Practice Test

7.SP.8a compound event as fraction of outcomes

Question 1 of 18

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 2 of 18

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 3 of 18

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 the card in a prize is 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.
Corrected from the DOE handbook: The explanation gave the wrong event the chance 1/100; one sentence 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

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.

What was corrected from the DOE handbook, and why

Explanation: The DOE’s explanation says “The probability that a randomly selected student will receive Card C17 is 1/100.” Only students who win a prize receive a card, so a randomly selected student’s chance of receiving Card C17 is the question’s own answer, 1/25 × 1/100. The 1/100 is the chance that the card in a prize is Card C17, and the site now says so. The answer is unchanged.

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. This is the DOE’s version, before the corrections listed in the analysis.

Question 4 of 18

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 5 of 18

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 6 of 18

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 7 of 18

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 8 of 18

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 9 of 18

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 10 of 18

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 11 of 18

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 12 of 18

2023 Test B Math, question 50

Question

$\{3, 4, 5, 6, 7\}$

A student made a list of all possible products of 2 different numbers in the set above. What fraction of the products are odd?

A $\frac{3}{20}$
B $\frac{3}{10}$
C $\frac{1}{2}$
D $\frac{3}{5}$

Show the answer

Correct answer: B

First, list the products:

$3\times 4=12$

$3\times 5=15$

$3\times 6=18$

$3\times 7=21$

$4\times 5=20$

$4\times 6=24$

$4\times 7=28$

$5\times 6=30$

$5\times 7=35$

$6\times 7=42$

Of the 10 products, 3 are odd. The fraction is $\frac{3}{10}$ .

Question analysis

A student lists every product of two different numbers from the set 3, 4, 5, 6, 7. Students find what fraction of those products are odd. It tests listing outcomes systematically without counting a pair twice, and reasoning about odd and even numbers.

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 Organized lists of outcomes are grade 7 work. Here students list the products of pairs from five numbers, without repeating a pair or pairing a number with itself, then count the ones that fit.
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
Something you must already know minor
Solution: 3 steps
Reading level: Question and choices: about grade 6, 28 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 count of odd products is easy, since only 3, 5 and 7 can make one; the total is where students go wrong. Counting ordered pairs, 5 × 4 = 20, while listing odd products once mixes two kinds of counting. Writing the 10 pairs in order, 3 with each larger number, then 4, and so on, keeps the total honest.

Much to keep track of moderate Working it out
Listing the products of 2 different numbers from five means tracking 10 pairs, with no pair repeated in reverse order and no number paired with itself. (Sweller, 1988; Carpenter, Just & Shell, 1990)
Something you must already know minor Working it out
The shortcut, that a product is odd only when both factors are odd, cuts the work to counting pairs from 3, 5 and 7, but it must be known. (Anderson & Krathwohl, 2001)

Strategy: List each pair of different numbers once, then find each product, or use the fact that a product is odd only when both factors are odd, and count.

Traps

Overlapping groups: Things in two groups at once are counted twice, or not subtracted.
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 Overlapping groups (King et al., 2004)
Reveals partial knowledge Expected pull: moderate
Counts ordered pairs for the total, 5 × 4 = 20, so each pair appears twice, but counts the odd products once, 3, giving 3/20.
C Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: weak
Assumes about half of all products are odd, as odd and even numbers come equally often, without listing the products or checking any.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a surface strategy Expected pull: moderate
Gives the share of the numbers in the set that are odd, 3 of 5, instead of the share of the products, never forming any products.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Fraction of a list: $\text{fraction} = \frac{\text{odd products}}{\text{all products}}$; The top counts the products that are odd. The bottom counts all the products in the list. A product is the answer when you multiply. The fraction tells what share of the list is odd.
Odd times odd: $\text{odd} \times \text{odd} = \text{odd}$; An odd number can’t be split into pairs, like 3 or 7. An even number can, like 4. A product is odd only when both numbers are odd. If either number is even, the product is even, because it splits into pairs.

Sort out the question
The set has 3, 4, 5, 6 and 7. The student multiplies 2 different numbers, every possible way. A product is the answer to a multiplication. Each pair counts once, since 3 × 4 and 4 × 3 give the same product. You need the fraction of the products that are odd.
List every pair once
Pair 3 with each bigger number. Then pair 4 with each bigger number, and so on. This way no pair is counted twice.
3 × 4 = 12 3 × 5 = 15 3 × 6 = 18 3 × 7 = 21
4 × 5 = 20 4 × 6 = 24 4 × 7 = 28
5 × 6 = 30 5 × 7 = 35
6 × 7 = 42
There are 10 products.
Count the odd products
The odd products are 15, 21 and 35. That makes 3. They come from the only pairs where both numbers are odd: 3 and 5, 3 and 7, and 5 and 7.
Write the fraction
Use the fraction formula.
$\text{fraction} = \frac{3}{10}$
Watch out: Counting 4 × 3 as well as 3 × 4 makes 20 pairs. Then $\frac{3}{20}$ , choice A, mixes two ways of counting. And $\frac{3}{5}$ , choice D, is the share of the numbers that are odd, not the products.
So the answer is B, $\frac{3}{10}$ .

Check your answer

Count without listing. Each of the 5 numbers can pair with 4 others, which makes 20. But that counts every pair twice, so there are 20 ÷ 2 = 10 pairs. Only the odd numbers 3, 5 and 7 make odd products, and they form 3 pairs.

More questions like this

The original DOE question

Question 107 as printed on page 205 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 205, question 107. Open page 205 of the handbook (PDF)

Question 13 of 18

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 14 of 18

2022 Test A Math, question 43

Question

A bag contains 15 marbles: 11 blue marbles and 4 red marbles. Sarah picks one marble randomly, returns it to the bag, and then randomly picks one marble again. What is the probability that she picks a red marble both times?

A $\frac{16}{225}$
B $\frac{16}{121}$
C $\frac{8}{15}$
D $\frac{121}{225}$

Show the answer

Correct answer: A

The probability of drawing a red marble at random from a bag of 15 marbles (11 blue and 4 red) is shown:

$P(\text{Red}) = \frac{4\text{ red marbles}}{15\text{ total marbles}} = \frac{4}{15}$

When the red marble is replaced in the bag, the probability of picking a red marble is unchanged, $P(\text{Red}) = \frac{4}{15}$ .

The two events are independent, so the multiplication rule of probability applies.

$P(\text{Red}) \bullet P(\text{Red}) = \frac{4}{15} \bullet \frac{4}{15} = \frac{16}{225}$

Question analysis

A bag holds 11 blue and 4 red marbles. Sarah picks one at random, puts it back, and picks again, and students find the probability that she picks red both times. It tests the probability of a compound event made of two independent picks; the bold word both and the replacement shape the calculation.

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 Finding the probability of two independent picks is grade 7 compound-event work. Because the marble is put back, the two picks are alike, and the numbers are small, which makes this 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: Something you must already know moderate
Turning words into math minor
Solution: 2 steps
Reading level: Question and choices: about grade 6, 42 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 two small fractions multiplied, so the question rests on knowing which rule applies. Adding the two chances gives more than either chance alone, which cannot be right for two reds in a row, and comparing red with blue instead of with all 15 marbles changes every denominator. A sense check helps: two reds in a row must be less likely than a single red, 4/15.

Something you must already know moderate Working it out
The multiplication rule for independent events is not given: because the marble goes back, each pick is 4 out of 15, and red both times means 4/15 × 4/15. (Anderson & Krathwohl, 2001)
Turning words into math minor Understanding the question
Two phrases carry the method: “returns it to the bag” keeps the second pick at 4 out of 15, and “red marble both times” asks for both events, not either. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Find the chance of red on a single pick out of all the marbles. Since the marble goes back, the second pick has the same chance; then use the rule for two independent events both happening.

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).
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 Misunderstands the concept (Smith, diSessa & Roschelle, 1993; De Bock et al., 1998; Ni & Zhou, 2005)
Reveals a misconception Expected pull: moderate
Uses red to blue, 4 to 11, as the chance of red instead of 4 out of all 15 marbles, then squares it, a part-to-part ratio.
C Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a misconception Expected pull: strong
Adds the two chances, 4/15 + 4/15 = 8/15, instead of multiplying, as if a red marble on either pick were enough.
D Answers a different question (King et al., 2004; Hegarty, Mayer & Monk, 1995)
Reveals a slip Expected pull: weak
Finds the chance of blue both times, (11/15)² = 121/225, answering for the wrong color with the right method.

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability of an event: $P = \frac{\text{number of ways it can happen}}{\text{total number of outcomes}}$; P is the probability, the chance that something happens, from 0 (never) to 1 (always). An outcome is one possible result, here one marble. Divide the number of results you want by the number of all results.
Two independent events both happening: $P(A \text{ and } B) = P(A) \times P(B)$; 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.

Sort out the question
The bag has 15 marbles: 11 blue and 4 red. Sarah picks one, puts it back, then picks again. You need the probability, or chance, that both picks are red. Putting the marble back matters. The bag is the same for the second pick.
Find the chance of red on one pick
$P = \frac{\text{number of ways it can happen}}{\text{total number of outcomes}}$
$P(\text{red}) = \frac{4}{15}$
There are 4 red marbles out of 15 in all. Watch out: compare red with all 15 marbles, not with the 11 blue ones. Using 4 out of 11 leads to $\frac{16}{121}$ , choice B.
See that the picks are independent
Sarah puts the marble back. So the bag again has 4 red marbles out of 15. The second pick has the same chance, 4/15. The first pick does not change the second one. That makes the two picks independent.
Multiply the two chances
$P(\text{red and red}) = \frac{4}{15} \times \frac{4}{15} = \frac{4 \times 4}{15 \times 15} = \frac{16}{225}$
To multiply fractions, multiply the tops and multiply the bottoms. Watch out: don’t add. $\frac{4}{15} + \frac{4}{15} = \frac{8}{15}$ , choice C, is a bigger chance than one red. And $\frac{121}{225}$ , choice D, is the chance of blue both times. So the answer is A, $\frac{16}{225}$ .

Check your answer

Two reds in a row should be less likely than one red. One red is $\frac{4}{15}$ , about 0.27. $\frac{16}{225}$ is about 0.07, much smaller. That makes sense. Adding gave $\frac{8}{15}$ , about 0.53, which is too big.

More questions like this

The original DOE question

Question 100 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 2023 Admissions, page 91, question 100. 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 15 of 18

2022 Test B Math, question 42

Question

At any given time, one person out of every 20 people has the common cold. Assume that the common cold can be caused by any one of 200 different viruses, each of which is equally likely to cause the cold. One of these 200 viruses is cold virus V144. What is the probability that a randomly selected person is suffering from a cold caused by virus V144?

A $\frac{1}{10}$
B $\frac{1}{20}$
C $\frac{1}{200}$
D $\frac{1}{4{,}000}$

Show the answer

Correct answer: D

To calculate the joint probability, you must first determine the probability of each independent event. The probability that a randomly selected person has a common cold is $\frac{1}{20}$ . The probability that a cold is caused by V144 is $\frac{1}{200}$ . The probability that a person has a cold and the cold is caused by the V144 virus is the product of those two probabilities, $\frac{1}{20}\times\frac{1}{200}=\frac{1}{4{,}000}$ .

Question analysis

One person in 20 has a common cold at any given time, and each cold comes from one of 200 equally likely viruses. Students find the probability that a randomly selected person has a cold caused by one particular virus, V144. It tests the probability of a compound event, which is hidden here inside a single phrase.

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 from organized lists or tree diagrams. Here the two stages are hidden in one phrase, a cold caused by one particular virus, and there are far too many outcomes to list.
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 moderate
Solution: 2 steps
Reading level: Question and choices: about grade 6, 68 words.
Corrected from the DOE handbook: A probability in the explanation that was stated for the wrong event 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

The arithmetic is one multiplication; the difficulty is seeing that the event has two stages. Each stated chance, 1 in 20 for a cold and 1 in 200 for the virus behind it, is listed as a choice on its own, and each answers only half the question. A chance that both stages happen must be smaller than either one alone, which leaves a single choice.

Turning words into math moderate Understanding the question
The single phrase “a cold caused by virus V144” holds two events, having a cold at all and that cold coming from V144, and they must be separated before any rule applies. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)
Something you must already know moderate Working it out
The stem never says that the chance of both stages is the product of their chances, so students must bring that rule to the 1 in 20 and the 1 in 200 themselves. (Anderson & Krathwohl, 2001)

Strategy: Split the event into two stages, having a cold and that cold coming from V144, write each stage's chance as a fraction, and combine them with the rule for events that must both happen.

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).

The wrong answers, classified

A Wrong operation (Hegarty, Mayer & Monk, 1995; King et al., 2004)
Reveals a surface strategy Expected pull: weak
Divides one chance by the other, 1/200 ÷ 1/20 = 20/200 = 1/10, instead of multiplying them, giving a chance larger than either stage.
B Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: moderate
Gives the chance that a person has a cold of any kind, 1 in 20, and stops before multiplying by the 1/200 chance that V144 caused it.
C Stops one step early (King et al., 2004; Gierl et al., 2017)
Reveals partial knowledge Expected pull: strong
Gives the chance that a cold is caused by V144, one virus out of 200, forgetting that only 1 person in 20 has a cold at all.

What was corrected from the DOE handbook, and why

Explanation: The handbook says “The probability that a person has a cold caused by V144 is 1/200.” A person having a cold caused by V144 is exactly what the question asks about, and the same paragraph goes on to give its probability as 1/4,000; as printed, the sentence gives that event the value of choice C. The 1/200 is the probability that a cold is caused by V144 (one of 200 equally likely viruses), so the site reads “The probability that a cold is caused by V144 is 1/200.” The working and the answer are unchanged: 1/20 × 1/200 = 1/4,000.

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 virus out of 200.
Chance that two events both happen: $P(A \text{ and } B) = P(A) \times P(B \text{ given } A)$; A is the first event and B is the second. P(A) is the chance of A. P(B given A) is the chance of B among the times A has happened. 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
One person in 20 has a cold. Every cold comes from one of 200 viruses, all equally likely. V144 is one of them. You need the chance that a random person has a cold caused by V144.
That phrase holds two events. First, the person has a cold. Second, that cold comes from V144.
Find the chance of a cold
The question says “one person out of every 20 people has the common cold.” So 1 person is the part with a cold, and 20 is the whole group.
$P(\text{cold}) = \frac{1}{20}$
Find the chance it is V144
Now think only about people who have a cold. Each cold comes from one of 200 viruses, all equally likely. V144 is 1 of the 200.
$P(\text{V144 given a cold}) = \frac{1}{200}$
This chance is only about people who already have a cold. So the second event depends on the first.
Multiply the two chances
Use the rule for two events that must both happen. Event A is a cold, and event B is V144 as its cause.
$P(A \text{ and } B) = P(A) \times P(B \text{ given } A)$
$P(\text{cold and V144}) = \frac{1}{20} \times \frac{1}{200} = \frac{1}{4{,}000}$
Multiply the tops: 1 × 1 = 1. Multiply the bottoms: 20 × 200 = 4,000.
Give the answer
Watch out: 1/20 is the chance of any cold, choice B. And 1/200 is the chance that a cold comes from V144, choice C. Each one answers only half the question. Dividing one by the other gives 1/10, choice A. So the answer is D, $\frac{1}{4{,}000}$ .

Check your answer

Picture 4,000 people. One in 20 has a cold, so $4{,}000 \div 20 = 200$ people have colds. Of those 200 colds, 1 in 200 comes from V144. That is 1 person out of 4,000. It is also smaller than both 1/20 and 1/200, as it should be.

More questions like this

The original DOE question

Question 99 as printed on page 209 of the DOE handbook. — As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2023 Admissions, page 209, question 99. Open page 209 of the handbook (PDF) This is the DOE’s version, before the corrections listed in the analysis.

Question 16 of 18

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 17 of 18

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 18 of 18

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.

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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.

3 × 4 = 12	3 × 5 = 15	3 × 6 = 18	3 × 7 = 21
4 × 5 = 20	4 × 6 = 24	4 × 7 = 28
5 × 6 = 30	5 × 7 = 35
6 × 7 = 42

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