 83 of the handbook (PDF)

Question 25 of 40

2023 Test A Math, question 53

Question

At West College, students are randomly assigned to one of 20 dormitories and one of 6 dining rooms. Kharleen likes 6 of the dormitories and 2 of the dining rooms. What is the probability that she is assigned to both a dormitory and a dining room that she likes?

A10%
B12%
C19%
D38%

Show the answer

Correct answer: A

Kharleen likes $\frac{6}{20} = \frac{3}{10}$ of the dormitories. She likes $\frac{2}{6} = \frac{1}{3}$ of the dining rooms. Multiply those fractions together to calculate the probability of Kharleen being assigned to both a dormitory and a dining room that she likes: $\frac{3}{10} \times \frac{1}{3} = \frac{1}{10} = 10\%$ .

Question analysis

This probability word problem says students at West College are randomly assigned one of 20 dormitories and one of 6 dining rooms, and Kharleen likes 6 dormitories and 2 dining rooms. It asks for the chance that she gets both a dormitory and a dining room she likes. Students combine two independent events and write the result as a percent.

NYS standards: 7.SP.8a compound event as fraction of outcomes
6.RP.3c percent as a rate per 100
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the chance that two independent events both happen is grade 7 compound probability. Each event's chance comes straight from the counts given, and joining them and writing the result as a percent is a typical grade 7 task.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
Turning words into math minor
A very close wrong answer minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 45 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The question turns on one rule, multiplying the chances of two independent events. Students who know it reduce the fractions first and finish in seconds; students who don't tend to combine the counts, and 6 × 2 = 12 reads naturally as 12%. The other two choices match no common working, so the real contest is between the product of the counts and the product of the chances.

Something you must already know moderate Working it out
The rule that the chance of two independent events both happening is the product of their chances, 3/10 × 1/3, is assumed; nothing in the stem says to multiply. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
The phrase “both a dormitory and a dining room that she likes” describes two separate random assignments that must both go her way, so the two chances are combined, not compared. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
A very close wrong answer minor Choosing the answer
The choice 12% sits only 2 points above 10%, so a rough sense that the answer is small cannot separate them; only computing 3/10 × 1/3 does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Write the chance of a liked dormitory and the chance of a liked dining room as fractions, multiply them because both must happen, and convert the product to a percent.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.
A formula you must know: Needs a formula the test does not give: area, circumference, volume, probability rules.

The wrong answers, classified

B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Multiplies the numbers of liked options, 6 × 2 = 12, and reads that product as 12%, without dividing by the 20 × 6 possible assignments.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No combination of 6/20 and 2/6 gives 19%, a value between the right answer and either single chance, so it mostly draws estimates and guesses.
D Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No common working gives 38%, which is larger than either single chance, 30% or about 33%, so it mostly catches guesses.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want. The bottom counts every outcome that could happen. Each outcome must be equally likely.
Probability of two separate events: $P(\text{A and B}) = P(\text{A}) \times P(\text{B})$; A and B are two events that don’t affect each other, like two separate random picks. P(A) is the chance of A, and P(B) is the chance of B. The chance that both happen is the two chances multiplied.

Sort out the question
Kharleen gets one of 20 dorms, or dormitories, where students live. She also gets one of 6 dining rooms. Each is picked at random. She likes 6 dorms and 2 dining rooms. You need the chance that she gets a dorm she likes and a dining room she likes.
Find the chance of a liked dorm
$P = \frac{\text{outcomes you want}}{\text{all outcomes}}$
$P(\text{dorm}) = \frac{6}{20} = \frac{3}{10}$
She likes 6 of the 20 dorms. Divide the top and bottom by 2 to simplify.
Find the chance of a liked dining room
Use the same formula for the dining rooms.
$P(\text{dining room}) = \frac{2}{6} = \frac{1}{3}$
She likes 2 of the 6 dining rooms. Divide the top and bottom by 2.
Multiply the two chances
The two picks are separate. One doesn’t change the other.
$P(\text{A and B}) = P(\text{A}) \times P(\text{B})$
$\frac{3}{10} \times \frac{1}{3} = \frac{3}{30} = \frac{1}{10}$
Both must happen, so the chance is smaller than either one alone.
Change to a percent
$\frac{1}{10} = \frac{10}{100} = 10\%$
Watch out: multiplying 6 × 2 = 12 and calling it 12% skips dividing by all the possible assignments. That is choice B. So the answer is A, 10%.

Check your answer

Count every possible assignment. There are $20 \times 6 = 120$ dorm and dining room pairs, all equally likely. She likes $6 \times 2 = 12$ of them. Then $\frac{12}{120} = \frac{1}{10}$ , which is 10%, the same answer.

More questions like this

The original DOE question

Question 110 as printed on page 90 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 90, question 110. Open page 90 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 26 of 40

2023 Test A Math, question 57

Question

Chanelle selects a marble from a container and then returns it to the container. When she does this 3 times, the probability of choosing a red marble all 3 times is $\frac{1}{216}$ . Based on this information, what is the probability of Chanelle choosing a red marble the first time she selects a marble?

A $\frac{1}{72}$
B $\frac{1}{36}$
C $\frac{1}{8}$
D $\frac{1}{6}$

Show the answer

Correct answer: D

Let $\frac{1}{x}$ be the probability of Chanelle selecting a red marble the first time. The probability of selecting a red marble three times is $\frac{1}{x} \times \frac{1}{x} \times \frac{1}{x}$ , which is given as $\frac{1}{216}$ . Since $216 = 6 \times 6 \times 6$ , the value of x is 6, and the probability of Chanelle selecting a red marble the first time is $\frac{1}{6}$ .

Question analysis

This probability question says Chanelle draws a marble and puts it back, three times, and the chance of red all three times is 1/216. It asks for the chance of red on her first draw. Students reverse the multiplication rule for repeated independent draws, which means finding a cube root, and dividing by 3 is the tempting shortcut.

NYS standards: 7.SP.8a compound event as fraction of outcomes
8.EE.2 square and cube roots
Grade level: Grade 8 content, by NYS standard
Compared with grade 7: Above grade 7 Multiplying probabilities for repeated draws with replacement is grade 7 compound probability. Running it backward, from three draws to one, needs a cube root, a grade 8 idea, though the numbers are friendly and the work is short once set up.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
No obvious method minor
Solution: 2 steps
Reading level: Question and choices: about grade 7, 55 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the three draws are seen as one chance cubed, the question is a cube root, and 216 is a perfect cube many students know. The risk lies in undoing the cube the wrong way: dividing by 3 because there are three draws, or stopping at 36 after one division by 6. A quick check, 1/6 × 1/6 × 1/6 = 1/216, confirms the answer.

Something you must already know moderate Working it out
Students must know that three draws with the marble returned each time multiply, p × p × p, and must recognize 216 as 6 × 6 × 6; the stem states neither. (Anderson & Krathwohl, 2001)
No obvious method minor Working it out
The usual direction runs from one draw to three; here the three-draw chance is given and the bold first asks for one draw, so students work backward and undo a cube. (Schoenfeld, 1985)

Strategy: Call the chance of red on one draw p, so three draws with replacement give p × p × p; then find the number whose cube equals the given probability.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Divides 216 by 3 because there are three draws, instead of finding the number that multiplied by itself three times makes 216.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Divides 216 by 6 to get 36 and stops one division short, keeping 1/36, which is the chance of two reds in a row, not one.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No working with 1/216 gives 1/8, the chance of three heads in three coin tosses, so it mostly attracts guesses at a familiar fraction.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of repeated picks: $P(\text{red 3 times}) = p \times p \times p$; p is the chance of red on one pick. Chanelle puts the marble back each time, so every pick has the same chance p, and one pick doesn’t change the next. The chance of red on all three picks is the three chances multiplied.

Sort out the question
Chanelle picks a marble and puts it back, 3 times. The chance of red all 3 times is $\frac{1}{216}$ . You need the chance of red on just the first pick. Call that chance p.
Write the three picks as a product
She puts each marble back, so the container is the same every time. Each pick has the same chance, p.
$P(\text{red 3 times}) = p \times p \times p$
$p \times p \times p = \frac{1}{216}$
Find the number that fits
You need a fraction that, multiplied by itself three times, makes $\frac{1}{216}$ . Its top is 1, since 1 × 1 × 1 = 1. For the bottom, find a number that times itself three times makes 216.
$6 \times 6 \times 6 = 36 \times 6 = 216$
So the bottom is 6, and p is $\frac{1}{6}$ .
Avoid the tempting shortcuts
Watch out: dividing 216 by 3 gives 72, choice A. Three picks means multiplying three times, not dividing by 3. Also, $\frac{1}{36}$ is the chance of red twice in a row, choice B.
So the answer is D, $\frac{1}{6}$ .

Check your answer

Multiply it out: $\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{216}$ . That matches the question. Choice A fails fast: $\frac{1}{72} \times \frac{1}{72}$ is already far smaller than $\frac{1}{216}$ .

More questions like this

The original DOE question

Question 114 as printed on page 91 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 91, question 114. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 27 of 40

2023 Test B Math, question 13

Question

Shiow-Chen has a box that contains 4 green cards, 7 yellow cards, and 6 red cards. She randomly draws one card at a time from the box and does not return the cards to the box after they have been drawn. The first card she draws is yellow. The second card she draws is green. What is the probability that the third card she draws will also be green?

A $\frac{1}{5}$
B $\frac{4}{17}$
C $\frac{1}{4}$
D $\frac{1}{3}$

Show the answer

Correct answer: A

After Shiow-Chen draws a yellow card and a green card, the remaining cards are 3 green, 6 yellow, and 6 red. The probability of drawing a green card from that group is:

$\frac{3}{3 + 6 + 6} = \frac{3}{15} = \frac{1}{5}$

Question analysis

Shiow-Chen's box holds 4 green, 7 yellow and 6 red cards, and she draws cards one at a time without putting them back. Her first card is yellow and her second is green. Students find the probability that the third card is also green, which means updating both the green count and the total first.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability is grade 6 content. The extra demand is tracking two earlier draws that were not put back, which changes the counts for the third draw, a step typical grade 7 problems often leave out.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of moderate
Turning words into math minor
Solution: 3 steps
Reading level: Question and choices: about grade 3, 70 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The probability itself is one fraction, but it has to be built from a box that two draws have already changed. Students who use the original 17 cards, or who compare green with the other colors instead of with all the cards, reach listed choices that look reasonable. Listing the counts after each draw, then putting the green cards over all the cards left, avoids both.

Much to keep track of moderate Working it out
Three color counts and the total must each be updated after two draws, 4 green, 7 yellow and 6 red becoming 3, 6 and 6 out of 15, before the third draw is considered. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Turning words into math minor Understanding the question
“Does not return the cards” and the two named draws are the conditions that change the box, and the stem states them in passing before asking about the third card. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Take the two drawn cards out of the box: lower the yellow and green counts by one each, recount the total, then put the green cards left over the cards left.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.

The wrong answers, classified

B Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Uses the full box, 4 green cards out of 17, as if neither of the first two cards had been taken out.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Updates the counts correctly but compares the 3 green cards with the 12 cards of other colors, 3/12 = 1/4, a part-to-part ratio.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the three colors as equally likely, giving green a 1-in-3 chance whatever the number of cards of each color.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number of results you want}}{\text{total number of results}}$; P is the probability, how likely something is, from 0 (never) to 1 (always). The top counts the results you want, here green cards. The bottom counts every possible result, here every card in the box. It works when each result is equally likely, as in a random draw.

Sort out the question
The box starts with 4 green, 7 yellow and 6 red cards. Shiow-Chen draws cards one at a time, and she does not put them back. The first card is yellow, and the second is green. You need the probability that the third card is green.
Take out the two cards
The cards are not put back, so the box changes. Lower the yellow count by 1 and the green count by 1.
Color At the start After two draws
Green 4 3
Yellow 7 6
Red 6 6
Total 17 15
Use the probability formula
For the third draw, the results you want are the 3 green cards left. The total is the 15 cards left.
$P = \frac{\text{number of results you want}}{\text{total number of results}}$
$P = \frac{3}{3 + 6 + 6} = \frac{3}{15}$
Watch out: using the full box gives $\frac{4}{17}$ , choice B. But two cards are gone.
Simplify and answer
Divide the top and bottom by 3.
$\frac{3}{15} = \frac{1}{5}$
Watch out: compare the green cards with all 15 cards, not with the 12 cards of other colors. That gives $\frac{1}{4}$ , choice C. And the three colors are not equally likely, so $\frac{1}{3}$ is wrong too. So the answer is A, $\frac{1}{5}$ .

Color	At the start	After two draws
Green	4	3
Yellow	7	6
Red	6	6
Total	17	15

Check your answer

17 cards minus the 2 drawn leaves 15. One fifth of 15 is 3, the number of green cards left. Also, the chance went down from $\frac{4}{17}$ . That makes sense, because a green card was taken out.

More questions like this

The original DOE question

Question 70 as printed on page 195 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 195, question 70. Open page 195 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 28 of 40

2023 Test B Math, question 53

Question

Micha is playing a game with five cards numbered 1 through 5. He will place the cards in a bag and draw one card at random three times, replacing the card each time. To win a prize, he must draw the number 5 all three times. What is the probability he will draw the number 5 all three times?

A $\frac{1}{125}$
B $\frac{3}{125}$
C $\frac{1}{15}$
D $\frac{3}{5}$

Show the answer

Correct answer: A

Each time Micha draws a card, the probability of drawing the number 5 is $\frac{1}{5}$ . To find the probability of drawing the number 5 three times, multiply:

$\frac{1}{5}\times\frac{1}{5}\times\frac{1}{5}=\frac{1}{125}$

Question analysis

Micha draws a card three times from five cards numbered 1 to 5, putting the card back each time, and wins only if he draws a 5 every time. Students find the probability that he wins. It tests the probability of independent events, where the chances for the separate draws have to be combined correctly.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: At grade 7 Finding the probability of a compound event is a grade 7 standard, and repeated draws with replacement are a standard case. Three equal factors make this a short, typical grade 7 task.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 3, 59 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

With replacement each draw is the same 1 in 5, so the work is one product, 1/5 × 1/5 × 1/5. The wrong choices mark three ways to mishandle repetition: adding the chances, multiplying 5 by 3 instead of cubing it, and multiplying by 3 for the three draws. A quick sense check helps, since needing three 5s in a row must be far less likely than needing one.

Something you must already know minor Working it out
The rule that the chance of independent events all happening is the product of their chances is assumed, as is that replacing the card keeps each draw at 1/5. (Anderson & Krathwohl, 2001)

Strategy: Find the probability of drawing a 5 on one draw, then, because the card goes back each time, multiply that probability by itself once for each draw.

Traps

Wrong operation: Adds instead of multiplying, divides the wrong way round, turns a ratio or rate upside down, or undoes a step in the wrong direction (subtracts what should be added back).
A faulty procedure: A systematic but wrong procedure: adding fractions straight across, applying an exponent to each term of a sum, dropping a sign when moving a term.
Overlapping groups: Things in two groups at once are counted twice, or not subtracted.

The wrong answers, classified

B Overlapping groups (King et al., 2004)
Reveals a misconception Expected pull: weak
Finds 1/125 for three 5s, then multiplies by 3 for the three draws, counting the single outcome 5, 5, 5 as if it could happen in three orders.
C A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals a misconception Expected pull: moderate
Multiplies 5 by 3 instead of raising it to the third power, so the denominator becomes 15 rather than 5 × 5 × 5 = 125.
D Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Adds the three probabilities, 1/5 + 1/5 + 1/5 = 3/5, as if drawing a 5 on any one draw were enough to win.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{results you want}}{\text{all possible results}}$; P is the probability, the chance something happens. It is a number from 0 to 1. The top counts the results you want. The bottom counts all the results that can happen. It works when every result is equally likely.
Chance of events in a row: $P(\text{A and B and C}) = P(\text{A}) \times P(\text{B}) \times P(\text{C})$; A, B and C are events, things that might happen, like drawing a 5. When they do not change each other’s chances, the chance that all of them happen is the product of their chances. Putting each card back keeps the draws from changing each other.

Sort out the question
The bag has 5 cards, numbered 1 to 5. Micha draws a card three times. He puts it back each time, so every draw starts with all 5 cards. He wins only if he draws a 5 all three times. You need the probability, or chance, of that.
Find the chance for one draw
One card out of 5 is a 5. Use the probability formula.
$P(5) = \frac{1}{5}$
Each draw has a 1 in 5 chance of being a 5.
Multiply for three draws
He needs a 5 on the first draw, and the second, and the third. The card goes back each time, so the draws don’t change each other. Use the rule for events in a row.
$P(\text{5 and 5 and 5}) = \frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{5 \times 5 \times 5} = \frac{1}{125}$
Give the answer
Watch out: Adding the chances gives $\frac{3}{5}$ , choice D. That would make winning likely, which makes no sense. Multiplying 5 × 3 instead of 5 × 5 × 5 gives $\frac{1}{15}$ , choice C.
So the answer is A, $\frac{1}{125}$ .

Check your answer

Count all the results. Each draw has 5 possible cards, so three draws have 5 × 5 × 5 = 125 possible orders. Only one of them is 5, 5, 5. That gives 1 out of 125. Needing three 5s should also be much less likely than needing one.

More questions like this

The original DOE question

Question 110 as printed on page 206 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 206, question 110. Open page 206 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 29 of 40

2023 Test B Math, question 57

Question

A box contains 14 red marbles, 10 black marbles, and 16 white marbles. How many more white marbles need to be added to the box so that the probability of randomly drawing a white marble is $\frac{3}{5}$ ?

Show the answer

Correct answer: D

Let w represent the number of additional white marbles.

Set the probability of drawing a white marble equal to $\frac{3}{5}$ and solve the equation for w:

$\frac{16+w}{14+10+16+w}=\frac{3}{5}$

$\frac{16+w}{40+w}=\frac{3}{5}$

$5(16+w)=3(40+w)$

$80+5w=120+3w$

$2w=40$

$w=20$

Question analysis

A box holds 14 red, 10 black and 16 white marbles, and students find how many white marbles must be added so that the chance of drawing a white one is 3/5. Working backward from a target probability to a number of marbles is the unusual step.

NYS standards: 6.SP.8a uniform probability models
7.RP.3 multistep ratio and percent problems
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 probability questions usually give the counts and ask for a probability. This runs the other way: the target probability is given, and the number to add changes both the white count and the total.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Variables instead of numbers minor
Many steps minor
Solution: 3 steps
Reading level: Question and choices: about grade 8, 34 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The tempting shortcut is to take 3/5 of the 40 marbles already in the box, which forgets that the new white marbles join the total. Students who set up the probability with the unknown in both the top and the bottom solve it in a few lines, and testing choices works too: adding 20 gives 36 white of 60, exactly 3/5. The other two choices are counts copied from the stem.

Turning words into math moderate Understanding the question
“How many more white marbles need to be added” puts the unknown in two places, (16 + w)/(40 + w), because each white marble added also raises the total. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Variables instead of numbers minor Understanding the question
The unknown number of added marbles has to be carried as a letter through a proportion, 5(16 + w) = 3(40 + w), before it can be found. (Koedinger & Nathan, 2004; Enright, Morley & Sheehan, 2002)
Many steps minor Working it out
Totaling the marbles, setting up the proportion, cross-multiplying and solving 80 + 5w = 120 + 3w make four steps, unless the choices are tested one by one. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

Strategy: Let w be the number of white marbles added, write the new probability with w in both the top and the bottom, set it equal to 3/5 and solve; or test each choice.

Traps

Percent of the wrong amount: Finding the whole from a part, or a percent change measured from the wrong starting amount.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Percent of the wrong amount (Parker & Leinhardt, 1995)
Reveals partial knowledge Expected pull: strong
Takes 3/5 of the 40 marbles now in the box, 24, and subtracts the 16 white ones, forgetting that added marbles raise the total.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 10, the number of black marbles in the stem, without setting up the new probability or testing it.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Picks 14, the number of red marbles in the stem, without checking that adding 14 white marbles gives 30 of 54, not 3/5.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P(\text{white}) = \frac{\text{white marbles}}{\text{all marbles}}$; P(white) is the probability, or chance, of drawing a white marble. The top counts the white marbles. The bottom counts every marble in the box. It works because each marble is equally likely to be drawn.
Cross multiplication: $\frac{a}{b} = \frac{c}{d} \text{ means } a \times d = b \times c$; a and c are the tops. b and d are the bottoms. When two fractions are equal, each top times the other bottom gives the same product. It works because you multiply both sides by both bottoms. Use it to clear fractions from an equation.

Sort out the box
The box has 14 red, 10 black and 16 white marbles.
$14 + 10 + 16 = 40$
That is 40 marbles. You will add some white marbles. Call that number w. You need w so that the chance of drawing white is $\frac{3}{5}$ .
Write the new probability
Adding w white marbles changes both counts. There will be 16 + w white marbles and 40 + w marbles in all. Use the probability formula.
$\frac{16 + w}{40 + w} = \frac{3}{5}$
Watch out: Taking $\frac{3}{5}$ of 40 gives 24, and 24 − 16 = 8, choice A. That forgets that the new marbles also make the box bigger.
Cross multiply
Fractions are hard to solve with. Clear them with cross multiplication.
$5 \times (16 + w) = 3 \times (40 + w)$
Then multiply each outside number by both terms inside the parentheses.
$80 + 5w = 120 + 3w$
Solve for w
Subtract 3w from both sides. Then subtract 80 from both sides.
$2w = 40$
Divide both sides by 2.
$w = 20$
You need to add 20 white marbles.
So the answer is D, 20.

Check your answer

Add 20 white marbles. Then there are 16 + 20 = 36 white marbles and 40 + 20 = 60 marbles in all. Divide the top and bottom by 12: $\frac{36}{60} = \frac{3}{5}$ . It works.

More questions like this

The original DOE question

Question 114 as printed on page 208 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 208, question 114. Open page 208 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 30 of 40

2018 Test A Math, question 16

Question

A box of colored pencils contains exactly 6 red pencils. The probability of choosing a red pencil from the box is $\frac{2}{7}$ . How many of the pencils in the box are not red?

Show the answer

Correct answer: B

Let $x$ be the total number of colored pencils in the box. Set up a proportion to find $x$ :

$\frac{2}{7}=\frac{6}{x}$

$2x=42$

$x=21$

If there are 6 red pencils, then the number of pencils that are not red is

21 − 6 = 15.

Question analysis

A probability word problem: a box holds exactly 6 red pencils, and the chance of drawing a red one is 2/7. Students work backward from the probability to the number of pencils in the box, then answer the question actually asked, which is about the pencils that are not red.

NYS standards: 6.SP.8a uniform probability models
6.RP.3 ratio and rate reasoning
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Simple probability is grade 6 content in New York. Reversing it to find a total is a step beyond plugging counts into a fraction, which puts this at about the level of typical grade 7 work.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 32 words.
Corrected from the DOE handbook: A spelling mistake in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Students have to treat a probability as a ratio and scale it up: 6 red pencils is three times the 2 in 2/7. The common loss is choosing the total number of pencils, which appears naturally halfway through the work and is a listed choice. Reading the 7 as a pencil count, or scaling by 6 instead of 3, leads to the other wrong answers.

Turning words into math moderate Understanding the question
"The probability of choosing a red pencil from the box is 2/7" has to become a part-to-whole ratio: 2 of every 7 pencils are red, so the 6 red pencils are 2 parts. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
The question assumes students know that a probability is favorable outcomes over all outcomes, here used in reverse to find the total from the red count. (Anderson & Krathwohl, 2001)

Strategy: Since 2 parts out of 7 are red, find how many pencils one part stands for, then count the parts that are not red.

Traps

Stops one step early: A choice is a number from partway through the solution.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Reads 2/7 as 2 red pencils among exactly 7 and takes 7 − 2 = 5, ignoring the stated 6 red pencils.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Solves 2/7 = 6/x to get 21 pencils in all and stops before subtracting the 6 red ones that the bold not excludes.
D A faulty procedure (Brown & Burton, 1978; Tatsuoka, 1983)
Reveals partial knowledge Expected pull: weak
Treats the 6 red pencils as one share of the 7 instead of two, so each share is 6 and the 5 non-red shares make 30.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “the number of pencils that are not read is 21 − 6 = 15”. “read” now reads “red”.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{favorable}}{\text{total}}$; P is the probability, the chance that something happens. Favorable is the number of ways to get what you want, here a red pencil. Total is the number of all possible picks, here every pencil in the box. It works when every pick is equally likely.

Sort out the question
A probability is a number that tells how likely something is. The box has exactly 6 red pencils. The chance of picking a red one is $\frac{2}{7}$ . You need the number of pencils that are not red.
Read the probability as parts
$\frac{2}{7}$ means 2 out of every 7 pencils are red. Picture the box split into 7 equal parts. Red pencils fill 2 of the parts. The other 5 parts are not red. Watch out: the box does not hold just 7 pencils. It has 6 red ones alone. Taking $7 - 2 = 5$ is choice A.
Find the size of one part
The 6 red pencils fill 2 parts. Divide to find one part.
$6 \div 2 = 3$
Each part is 3 pencils. So the whole box has $7 \times 3 = 21$ pencils.
Check with the probability formula
Put the counts into the formula to make sure.
$P = \frac{\text{favorable}}{\text{total}} = \frac{6}{21} = \frac{2}{7}$
Divide the top and bottom of 6/21 by 3 to get 2/7. It matches.
Count the pencils that are not red
$21 - 6 = 15$
That is the same as 5 parts of 3 pencils each. Watch out: 21 is all the pencils, choice C. The question asks only for the ones that are not red. So the answer is B, 15.

Check your answer

Red and not red should make the whole box: $6 + 15 = 21$ . Red is less likely than not red, since 2/7 is less than half. 6 red against 15 not red fits that.

More questions like this

The original DOE question

Question 73 as printed on page 84 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 84, question 73. Open page 84 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 31 of 40

2018 Test A Math, question 21

Question

In a sample of 10 cards, 4 are red and 6 are blue. If 2 cards are selected at random from the sample, one at a time without replacement, what is the probability that both cards are not blue?

A $\frac{2}{15}$
B $\frac{4}{25}$
C $\frac{3}{10}$
D $\frac{1}{3}$

Show the answer

Correct answer: A

The probability that both cards are not blue is the same as the probability that both cards are red. There are 4 red cards out of the 10, so the probability of the first card being red is

$\frac{4}{10}$

Now there are 9 cards left, and 3 of those are red, so the probability of the second card being red is

$\frac{3}{9}$

Multiply the two probabilities to find the probability that both cards are red (not blue):

$\left (\frac{4}{10} \right )\left ( \frac{3}{9} \right )=\frac{12}{90}=\frac{2}{15}$

Question analysis

This probability question draws 2 cards, one at a time and without replacement, from 10 cards: 4 red and 6 blue. Students find the chance that neither card drawn is blue. The second draw has one fewer card, and the word not changes which color to count.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability is usually found with a list or tree of equally likely outcomes. Dependent draws without replacement, plus a negative phrasing to translate, go beyond that typical work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Something you must already know moderate
A very close wrong answer moderate
Turning words into math minor
An answer in an unexpected form minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 39 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two ideas stack here: restating not blue as red, and changing the second draw's numbers after a card is removed. Students who know to multiply but forget the removal reach a listed choice, and so do students who compute both blue. Writing the second fraction explicitly as 3 red out of 9 cards, before multiplying, is what separates right from nearly right.

Something you must already know moderate Working it out
Multiplying the first draw's chance by the second draw's, with the second adjusted for the card already taken, is a rule the test assumes and does not give. (Anderson & Krathwohl, 2001)
A very close wrong answer moderate Choosing the answer
2/15 and 4/25 are about 0.13 and 0.16, too close to tell apart by estimate; only the second draw's fraction, 3 red of 9 cards rather than 4 of 10, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Turning words into math minor Understanding the question
"Both cards are not blue" has to be restated as both cards red, and the bold not also invites computing the chance that both are blue. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
An answer in an unexpected form minor Choosing the answer
The product comes out as 12/90, and it has to be reduced to lowest terms before it matches any of the listed fractions. (Enright, Morley & Sheehan, 2002)

Strategy: Restate not blue as red, then multiply the chance of red on the first draw by the chance of red on the second, with one card fewer.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

B Without replacement (Fischbein & Schnarch, 1997)
Reveals partial knowledge Expected pull: strong
Multiplies 4/10 × 4/10 = 16/100 = 4/25, as if the first red card were put back before the second draw.
C Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
No standard working gives 3/10: it is not the chance of red or of blue on one draw or on two, so it mostly catches guesses.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Computes the chance that both cards are blue, 6/10 × 5/9 = 30/90 = 1/3, missing the bold word not.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{favorable}}{\text{total}}$; P is the probability, the chance that something happens. Favorable is the number of ways to get what you want, here a red card. Total is the number of all possible picks, here all the cards you can pick from. It works when every pick is equally likely.
Two events in a row: $P(\text{A and B}) = P(\text{A}) \times P(\text{B after A})$; A is the first event and B is the second. P( ) means the probability of what is inside. The chance that both happen is the first chance times the second. For the second chance, count only what is left after the first event.

Sort out the question
There are 10 cards: 4 red and 6 blue. You pick 2 cards “one at a time without replacement”. That means the first card is not put back. “Both cards are not blue” means both are red, since red is the only other color. You need the probability, or chance, of that.
Find the chance the first card is red
4 of the 10 cards are red. Each card is equally likely to be picked. Use the probability formula.
$P(\text{red first}) = \frac{4}{10}$
Find the chance the second card is red
One red card is gone now, and it is not put back. So 9 cards are left, and 3 of them are red.
$P(\text{red second}) = \frac{3}{9}$
Watch out: using 4/10 again acts as if the card went back. That gives 4/25, choice B.
Multiply the two chances
Both cards must be red. So multiply the first chance by the second. To multiply fractions, multiply the tops, then multiply the bottoms.
$\frac{4}{10} \times \frac{3}{9} = \frac{4 \times 3}{10 \times 9} = \frac{12}{90}$
Simplify the fraction
None of the choices is 12/90, so simplify it. Divide the top and the bottom by 6, the biggest number that goes into both.
$\frac{12 \div 6}{90 \div 6} = \frac{2}{15}$
Watch out: the chance that both are blue is $\frac{6}{10} \times \frac{5}{9} = \frac{1}{3}$ , choice D. The question says not blue. So the answer is A, $\frac{2}{15}$ .

Check your answer

Count the ways. There are $10 \times 9 = 90$ ways to pick two cards in order. Red then red has $4 \times 3 = 12$ ways. That gives 12/90, or 2/15. The answer is small, which makes sense, because there are fewer red cards than blue ones.

More questions like this

The original DOE question

Question 78 as printed on page 85 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 85, question 78. Open page 85 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 32 of 40

2018 Test A Math, question 42

Question

A box contains 5 strawberry candies, 3 banana candies, and 2 orange candies. If Braden selects 2 candies at random from this box, without replacement, what is the probability that both candies are not banana?

A $\frac{1}{15}$
B $\frac{9}{100}$
C $\frac{7}{15}$
D $\frac{49}{100}$

Show the answer

Correct answer: C

The total number of candies in the box is

5 + 3 + 2 = 10

The number of candies that are not banana is

5 + 2 = 7

The probability of the first candy not being banana is

$\frac{7}{10}$

Now, out of 9 candies, there are 6 candies left that are not banana. The probability of the second candy not being banana is

$\frac{6}{9}$

Multiply these two probabilities to get the solution:

$(\frac{7}{10})(\frac{6}{9})=\frac{42}{90}=\frac{7}{15}$

Question analysis

This probability question has a box of strawberry, banana and orange candies and asks the chance that two candies drawn without replacement are both not banana. Students must count the non-banana candies, find the chance for each draw, and multiply. The second draw changes because one candy is already gone.

NYS standards: 7.SP.8a compound event as fraction of outcomes
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability usually uses lists or tree diagrams for independent events such as coins and number cubes. Here the draws are dependent, and students must also work with the complement, 'not banana'.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Combines ideas moderate
A very close wrong answer moderate
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 9, 35 words.
Corrected from the DOE handbook: A stray punctuation mark in the explanation was removed.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Two phrases in the stem change the calculation, "not" and "without replacement", and each has a choice for the student who misses it. The cleanest approach treats strawberry and orange together as one group of 7 and then works through the two draws in order. Students who know independent probability well are the most likely to square a fraction out of habit.

Combines ideas moderate Working it out
Two ideas meet: counting the 7 non-banana candies as the favorable group, and reducing both counts by one for the dependent second draw. (Embretson, 1998; Daniel & Embretson, 2010)
A very close wrong answer moderate Choosing the answer
The choices 49/100 and 7/15 are about 0.49 and 0.47, so estimating cannot separate them; only noticing that the second draw is 6 of 9, not 7 of 10, does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Error-prone arithmetic minor Working it out
The product 7/10 × 6/9 = 42/90 must be simplified to 7/15 to match a choice, a reduction by 6 that is easy to stall on. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Count the candies that are not banana, find the chance the first draw is one of them, reduce both counts by one for the second draw, and multiply.

Traps

Without replacement: In a second pick there is one fewer item, which changes the probability.
Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: moderate
Misses the bold "not" and finds the chance that both candies are banana, 3/10 × 2/9 = 6/90 = 1/15.
B Without replacement (Fischbein & Schnarch, 1997)
Reveals a misconception Expected pull: weak
Squares 3/10, finding two bananas as if the first candy were put back, which misses both "not" and "without replacement".
D Without replacement (Fischbein & Schnarch, 1997)
Reveals a misconception Expected pull: strong
Squares 7/10 as if the first candy were returned, keeping 7 of 10 for the second draw instead of 6 of 9.

What was corrected from the DOE handbook, and why

Explanation: The handbook reads “The probability of the second candy not being banana is. 6/9.” The period after “is” was removed.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{f}{t}$; P is the probability, the chance that something happens. f is the number of favorable outcomes, the ones you want. t is the total number of outcomes. It works when every outcome is equally likely, like picking one item at random.
Two picks in a row: $P(\text{both}) = P(\text{first}) \times P(\text{second})$; To find the chance that two things both happen, multiply their chances. For the second chance, use what is left after the first pick. You multiply because you want a fraction of a fraction: the second pick must work in the share of times the first one did.

Sort out what you know
The box has 5 strawberry, 3 banana and 2 orange candies. Braden takes 2 candies. “Without replacement” means he keeps the first candy out. He does not put it back.
You need the chance that neither candy is banana. Watch out: the bold “not” matters. The chance that both are banana is $\frac{1}{15}$ , choice A.
Count the candies
$5 + 3 + 2 = 10$
There are 10 candies. The ones that are not banana are the strawberry and orange ones:
$5 + 2 = 7$
Find the chance for the first candy
7 of the 10 candies are not banana. Use the probability formula:
$P = \frac{7}{10}$
So the first candy is not banana 7 times out of 10.
Find the chance for the second candy
Say the first candy was not banana. Now 9 candies are left, and 6 of them are not banana:
$P = \frac{6}{9}$
Watch out: the first candy is not put back. Using $\frac{7}{10}$ again gives $\frac{49}{100}$ , choice D.
Multiply the two chances
$\frac{7}{10} \times \frac{6}{9} = \frac{42}{90}$
Simplify by dividing the top and bottom by 6:
$\frac{42 \div 6}{90 \div 6} = \frac{7}{15}$
So the answer is C, $\frac{7}{15}$ .

Check your answer

Count pairs instead. From 10 candies there are 10 × 9 ÷ 2 = 45 different pairs. Pairs with no banana come from the 7 others: 7 × 6 ÷ 2 = 21. And 21 out of 45 is 7/15. It matches.

More questions like this

The original DOE question

Question 99 as printed on page 89 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 89, question 99. Open page 89 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 33 of 40

2018 Test A Math, question 51

Question

A sports store has a container of handballs: 4 blue, 5 red, 8 yellow, 9 white, and 11 green. If one ball is picked from the container at random, what is the probability that it will be yellow?

A $\frac{1}{37}$
B $\frac{1}{8}$
C $\frac{8}{37}$
D $\frac{8}{29}$

Show the answer

Correct answer: C

The total number of handballs in the container is

$4+5+8+9+11=37$

Since there are 8 yellow handballs, the probability of selecting a yellow handball is

$\frac{8}{37}$

Question analysis

This probability question describes a container of handballs in five colors and asks the chance that one ball picked at random is yellow. Students must add all the balls to get the total and compare the yellow count to it. Using the wrong total, such as only the balls that are not yellow, is the easy mistake.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding the probability of a single event as favorable outcomes over total outcomes is grade 6 content. This item takes one addition and one fraction, so it is easier than typical grade 7 compound probability work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 5, 37 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-event probability: add the five colors to get 37 balls and put the 8 yellow ones over that total. The only real decision is the denominator, and the wrong choices offer the non-yellow count, the yellow count, and a numerator of 1. Students who say "favorable over total" to themselves avoid all three.

A very close wrong answer minor Choosing the answer
The choices 8/37 and 8/29 share the numerator 8 and differ only in the denominator, all 37 balls against the 29 that are not yellow. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Add up every ball in the container to get the total number of outcomes, then put the number of yellow balls over that total.

Traps

Misunderstands the concept: A wrong belief about a concept: that percents can be added, that area grows like length, that averages can be averaged, that a bigger denominator means a bigger fraction.
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the 37 balls correctly but puts 1 on top, giving the chance of picking one particular ball rather than any yellow one.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Puts 1 over the 8 yellow balls, a fraction built from the yellow count alone that ignores the other 29 balls.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: strong
Compares the 8 yellow balls with the 29 that are not yellow, a part-to-part ratio, instead of with all 37 balls.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{f}{t}$; P is the probability, the chance that something happens. f is the number of favorable outcomes, the ones you want. t is the total number of outcomes. It works when every outcome is equally likely, like picking one item at random.

Sort out what you know
The container has 4 blue, 5 red, 8 yellow, 9 white and 11 green balls. One ball is picked at random, so every ball has the same chance. You need the chance it is yellow.
Count all the balls
Add up every color, because any ball could be picked:
$4 + 5 + 8 + 9 + 11 = 37$
There are 37 balls in all. That is the total number of outcomes.
Put yellow over the total
8 of the balls are yellow. Those are the favorable outcomes:
$P = \frac{8}{37}$
Watch out: $\frac{8}{29}$ compares yellow with the 29 balls that are not yellow. Probability compares the part with all the balls, so choice D is wrong. So the answer is C, $\frac{8}{37}$ .

Check your answer

8 out of 37 is about 0.22, a little more than 1 in 5. That makes sense, since yellow is a fair share of the balls. Choice A, 1/37, is the chance of picking one certain ball, not any yellow one.

More questions like this

The original DOE question

Question 108 as printed on page 91 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 91, question 108. Open page 91 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 34 of 40

2018 Test B Math, question 13

Question

DESSERT CHOICES

Dessert	Number of Times Ordered
Cookies	42
Pie	23
Cake	47
Ice Cream	48

The table above shows the number of times that different desserts were ordered at a restaurant. Based on this information, what is the probability of a customer ordering ice cream as a dessert?

A25%
B30%
C40%
D48%

Show the answer

Correct answer: B

The total number of desserts ordered is

42 + 23 + 47 + 48 = 160

The probability that ice cream was chosen is

$\frac{48}{160}=\frac{3}{10}=30\%$ .

Question analysis

This probability question gives a table of how many times four desserts were ordered at a restaurant and asks how likely a customer is to order ice cream, as a percent. Students total the orders and divide. The trap is using a raw count, or the wrong total, in place of the full number of desserts ordered.

NYS standards: 6.SP.8b probability models from observed frequencies
6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a probability from observed frequencies is grade 6 content in New York. Here it takes one sum, one division and a conversion to a percent with friendly numbers, lighter than typical grade 7 work.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Reading a figure, table or graph minor
Error-prone arithmetic minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 45 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a direct relative-frequency question with friendly numbers, and most students finish it quickly. The risk is using a number from the table as if it were already a percent, or treating the four desserts as equally likely. Remembering that a probability compares one count with the total of all counts prevents both mistakes.

Reading a figure, table or graph minor Understanding the question
The four counts sit in a table, and the probability needs all of them: the ice cream row gives the numerator and the sum of the column gives the denominator. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)
Error-prone arithmetic minor Working it out
Adding 42 + 23 + 47 + 48 by hand and then simplifying 48/160 to 3/10 are small steps where a careless total spoils the percent. (Brown & Burton, 1978; Junker & Sijtsma, 2001)

Strategy: Add the four counts in the table to get all the orders, divide the ice cream count by that total, and convert the fraction to a percent.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the four desserts as equally likely, one chance in four, ignoring how often each was actually ordered.
C Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Divides 48 by the 112 orders of the other desserts instead of all 160, getting about 43%, and takes the nearest choice.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: strong
Reads the 48 ice cream orders as 48 percent without dividing by the total number of desserts ordered.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want, here ice cream orders. The bottom counts every outcome, here all dessert orders. It works because each order counts the same.
Fraction to percent: $\text{percent} = \text{fraction} \times 100$; Percent means out of 100. Multiply a fraction or a decimal by 100 to write it as a percent. For example, 3/10 times 100 is 30, so 3/10 is 30%.

Sort out the table
The table shows how many times each dessert was ordered. Ice cream was ordered 48 times.
You need the chance that a customer orders ice cream. The orders in the table show how often each dessert is picked.
Find the total orders
Add all four rows of the table.
$42 + 23 + 47 + 48 = 160$
There were 160 dessert orders in all. This is the bottom of the fraction, because it counts every order.
Write the probability
Put the counts into the probability formula. The top is the 48 ice cream orders. The bottom is all 160 orders.
$P = \frac{\text{outcomes you want}}{\text{all outcomes}} = \frac{48}{160}$
Simplify the fraction. Both numbers can be divided by 16.
$\frac{48 \div 16}{160 \div 16} = \frac{3}{10}$
Change to a percent
$\frac{3}{10} \times 100 = 30$
The probability is 30%.
Watch out: 48 orders is not 48%, choice D. You must divide by the total. Also, the four desserts are not equally popular. So the chance is not 1 out of 4, or 25%, choice A.
So the answer is B, 30%.

Check your answer

Work backward: $0.30 \times 160 = 48$ . That matches the ice cream row. Also, 48 is a bit less than a third of 160, and 30% is a bit less than a third.

More questions like this

The original DOE question

Question 70 as printed on page 155 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 155, question 70. Open page 155 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 35 of 40

2018 Test B Math, question 16

Question

A cooler contains three types of beverages: 5 bottles of apple juice, 3 bottles of grape juice, and 6 bottles of orange juice. What is the probability that a bottle chosen at random from this cooler is not apple juice?

A $\frac{1}{9}$
B $\frac{5}{14}$
C $\frac{9}{14}$
D $\frac{2}{3}$

Show the answer

Correct answer: C

The total number of bottles of juice in the cooler is

5 + 3 + 6 = 14

The number of bottles of juice that are not apple juice (grape juice and orange juice) is

3 + 6 = 9

So the probability is $\frac{9}{14}$ .

Question analysis

This probability question describes a cooler holding apple, grape and orange juice bottles in different numbers and asks for the chance that a randomly chosen bottle is not apple juice. Students count favorable bottles over all bottles. The word not is the trap, along with counting kinds of drink instead of bottles.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Simple probability with equally likely outcomes is grade 6 content in New York. This needs one count and one fraction, and the only extra demand is reading the word not, so it sits below grade 7.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 41 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The counting is easy, 14 bottles in all, so the question mainly tests careful reading. The word “not” turns the event around, and the probability of the opposite event is waiting among the choices. The other wrong choices catch students who count kinds of juice instead of bottles, or who build a fraction with the wrong total.

Turning words into math minor Understanding the question
The event “not apple juice” must be turned into grape plus orange, 3 + 6 bottles over all 14, so the word “not” decides which counts go in the numerator. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Add all the bottles for the denominator, then count only the grape and orange bottles for the numerator, or subtract the chance of apple juice from 1.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Puts 1 over the 9 bottles that are not apple juice, as if one particular bottle had to be picked.
B Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: strong
Finds the probability of choosing apple juice, 5 out of 14, missing the word “not” in the question.
D Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Counts two of the three kinds of juice as not apple and treats the kinds as equally likely, ignoring how many bottles each has.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{outcomes you want}}{\text{all outcomes}}$; P is the probability, the chance that something happens. The top counts the outcomes you want, here bottles that are not apple. The bottom counts all outcomes, here every bottle. It works when every bottle has the same chance.
Probability of “not”: $P(\text{not } A) = 1 - P(A)$; A is an event, such as picking apple juice. P(A) is its chance. Every bottle either is apple or is not, so the two chances add to 1. Subtract from 1 to get the chance that A does not happen.

Sort out the question
The cooler has 5 apple, 3 grape and 6 orange bottles. One bottle is picked at random. That means every bottle has the same chance.
You need the chance that it is not apple juice.
Count all the bottles
$5 + 3 + 6 = 14$
There are 14 bottles. This is the bottom of the probability fraction, since any of them could be picked.
Count the bottles that are not apple
The bottles that are not apple are the grape and orange ones.
$3 + 6 = 9$
There are 9 such bottles. This is the top of the fraction.
Write the probability
$P = \frac{\text{outcomes you want}}{\text{all outcomes}} = \frac{9}{14}$
This fraction cannot be simplified. 9 and 14 share no factor except 1.
Watch out: $\frac{5}{14}$ is the chance of picking apple juice. That is choice B. Read the word “not” carefully.
So the answer is C, $\frac{9}{14}$ .

Check your answer

Use the “not” rule. The chance of apple is $\frac{5}{14}$ . So the chance of not apple is $1 - \frac{5}{14} = \frac{9}{14}$ . Both ways give the same answer.

More questions like this

The original DOE question

Question 73 as printed on page 156 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 156, question 73. Open page 156 of the handbook (PDF)

Question 36 of 40

2018 Test B Math, question 32

Question

PET SURVEY

Number of Pets	Number of Students
0	12
1	16
2	7
3 or more	5

Amy surveyed students at her school about the number of pets they have. What is the probability that a student who participated in the survey has at least 2 pets?

A $\frac{7}{40}$
B $\frac{1}{12}$
C $\frac{1}{8}$
D $\frac{3}{10}$

Show the answer

Correct answer: D

First, add the number of students for each category to find out how many total students were in the survey:

12 + 16 + 7 + 5 = 40

The number of students who had at least 2 pets are the ones who have 2 pets (7) plus the ones who have 3 or more (5). The total number of students with at least 2 pets is

7 + 5 = 12

The probability of a student in the survey having at least two pets is:

$\frac{12}{40}=\frac{3}{10}$

Question analysis

Amy's pet survey table shows how many pets each student has, and the question asks for the probability that a randomly chosen student has at least 2 pets. It tests finding a probability from survey data. The tricky part is reading "at least 2" correctly and using the right total as the denominator.

NYS standards: 6.SP.8a uniform probability models
6.SP.5a number of observations
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Simple probability from a frequency table is grade 6 work in New York. The only extra demand is interpreting "at least" and adding a column, two short steps a grade 7 student should handle easily.
Estimated difficulty: Straightforward: two steps or one careful reading.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math minor
Reading a figure, table or graph minor
Solution: 3 steps
Reading level: Question and choices: about grade 5, 43 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The arithmetic is light, so the question is decided by the phrase "at least 2" and by remembering that the denominator is everyone surveyed. Students who include only one of the two qualifying rows reach an answer with the right denominator, which makes the mistake hard to notice. Rereading the condition before choosing is the main safeguard.

Turning words into math minor Understanding the question
"At least 2 pets" has to be turned into two rows of the table, the 2-pet row and the 3-or-more row, not one. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The denominator is not printed anywhere; it has to be built by adding all four rows of the table, 12 + 16 + 7 + 5 = 40. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Total the students in every row to get the denominator, then add the counts for each row that satisfies "at least 2" and simplify the fraction.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".

The wrong answers, classified

A Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Reads "at least 2 pets" as exactly 2, counting only the 7 students in that row and leaving out the 5 with 3 or more.
B Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Writes 1 over 12, apparently from the 12 qualifying students or the 12 in the 0-pet row, with no total in the denominator.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Reads "at least 2" as more than 2, counting only the 5 students with 3 or more pets: 5/40 = 1/8.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number you want}}{\text{total number}}$; P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

Sort out the question
The table shows how many students have each number of pets. You pick one student who took the survey. You need the chance that the student has “at least 2 pets”.
At least 2 means 2 or more. So two rows count: the 2 pets row and the 3 or more row.
Find the total number of students
The bottom of the fraction is everyone in the survey. Add all four rows.
$12 + 16 + 7 + 5 = 40$
There are 40 students in all.
Count the students you want
Add the 2 pets row and the 3 or more row.
$7 + 5 = 12$
So 12 students have at least 2 pets.
Watch out: using only the 7 students with exactly 2 pets gives $\frac{7}{40}$ , choice A. Using only the 5 with 3 or more gives $\frac{5}{40} = \frac{1}{8}$ , choice C.
Write and simplify the probability
Use the probability formula.
$P = \frac{12}{40}$
Divide the top and bottom by 4, since 4 goes into both.
$\frac{12}{40} = \frac{12 \div 4}{40 \div 4} = \frac{3}{10}$
So the answer is D, $\frac{3}{10}$ .

Check your answer

The students with fewer than 2 pets are $12 + 16 = 28$ . Then $28 + 12 = 40$ , so every student is counted once. Also, 12 out of 40 is a bit less than a third. So is $\frac{3}{10}$ .

More questions like this

The original DOE question

Question 89 as printed on page 159 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 159, question 89. Open page 159 of the handbook (PDF)

Question 37 of 40

2018 Test B Math, question 46

Question

In a sample of 50 cars at a local dealership, there are 12 red cars and 10 cars with backup cameras. Of the 12 red cars, 4 have backup cameras. If a car is selected at random from the given sample, what is the probability that both of the following are true: the car is not red and does not have a backup camera?

A $\frac{3}{5}$
B $\frac{16}{25}$
C $\frac{19}{25}$
D $\frac{4}{5}$

Show the answer

Correct answer: B

Create a chart using the given information and use subtraction to figure out how many cars are not red and do not have a back-up camera:

	Red	Not Red	TOTAL
Back-up Camera	4	6 (10-4)	10
No back-up Camera		32 (38-6)
TOTAL	12	38 (50-12)	50

The probability of selecting a car that meets both conditions from the total of 50 cars at the dealership is $\frac{32}{50}=\frac{16}{25}$ .

Question analysis

In a sample of 50 cars, 12 are red, 10 have backup cameras, and 4 of the red cars have cameras. Students find the probability that a randomly chosen car is neither red nor equipped with a camera. It tests compound probability with overlapping groups, and the overlap between red cars and camera cars is what makes it tricky.

NYS standards: 7.SP.8a compound event as fraction of outcomes
7.SP.8b sample spaces for compound events
Grade level: Grade 7 content, by NYS standard
Compared with grade 7: Above grade 7 Grade 7 compound probability usually uses a ready-made table or tree. Here students must build the two-way table from words, handle an overlap, and use a "not this and not that" condition, several steps beyond typical work.
Estimated difficulty: Hard: combines ideas, needs a plan, and has more than one tempting wrong answer.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Much to keep track of major
Many steps moderate
A very close wrong answer moderate
Complex sentences minor
Solution: 4 steps
Reading level: Question and choices: about grade 6, 63 words.
Corrected from the DOE handbook: A grammar slip in the explanation was corrected.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is an overlap problem dressed as probability: the hard part is counting the cars that avoid both features, not the division by 50. A two-way table makes it routine, but students working in their heads often keep only one of the two conditions. The question rewards placing the 4 red cars with cameras in the table before doing any subtraction.

Much to keep track of major Working it out
Three overlapping counts, 12 red, 10 with cameras and 4 that are both, must be sorted into four groups before the wanted one, neither red nor with a camera, can be found. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Many steps moderate Working it out
Find the non-red cars with cameras, 10 − 4 = 6, and the non-red cars, 50 − 12 = 38, subtract to get 32, then simplify 32/50. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)
A very close wrong answer moderate Choosing the answer
19/25, or 38/50, has the same denominator as the right 16/25 and sits just above it; only removing the 6 non-red cars that have cameras from the 38 separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)
Complex sentences minor Understanding the question
The last sentence stacks "both of the following are true", a colon and two negatives, "not red" and "does not have"; untangling it is reading work, not extra math. (Abedi & Lord, 2001; Freedle & Kostin, 1993; Messick, 1989)

Strategy: Draw a two-way table with red and not red across, camera and no camera down, fill in the given counts, and complete the missing cells by subtraction.

Traps

Misreads the question: Misses or misreads a condition or a word in the question: "at least", "each", "remaining", "not".
Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

A Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Finds the 6 non-red cars with cameras and divides by the 10 camera cars, 6/10 = 3/5, a different probability from the one asked.
C Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Uses only the 38 cars that are not red, 38/50, and ignores the second condition that the car also lacks a backup camera.
D Misreads the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: moderate
Counts only the 40 cars without backup cameras, 40/50, dropping the requirement that the car must not be red as well.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “a car that meet both conditions”; the site says “a car that meets both conditions.”

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number you want}}{\text{total number}}$; P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

Sort out what you know
There are 50 cars. Of these, 12 are red and 10 have backup cameras. Of the red cars, 4 have cameras. You pick one car at random. You need the chance that “both of the following are true”: the car is not red, and it has no camera.
Set up a two-way table
A two-way table sorts each car by two things at once: its color and its camera. Put in the numbers you know.
Red Not red Total
Camera 4 ? 10
No camera ?
Total 12 ? 50
Fill in the missing numbers
Not red with a camera: $10 - 4 = 6$ cars.
All cars that are not red: $50 - 12 = 38$ cars.
Not red with no camera: $38 - 6 = 32$ cars.
Watch out: the 38 cars that are not red include 6 with cameras. Using 38 gives $\frac{38}{50} = \frac{19}{25}$ , choice C. Using all 40 cars with no camera gives $\frac{40}{50} = \frac{4}{5}$ , choice D.
Write the probability
Use the probability formula. There are 32 cars you want out of 50 cars in all.
$P = \frac{32}{50}$
Divide the top and bottom by 2.
$\frac{32}{50} = \frac{16}{25}$
So the answer is B, $\frac{16}{25}$ .

	Red	Not red	Total
Camera	4	?	10
No camera		?
Total	12	?	50

Check your answer

Finish the table. Red cars with no camera: $12 - 4 = 8$ . Now add the four inside boxes: $4 + 6 + 8 + 32 = 50$ . That is every car, so the numbers fit. Also, the cars with no camera are $8 + 32 = 40$ , and $50 - 10 = 40$ too.

More questions like this

The original DOE question

Question 103 as printed on page 162 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 162, question 103. Open page 162 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 38 of 40

2018 Test B Math, question 49

Question

ANIMAL CARDS

Number of Cards	Picture on Card
8	cat
6	dog
5	bird
4	fish
1	horse

The cards in the table above are mixed in a box. Which animal pictured on a card has exactly a 1 in 4 chance of being picked at random from the box?

Acat
Bdog
Cfish
Dhorse

Show the answer

Correct answer: B

The total number of cards in the box is

8 + 6 + 5 + 4 + 1 = 24

Set up a proportion to figure out which card has exactly a 1 in 4 chance of being picked at random.

$\frac{x}{24}=\frac{1}{4}$

$x=6$

The dog card has a 1 in 4 chance of being randomly selected.

Question analysis

A table lists picture cards for five animals, from 8 cat cards down to 1 horse card, all mixed in one box. Students decide which animal has exactly a 1 in 4 chance of being drawn at random. It tests simple probability as a fraction of the total, and the numbers in the table invite matching the wrong count.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Simple probability from a list of counts is grade 6 work. This takes one sum and one division with small whole numbers, a shorter and more routine task than typical grade 7 probability problems.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Turning words into math minor
Reading a figure, table or graph minor
Solution: 2 steps
Reading level: Question and choices: about grade 1, 47 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

Once the 24 cards are totaled, one fourth of them is 6 and the answer is a lookup. The wrong choices mostly attract students who match the numbers 1 and 4 in the phrase to the horse and fish rows, or who assume the most common card is the one asked about. Little time is needed, so the main risk is skipping the total.

Turning words into math minor Understanding the question
"Exactly a 1 in 4 chance" has to be turned into a count, one fourth of the 24 cards, rather than matched to a 1 or a 4 in the table. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Reading a figure, table or graph minor Understanding the question
The counts sit in the table's first column with the animals in the second, and the total of 24 has to be added up from all five rows. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Add the cards to find the total in the box, find one-fourth of that total, and look for the animal with exactly that many cards.

Traps

Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Little reason to choose it (Haladyna & Downing, 1993; Rodriguez, 2005)
Reveals a surface strategy Expected pull: weak
Picks the most common card, but cats are 8 of 24 cards, a 1 in 3 chance, and no reading of the table gives them 1 in 4.
C Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Matches the 4 in "1 in 4" to the 4 fish cards, confusing a number of cards with a probability.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: weak
Matches the 1 in "1 in 4" to the single horse card instead of comparing each count with the total of 24.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number you want}}{\text{total number}}$; P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

Sort out the question
The table shows 8 cat cards, 6 dog, 5 bird, 4 fish and 1 horse. All the cards are mixed in a box. You need the animal with “exactly a 1 in 4 chance” of being picked. A 1 in 4 chance means a probability of $\frac{1}{4}$ .
Find the total number of cards
Add every row of the table. This total is the bottom of the probability fraction.
$8 + 6 + 5 + 4 + 1 = 24$
There are 24 cards in the box.
Find one fourth of the cards
The animal you want must have one fourth of all the cards.
$24 \div 4 = 6$
So you need the animal with exactly 6 cards. The dog has 6 cards.
$P = \frac{6}{24} = \frac{1}{4}$
Rule out the other choices
Watch out: do not match the numbers 1 and 4 to the table. Fish has 4 cards, but its chance is $\frac{4}{24} = \frac{1}{6}$ . So C is wrong. Horse has 1 card, a chance of $\frac{1}{24}$ , so D is wrong. Cat has $\frac{8}{24} = \frac{1}{3}$ , so A is wrong.
Give the answer
Only the dog has exactly one fourth of the cards. So the answer is B, dog.

Check your answer

Four groups of 6 cards make 24, the whole box. So 6 cards is one of four equal shares. That is a 1 in 4 chance. The dog cards are exactly one of those shares.

More questions like this

The original DOE question

Question 106 as printed on page 163 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 106. Open page 163 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 39 of 40

2018 Test B Math, question 52

Question

A ball is selected at random from a box that contains 7 black balls, 14 green balls, and 21 red balls. What is the probability that the ball selected is black?

A $\frac{1}{6}$
B $\frac{1}{5}$
C $\frac{1}{3}$
D $\frac{5}{6}$

Show the answer

Correct answer: A

The total number of balls in the box is

$7+14+21=42$

The probability that the ball is black is

$\frac{7}{42}=\frac{1}{6}$

Question analysis

Students find the probability of drawing a black ball at random from a box with 7 black, 14 green and 21 red balls. It tests simple probability as a part of a whole. It is a short question, but the denominator must be chosen with care.

NYS standards: 6.SP.8a uniform probability models
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: Below grade 7 Finding a simple probability from given counts is grade 6 work in New York. This one is a single fraction to simplify, a routine task well below the compound-event problems of grade 7.
Estimated difficulty: Routine: one familiar step, little to misread.
Thinking it demands: Depth of knowledge 1, recall and reproduction · mainly apply
What makes it hard: Something you must already know minor
A very close wrong answer minor
Solution: 2 steps
Reading level: Question and choices: about grade 4, 32 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

This is a one-formula probability question with friendly numbers, since 7 divides 42 evenly. The wrong answers come from comparing black balls with the other balls only, from treating the three colors as equally likely, or from finding the chance of not drawing black. Students who write the total first rarely miss it.

Something you must already know minor Working it out
The probability of an outcome, favorable balls over all balls, is assumed; nothing in the stem says to add 7 + 14 + 21 for the denominator. (Anderson & Krathwohl, 2001)
A very close wrong answer minor Choosing the answer
1/5 and the right 1/6 are neighboring unit fractions; only counting the 7 black balls in the denominator, 7/42 rather than 7/35, separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Add all the balls to get the total, put the number of black balls over that total, and simplify the fraction.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.

The wrong answers, classified

B Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Compares the 7 black balls with the 35 others, 7/35 = 1/5, a part-to-part ratio instead of part to whole.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Treats the three colors as equally likely outcomes, so black gets a 1 in 3 chance however many balls each color has.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds the probability that the ball is not black, 35/42 = 5/6, the complement of the event asked about.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability: $P = \frac{\text{number you want}}{\text{total number}}$; P is the probability, the chance that something happens. The top counts the results you want. The bottom counts every possible result. Use it when each result is equally likely, like picking one at random.

Sort out what you know
The box has 7 black balls, 14 green balls and 21 red balls. One ball is picked at random. That means every ball has the same chance. You need the probability, or chance, that it is black.
Find the total number of balls
The bottom of the fraction is every ball in the box.
$7 + 14 + 21 = 42$
There are 42 balls in all.
Write and simplify the probability
Use the probability formula. There are 7 black balls out of 42.
$P = \frac{7}{42} = \frac{7 \div 7}{42 \div 7} = \frac{1}{6}$
Watch out: comparing black with the other 35 balls gives $\frac{7}{35} = \frac{1}{5}$ , choice B. Saying 1 of 3 colors gives $\frac{1}{3}$ , choice C. But the colors have different numbers of balls.
Give the answer
The chance of picking a black ball is 1 out of 6. So the answer is A, $\frac{1}{6}$ .

Check your answer

Notice that 14 is 2 sevens and 21 is 3 sevens. So for every 1 black ball there are 2 green and 3 red. That is 1 black out of every 6 balls. Choice D, $\frac{5}{6}$ , is the chance of not black.

More questions like this

The original DOE question

Question 109 as printed on page 163 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 163, question 109. Open page 163 of the handbook (PDF)

Question 40 of 40

2018 Test B Math, question 57

Question

Carlos tossed a paper cup in the air 50 times and found that the probability of it landing on its side was 72%. If he tosses the cup in the air 150 more times, what is the total number of times he can expect the cup to land on its side?

A72
B108
C144
D158

Show the answer

Correct answer: C

The probability of the cup landing on its side is 72%. Carlos tossed the cup a total of 200 times (50 + 150). The number of times the cup lands on its side is 72% of 200:

0.72 × 200 = 144.

Question analysis

Carlos tossed a paper cup 50 times and found it landed on its side 72% of the time. If he tosses it 150 more times, students predict the total number of side landings he can expect. It tests using a probability to predict outcomes, and the word total decides which tosses to count.

NYS standards: 6.SP.7 long-run relative frequency
6.RP.3c percent as a rate per 100
Grade level: Grade 6 content, by NYS standard
Compared with grade 7: At grade 7 Predicting outcomes from a probability is grade 6 content, and the arithmetic is a single percent. What lifts it to grade 7 level is deciding exactly which tosses the word "total" refers to.
Estimated difficulty: Moderate: several steps, or a close reading with one strong distractor.
Thinking it demands: Depth of knowledge 2, skills and concepts · mainly apply
What makes it hard: Turning words into math moderate
Something you must already know minor
Solution: 2 steps
Reading level: Question and choices: about grade 8, 48 words.
Students, last 30 days: Not enough answers yet to show (it takes 10).

Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

The computation is one multiplication, so the question is decided by what "total" covers. Students who focus on the 150 new tosses answer for part of the experiment, and students who add the original 50 tosses as if every one landed on its side overshoot. Reading the last sentence twice, and noticing the word total, is the safeguard.

Turning words into math moderate Understanding the question
"150 more times" and "the total number of times" together mean all 200 tosses, so the expected count is 72% of 200, not of 150. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know minor Working it out
Students must know that an expected count is the probability times the number of trials; the stem never says to multiply. (Anderson & Krathwohl, 2001)

Strategy: Decide which tosses the word "total" covers, then multiply that number of tosses by 0.72 to get the expected count.

Traps

Answers a different question: A choice is right for another quantity: the other unknown, the rest instead of the part, the part instead of the whole.
Numbers taken from the question: Combines the numbers in the question without a model of the situation, or is simply a number that appears in the question.

The wrong answers, classified

A Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Reads the 72% as 72 side landings, as if there had been 100 tosses, without applying the percent to the actual tosses.
B Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: strong
Takes 72% of only the 150 new tosses, 0.72 × 150 = 108, and leaves out the side landings expected in the first 50.
D Numbers taken from the question (Hegarty, Mayer & Monk, 1995)
Reveals partial knowledge Expected pull: weak
Adds all 50 first tosses to the 108 expected from the next 150, as if every early toss had landed on its side.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Expected count: $\text{expected count} = P \times n$; P is the probability, the chance of the result on one try, written as a decimal. n is the number of tries. Multiplying tells how many times you can expect the result. It is a good guess, not a promise.

Sort out what you know
Carlos tossed the cup 50 times. It landed on its side 72% of the time. He will toss it 150 more times. You need “the total number of times” it should land on its side. Total means all the tosses, the first 50 and the next 150.
Count all the tosses
Add the first tosses and the new ones.
$50 + 150 = 200$
There are 200 tosses in all.
Watch out: using only the 150 new tosses gives $0.72 \times 150 = 108$ , choice B.
Change the percent to a decimal
Percent means out of 100. So 72% is 72 out of 100. Move the decimal point two places left.
$72\% = \frac{72}{100} = 0.72$
Multiply to find the expected count
Use the expected count formula with 0.72 and 200 tosses.
$0.72 \times 200 = 144$
He can expect the cup to land on its side 144 times.
Watch out: 72 is the count for 100 tosses, not 200. That is choice A.
Give the answer
All 200 tosses count, so the expected total is 144. So the answer is C, 144.

Check your answer

Split it up. In the first 50 tosses, $0.72 \times 50 = 36$ . In the next 150, $0.72 \times 150 = 108$ . Then $36 + 108 = 144$ . Also, 144 is a bit less than three quarters of 200, which is 150. That fits 72%.

More questions like this

The original DOE question

Question 114 as printed on page 164 of the DOE handbook. — As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 164, question 114. Open page 164 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).