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Question 1 of 11

2026 Practice Test D Math, question 38

Question

Melissa has a bag of marbles that are all the same size. The bag contains three red (R) marbles, three green (G) marbles, and two white (W) marbles. For an experiment, she will take two marbles out of the bag without looking. Which list shows the sample space for Melissa’s experiment?

  • AR, G, W
  • BRG, RW, GW
  • CR, R, R, G, G, G, W, W
  • DRR, RG, RW, GG, GW, WW
Show the answer

Correct answer: D

The sample space of a random experiment is the collection of all possible outcomes. For this experiment, R represents a red marble, G represents a green marble, and W represents a white marble.

Since two marbles are being drawn, the possible outcomes for the first marble drawn are R, G, W.

List the possible results for each initial outcome combined with the second choice of marble. For example, if the first marble and second marble are both red, the sample space will include RR (red, red).

RR, RG, RW

GR, GG, GW

WR, WG, WW

Since order does not matter in this selection, remove any pairing that presents a duplicate choice, such as RG and GR—only one of the two pairings will be included in the sample space. Those are removed and leave a sample space of:

RR, RG, RW, GG, GW, WW

Question analysis

A bag holds three red, three green and two white marbles, and Melissa will take out two without looking. This probability question asks which list shows the sample space for her experiment. Students must know what a sample space is and what one outcome of drawing two marbles looks like, then decide which color pairs are possible.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Representing the sample space of a compound event is a named grade 7 skill. Nothing is computed here; the demand is knowing what one outcome of drawing two marbles looks like and which pairs the bag allows, much as in grade 7 classwork.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 3, 71 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 tests a definition more than a skill, and three choices describe something real: the colors of one marble, the contents of the bag, and the pairs of different colors. The strongest pull comes from treating the colors as three single objects, which leaves out the pairs of one color although the bag holds three reds, three greens and two whites. Asking what Melissa could be holding after the draw settles it.

  1. Something you must already know moderate Working it out

    The term sample space is not defined in the question: students must know it means every possible result of the whole experiment, here a pair of colors drawn together. (Anderson & Krathwohl, 2001)

  2. A very close wrong answer moderate Choosing the answer

    The three mixed pairs, RG, RW and GW, form a complete-looking list that is the right list minus RR, GG and WW, so only asking whether two marbles of one color can come out together separates them. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

Strategy: Describe one possible result of taking out two marbles, then build the list systematically, starting from each color and pairing it with every color that could come out with it.

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.
  • Without replacement: In a second pick there is one fewer item, which changes the probability.
  • 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Sample space
sample space=the list of every possible outcome\text{sample space} = \text{the list of every possible outcome}
An outcome is one possible result of an experiment. The sample space lists every outcome once. For this experiment, one outcome is the pair of colors Melissa holds after she takes out two marbles.

  1. Sort out the question

    The bag holds 3 red (R), 3 green (G) and 2 white (W) marbles. Melissa takes out 2 marbles without looking.

    You need the sample space: every possible result of taking out 2 marbles. So each outcome must name two colors, like RG for one red and one green.

  2. List pairs by the first color

    Start with each color the first marble could be. Then pair it with each color the second could be:

    • First red: RR, RG, RW
    • First green: GR, GG, GW
    • First white: WR, WG, WW

    That gives 9 pairs.

  3. Remove pairs that repeat

    Melissa ends up holding two marbles. One red and one green is the same result whichever came out first. So GR is the same as RG. Cross out GR, WR and WG. Six pairs are left:

    RR, RG, RW, GG, GW, WW

  4. Check the same-color pairs

    Can both marbles be the same color? Yes. The bag has 3 reds, 3 greens and 2 whites, so RR, GG and WW can all happen.

    Watch out: choice B leaves out RR, GG and WW. That would only be right if the bag had one marble of each color. Choice A is the colors of one marble. Choice C lists the marbles in the bag, not results.

    So the answer is D, RR, RG, RW, GG, GW, WW.

Check your answer

Ask what Melissa could be holding. Two reds? Yes, there are 3. Two whites? Yes, there are exactly 2. A green and a white? Yes. Every pair in D can happen, and no possible pair is missing. Choice B misses three pairs that can happen.

More questions like this

The original DOE question
Question 38 as shown on the DOE’s digital test, page 111 of its printable version.
As printed in the DOE’s printable version of 2026 Practice Test D Math, page 111, question 38.

Question 2 of 11

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?

  • A82,187\frac{8}{2{,}187}
  • B1126\frac{1}{126}
  • C2566,561\frac{256}{6{,}561}
  • D169\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 49\frac{4}{9}.

The probabilities of selecting blue for the next three selections are 38\frac{3}{8}, 27\frac{2}{7}, and 16\frac{1}{6}.

49×38×27×16=243,024=1126\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.
What makes it hard
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.

  1. 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)

  2. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of wanted outcomestotal number of outcomesP = \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 then B)=P(A)×P(B after A)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.

  1. 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.

  2. 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.

    PickBlue disks leftDisks leftChance of blue
    1st4949\frac{4}{9}
    2nd3838\frac{3}{8}
    3rd2727\frac{2}{7}
    4th1616\frac{1}{6}

  3. Multiply the four chances

    All four picks must go right, one after another. Use the rule for events in a row and multiply.

    49×38×27×16=4×3×2×19×8×7×6=243,024\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}

  4. Simplify the fraction

    Divide the top and the bottom by 24. That keeps the value the same.

    24÷243,024÷24=1126\frac{24 \div 24}{3{,}024 \div 24} = \frac{1}{126}

    Watch out: Using 49\frac{4}{9} for every pick gives 2566,561\frac{256}{6{,}561}, choice C. That is right only if each disk goes back in. Keeping 9 on the bottom each time gives 82,187\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, 1126\frac{1}{126}.

Check your answer

Cancel before you multiply. On top, 4×2=84 \times 2 = 8 cancels the 8 on the bottom. That leaves 39×7×6=3378\frac{3}{9 \times 7 \times 6} = \frac{3}{378}. Divide the top and bottom by 3 to get 1126\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 3 of 11

2024 Test B Math, question 45

Question

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

  • A527\frac{5}{27}
  • B27\frac{2}{7}
  • C827\frac{8}{27}
  • D310\frac{3}{10}
Show the answer

Correct answer: C

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

2x+3x+5x=302x+3x+5x=30

10x=3010x=30

x=3x=3

Red: 2x=2(3)=6\text{Red: }2x=2(3)=6

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

Blue: 3x=3(3)=9\text{Blue: }3x=3(3)=9

If one paper clip of each color is removed (3 total), there will be 8 blue paper clips out of 27 in the box. The probability that the next paper clip chosen will be blue is 827\frac{8}{27}.

Question analysis

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

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

Why it is hard

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

  1. Many steps moderate Working it out

    The ratio 2:5:3 must become counts of 6, 15 and 9 clips, one clip of each color must come out, and only then can blue's 8 be put over 27. (Daniel & Embretson, 2010; Enright, Morley & Sheehan, 2002)

  2. A very close wrong answer moderate Choosing the answer

    3/10 = 0.3 and 8/27, about 0.296, are almost equal, so an estimate cannot separate them; only subtracting the three removed clips does. (Embretson & Wetzel, 1987; Gorin & Embretson, 2006; Ascalon et al., 2007)

  3. Turning words into math minor Understanding the question

    The removal of one clip of each color, not replaced, happens before the draw, so it changes both the blue count and the total in the box. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The box has 30 clips. The ratio red : white : blue = 2 : 5 : 3 means that for every 2 red clips, there are 5 white clips and 3 blue clips.

    Then one clip of each color is taken out and not put back. You need the chance that the next clip picked is blue.

  2. Find the size of one part

    The ratio has 2+5+3=102 + 5 + 3 = 10 parts in all. The 30 clips are shared among those 10 parts.

    one part=3010=3\text{one part} = \frac{30}{10} = 3

    Each part is 3 clips.

  3. Count the clips of each color

    Multiply each ratio number by the size of one part.

    red=2×3=6\text{red} = 2 \times 3 = 6

    white=5×3=15\text{white} = 5 \times 3 = 15

    blue=3×3=9\text{blue} = 3 \times 3 = 9

    Check: 6 + 15 + 9 = 30 clips. That matches the total.

  4. Take out one of each color

    One red, one white and one blue clip come out. Each color drops by 1, and the total drops by 3.

    ColorBeforeAfter
    Red65
    White1514
    Blue98
    All3027

  5. Find the chance of blue

    P(blue)=827P(\text{blue}) = \frac{8}{27}

    Watch out: 930=310\frac{9}{30} = \frac{3}{10}, choice D, ignores the 3 clips taken out. 527\frac{5}{27}, choice A, is the chance of red. Taking 1 from each ratio number, as if the box held only 10 clips, gives 27\frac{2}{7}, choice B.

    So the answer is C, 827\frac{8}{27}.

Check your answer

Add the clips left: 5 + 14 + 8 = 27. That matches. Also, 827\frac{8}{27} is about 0.296, a bit less than blue’s share of the full box, 0.3. That makes sense, since one blue clip came out.

More questions like this

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

Question 4 of 11

2024 Test B Math, question 47

Question

Three different colors of paint are available to paint the window frame, the door, and the wall of a house. If no two parts of the house may be painted the same color, how many different ways are there to paint the three parts of the house?

  • A3
  • B6
  • C9
  • D27
Show the answer

Correct answer: B

One method of solving the problem is to use a table to represent the sample space. Let the three colors be red, blue, and green.

PAINT COLORS

Window Frame Door Wall
Red Blue Green
Red Green Blue
Blue Red Green
Blue Green Red
Green Red Blue
Green Blue Red

There are 6 possible combinations of colors for the window frame, the door, and the wall where no two parts are painted the same color.

Question analysis

Three paint colors are available for a house's window frame, door and wall, and no two of the three parts may be painted the same color. Students count the different ways to paint them. It tests counting arrangements with a sample space, and the rule against repeating a color is what shapes the count.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Counting arrangements with an organized list is grade 7 work, and this one has only three parts and three colors. It is typical practice, with the no-repeat rule as the one condition to keep in mind.
Estimated difficulty
Straightforward: two steps or one careful reading.
Solution
2 steps
Reading level
Question and choices: about grade 8, 47 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

With three colors and three parts, the six arrangements can be listed in well under a minute, so the question rewards either a quick organized list or the counting principle. The main loss comes from ignoring the rule that no two parts match, which turns 3 × 2 × 1 into 3 × 3 × 3. The other wrong choices simply combine the two 3s in the stem.

  1. Something you must already know minor Working it out

    The counting principle is assumed: 3 color choices for the window frame, 2 left for the door, then 1 for the wall, multiplied together. (Anderson & Krathwohl, 2001)

  2. Turning words into math minor Understanding the question

    The rule that "no two parts of the house may be painted the same color" means each color is used exactly once, which removes one choice at every step. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

Strategy: Count the color choices for the window frame, then for the door with one color already used, then for the wall, and multiply; or list the arrangements.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle
total ways=a×b×c\text{total ways} = a \times b \times c
a is the number of choices for the first part, b for the second and c for the third. Multiply them to get every possible way. It works because each choice for one part can go with each choice for the next part.

  1. Sort out the question

    There are 3 colors and 3 parts to paint: the window frame, the door and the wall. No two parts may be the same color. So each color is used exactly once.

    You need the number of different ways to paint the three parts.

  2. Count the choices for each part

    Paint the parts in order. Each color you use is gone for the next part.

    • Window frame: any of the 3 colors, so 3 choices.
    • Door: one color is used, so 2 choices are left.
    • Wall: only 1 color is left, so 1 choice.

  3. Multiply the choices

    Use the counting principle. Each choice for the window frame goes with each choice for the door, and so on. So you multiply.

    total ways=3×2×1=6\text{total ways} = 3 \times 2 \times 1 = 6

    Watch out: 3×3×3=273 \times 3 \times 3 = 27, choice D, lets a color be used twice. 3×3=93 \times 3 = 9, choice C, just multiplies the numbers in the question.

    So the answer is B, 6.

Check your answer

List them. Use R for red, B for blue and G for green, in the order frame, door, wall: RBG, RGB, BRG, BGR, GRB and GBR. That is 6 ways, and each one uses all three colors once.

More questions like this

The original DOE question
Question 104 as printed on page 212 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2025 Admissions, page 212, question 104. Open page 212 of the handbook (PDF) The handbook letters this question’s choices E–H; here they are A–D (E = A, F = B, G = C, H = D).

Question 5 of 11

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?

  • A112\frac{1}{12}
  • B110\frac{1}{10}
  • C15\frac{1}{5}
  • D710\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 36\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 15\frac{1}{5}.

Then P (black shirt first) • P (red shirt second, given black shirt first)
=36∙15=330=110=\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.
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.

  1. 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)

  2. 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)

  3. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=number of wanted outcomestotal number of outcomesP = \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 then B)=P(A)×P(B after A)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.

  1. 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.

  2. Find the chance of black first

    3 of the 6 shirts are black. Each shirt is equally likely to be picked.

    P(black first)=36P(\text{black first}) = \frac{3}{6}

    That is one half.

  3. 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(red second)=15P(\text{red second}) = \frac{1}{5}

    Watch out: The bottom is 5, not 6. Using 6 gives 112\frac{1}{12}, choice A, as if the shirt went back.

  4. 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(black, then red)=36×15=330=110P(\text{black, then red}) = \frac{3}{6} \times \frac{1}{5} = \frac{3}{30} = \frac{1}{10}

    Watch out: Adding the chances gives 710\frac{7}{10}, choice D. Stopping at 15\frac{1}{5} gives choice C.

    So the answer is B, 110\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 6×5=306 \times 5 = 30 pairs. Pairs with black then red: 3 black shirts times 1 red shirt is 3. So the chance is 330=110\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 6 of 11

2023 Test B Math, question 13

Question

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

  • A15\frac{1}{5}
  • B417\frac{4}{17}
  • C14\frac{1}{4}
  • D13\frac{1}{3}
Show the answer

Correct answer: A

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

33+6+6=315=15\frac{3}{3 + 6 + 6} = \frac{3}{15} = \frac{1}{5}

Question analysis

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

NYS standards
Grade level
Grade 6 content, by NYS standard
Compared with grade 7
At grade 7 Simple probability is grade 6 content. The extra demand is tracking two earlier draws that were not put back, which changes the counts for the third draw, a step typical grade 7 problems often leave out.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
3 steps
Reading level
Question and choices: about grade 3, 70 words.
Students, last 30 days
Not enough answers yet to show (it takes 10).
Why it is hard, traps and wrong answers (shows the answer)

Why it is hard

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

  1. Much to keep track of moderate Working it out

    Three color counts and the total must each be updated after two draws, 4 green, 7 yellow and 6 red becoming 3, 6 and 6 out of 15, before the third draw is considered. (Sweller, 1988; Carpenter, Just & Shell, 1990)

  2. Turning words into math minor Understanding the question

    “Does not return the cards” and the two named draws are the conditions that change the box, and the stem states them in passing before asking about the third card. (Hegarty, Mayer & Monk, 1995; Koedinger & Nathan, 2004)

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

Traps

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

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

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

  1. Sort out the question

    The box starts with 4 green, 7 yellow and 6 red cards. Shiow-Chen draws cards one at a time, and she does not put them back. The first card is yellow, and the second is green. You need the probability that the third card is green.

  2. Take out the two cards

    The cards are not put back, so the box changes. Lower the yellow count by 1 and the green count by 1.

    ColorAt the startAfter two draws
    Green43
    Yellow76
    Red66
    Total1715

  3. Use the probability formula

    For the third draw, the results you want are the 3 green cards left. The total is the 15 cards left.

    P=number of results you wanttotal number of resultsP = \frac{\text{number of results you want}}{\text{total number of results}}

    P=33+6+6=315P = \frac{3}{3 + 6 + 6} = \frac{3}{15}

    Watch out: using the full box gives 417\frac{4}{17}, choice B. But two cards are gone.

  4. Simplify and answer

    Divide the top and bottom by 3.

    315=15\frac{3}{15} = \frac{1}{5}

    Watch out: compare the green cards with all 15 cards, not with the 12 cards of other colors. That gives 14\frac{1}{4}, choice C. And the three colors are not equally likely, so 13\frac{1}{3} is wrong too. So the answer is A, 15\frac{1}{5}.

Check your answer

17 cards minus the 2 drawn leaves 15. One fifth of 15 is 3, the number of green cards left. Also, the chance went down from 417\frac{4}{17}. That makes sense, because a green card was taken out.

More questions like this

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

Question 7 of 11

2023 Test B Math, question 47

Question

Four slips of paper numbered 1, 2, 3, and 4 were placed in a bag:

  • One slip was selected from the bag, and the number recorded.
  • Without replacement, a second slip was chosen.
  • This process continued until no more slips remained in the bag.
  • One part of a tree diagram representing the sample space is shown.

A partial tree diagram. The top number, 1, branches down to 2, 3, and 4. Below 2 are 3 and 4, below 3 are 2 and 4, and below 4 are 2 and 3. The 4 under the 2 has one branch leading down to an empty box.

Which number should be placed in the box to complete this part of the tree diagram?

  • A1
  • B2
  • C3
  • D4
Show the answer

Correct answer: C

The “without replacement” statement means that the bag contained four slips for the first draw, three slips for the second draw, two slips for the third draw, and one slip for the last draw. Following the branch from the top and navigating down to the box, the 1 was chosen first, the 2 was chosen second, and the 4 was chosen third. The only remaining number is 3. The element would be 1243.

Question analysis

Four slips numbered 1 to 4 are drawn from a bag one at a time without replacement, and part of a tree diagram of the outcomes is shown. Students decide which number belongs in the empty box at the end of one branch. It tests reading a tree diagram and understanding what drawing without replacement means.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
Below grade 7 Tree diagrams for compound events are a grade 7 topic, but here the tree is already drawn and only one box is empty. Following a single branch to that box is shorter than typical grade 7 work with sample spaces.
Estimated difficulty
Routine: one familiar step, little to misread.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 5, 68 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

Only one branch matters, and once it is read as the slips drawn so far, 1, 2 and 4, the last slip left in the bag is the only one missing. Errors come from not counting the top 1 as a draw, from forgetting that a drawn slip stays out, or from copying a nearby number in the diagram instead of following the path.

  1. Reading a figure, table or graph minor Understanding the question

    The box hangs below the branch 1, then 2, then 4, and students must read that path from the top as the order of the first three draws. (Lesh, Post & Behr, 1987; Kirsch & Mosenthal, 1990)

Strategy: Follow the branch from the top down to the box, list the slips already drawn on that path, and name the one still left in the bag.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

  1. Sort out the question

    The bag holds four slips: 1, 2, 3 and 4. Slips are drawn one at a time. “Without replacement” means a drawn slip does not go back in the bag. So each number can be drawn only once.

  2. Read the tree diagram

    A tree diagram shows the orders the draws can happen in. Each row going down is one draw. The 1 at the top is the first slip drawn. The numbers below it are the choices for the second draw, and so on.

  3. Follow the path to the box

    Start at the top and go down the lines to the box.

    • First draw: 1
    • Second draw: 2
    • Third draw: 4
    • Fourth draw: the box

    Slips 1, 2 and 4 are already out of the bag.

  4. Find the slip that is left

    The bag started with 1, 2, 3 and 4. Only slip 3 has not been drawn. So it must be the fourth draw.

    Watch out: The 1 at the top counts as a draw, so 1 is not left. That trap is choice A. Slip 2 was already drawn, so choice B can’t be right. And 4 is just the number above the box, choice D.

    So the answer is C, 3.

Check your answer

Read the whole path: 1, 2, 4, 3. Each number from 1 to 4 shows up exactly once. That is just what should happen when no slip goes back in the bag.

More questions like this

The original DOE question
Question 104 as printed on page 204 of the DOE handbook.
As printed in the DOE’s NYC Guide to the Specialized High Schools Admissions Test (SHSAT) for 2024 Admissions, page 204, question 104. Open page 204 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 8 of 11

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?

  • A215\frac{2}{15}
  • B425\frac{4}{25}
  • C310\frac{3}{10}
  • D13\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

410\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

39\frac{3}{9}

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

(410)(39)=1290=215\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.
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.

  1. 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)

  2. 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)

  3. 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)

  4. 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

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Probability
P=favorabletotalP = \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(A and B)=P(A)×P(B after A)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.

  1. 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.

  2. 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(red first)=410P(\text{red first}) = \frac{4}{10}

  3. 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(red second)=39P(\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.

  4. 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.

    410×39=4×310×9=1290\frac{4}{10} \times \frac{3}{9} = \frac{4 \times 3}{10 \times 9} = \frac{12}{90}

  5. 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.

    12÷690÷6=215\frac{12 \div 6}{90 \div 6} = \frac{2}{15}

    Watch out: the chance that both are blue is 610×59=13\frac{6}{10} \times \frac{5}{9} = \frac{1}{3}, choice D. The question says not blue. So the answer is A, 215\frac{2}{15}.

Check your answer

Count the ways. There are 10×9=9010 \times 9 = 90 ways to pick two cards in order. Red then red has 4×3=124 \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 9 of 11

2018 Test A Math, question 34

Question

There are 6 different cookies on a plate. Aiden will choose 2 of these cookies to pack in his lunch. How many different pairs of 2 cookies can he choose from the 6?

  • A12
  • B15
  • C30
  • D36
Show the answer

Correct answer: B

Create a list of the possible pairs. Let the cookies be named A, B, C, D, E, and F.

AB, AC, AD, AE, AF

BC, BD, BE, BF

CD, CE, CF

DE, DF

EF

There are a total of 15 possible pairs of cookies that Aiden can choose.

Question analysis

This counting question asks how many different pairs of cookies Aiden can choose from six different cookies on a plate. Students must count combinations, where taking the same two cookies in a different order is not a new pair. The challenge is that the order of picking does not matter, so each pair must be counted once.

NYS standards
  • 7.SP.8b sample spaces for compound events
Grade level
Grade 7 content, by NYS standard
Compared with grade 7
At grade 7 Grade 7 students list the outcomes of compound events with organized lists and tree diagrams. Counting pairs is close to that work, but students must notice that the order of choosing does not matter.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
What makes it hard
Solution
2 steps
Reading level
Question and choices: about grade 4, 29 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 principle are the most at risk, because it counts each pair twice. The question rewards noticing that a pair has no order, either by listing systematically or by halving 6 × 5. The list of 15 pairs takes under a minute, so there is no time pressure to justify a shortcut.

  1. Much to keep track of moderate Working it out

    Listing all pairs by hand means tracking which have been written so none is repeated or skipped, which an organized list such as AB to AF handles. (Sweller, 1988; Carpenter, Just & Shell, 1990)

Strategy: Name the cookies with letters and list the pairs systematically: the first cookie with every later one, then the second with every later one, and so on.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Number of pairs
n×(n−1)2\frac{n \times (n - 1)}{2}
n is how many things you choose from. Each of the n things can pair with the n − 1 others. That counts every pair twice, once from each side, so you divide by 2.

  1. Sort out the question

    There are 6 different cookies. Aiden picks 2 of them. A pair is just two cookies together.

    Order does not matter. Picking cookie 1 then cookie 2 gives the same pair as cookie 2 then cookie 1. You need to count the different pairs.

  2. List the pairs in order

    Number the cookies 1 to 6. Pair cookie 1 with every cookie after it. Then pair cookie 2 with every cookie after it, and so on. Only pairing with later numbers stops you from listing a pair twice.

    Start withPairsHow many
    11-2, 1-3, 1-4, 1-5, 1-65
    22-3, 2-4, 2-5, 2-64
    33-4, 3-5, 3-63
    44-5, 4-62
    55-61

  3. Add up the pairs

    Each row has one fewer pair than the row above it. That is because one fewer cookie comes after it. Add the last column of the table:

    5+4+3+2+1=155 + 4 + 3 + 2 + 1 = 15

    There are 15 different pairs.

  4. Check with the pairs formula

    Put n = 6 into the formula:

    6×52=302=15\frac{6 \times 5}{2} = \frac{30}{2} = 15

    Watch out: 6 × 5 = 30 counts each pair twice, once for each order. Stopping there gives choice C. And 6 × 6 = 36, choice D, lets Aiden pick the same cookie twice. So the answer is B, 15.

Check your answer

Think of handshakes. Each of 6 people shakes hands with the 5 others. That looks like 30 handshakes, but each handshake has two people in it, so it was counted twice. 30 ÷ 2 = 15, which matches the list.

More questions like this

The original DOE question
Question 91 as printed on page 87 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 87, question 91. Open page 87 of the handbook (PDF)

Question 10 of 11

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?

  • A115\frac{1}{15}
  • B9100\frac{9}{100}
  • C715\frac{7}{15}
  • D49100\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

710\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

69\frac{6}{9}

Multiply these two probabilities to get the solution:

(710)(69)=4290=715(\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.
What makes it hard
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.

  1. 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)

  2. 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)

  3. 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

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=ftP = \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(both)=P(first)×P(second)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.

  1. 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 115\frac{1}{15}, choice A.

  2. Count the candies

    5+3+2=105 + 3 + 2 = 10

    There are 10 candies. The ones that are not banana are the strawberry and orange ones:

    5+2=75 + 2 = 7

  3. Find the chance for the first candy

    7 of the 10 candies are not banana. Use the probability formula:

    P=710P = \frac{7}{10}

    So the first candy is not banana 7 times out of 10.

  4. 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=69P = \frac{6}{9}

    Watch out: the first candy is not put back. Using 710\frac{7}{10} again gives 49100\frac{49}{100}, choice D.

  5. Multiply the two chances

    710×69=4290\frac{7}{10} \times \frac{6}{9} = \frac{42}{90}

    Simplify by dividing the top and bottom by 6:

    42÷690÷6=715\frac{42 \div 6}{90 \div 6} = \frac{7}{15}

    So the answer is C, 715\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 11 of 11

2018 Test B Math, question 1

Question

How many 5-digit numbers can be created using the digits 2, 3, 5, 7, and 8 without repeating any digits within that 5-digit number?

Grid-in question: work out the answer and enter it as a number.

Show the answer

Correct answer: 120

There are 5 choices for the first digit, 4 choices for the second digit, 3 choices for the third digit, 2 choices for the fourth digit, and 1 choice for the final digit.

The total number of possibilities is

5 × 4 × 3 × 2 × 1 = 120

Question analysis

This grid-in counting question asks how many different 5-digit numbers can be made from five given digits when no digit may repeat. It tests the counting principle for arrangements rather than arithmetic. The catch is the no-repeat rule: each place has one fewer digit to choose from, and ignoring that gives a far larger count.

NYS standards
  • 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 probability work builds sample spaces with lists and tree diagrams for two or three stages. This needs the multiplication principle with a shrinking number of choices across five places, which listing cannot reach.
Estimated difficulty
Moderate: several steps, or a close reading with one strong distractor.
Solution
2 steps
Reading level
Question and choices: about grade 11, 19 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 tiny; the difficulty is knowing that arrangements are counted by multiplying, and that the no-repeat rule removes one digit for each later place. Students who have met the counting principle finish in seconds, while those who start listing numbers run out of time or guess a product they half remember. With no choices to check against, the 3,125 that comes from ignoring the rule looks as convincing as the right count.

  1. Something you must already know moderate Working it out

    Nothing in the stem names the counting principle: students must already know to multiply the digits left for each place, 5 × 4 × 3 × 2 × 1, because listing 5-digit numbers by hand cannot finish. (Anderson & Krathwohl, 2001)

  2. An answer in an unexpected form minor Choosing the answer

    As a grid-in there are no choices to compare with, so a count of 3,125 or 25 cannot be caught by noticing that it is missing from a list. (Enright, Morley & Sheehan, 2002)

Strategy: Fill the five places from left to right, write how many digits are still available for each place, and multiply those five counts together.

Traps

The wrong answers, classified

How the questions are analyzed, and the research behind it

Worked solution, step by step

Formulas used

Counting principle
total ways=a×b×c×…\text{total ways} = a \times b \times c \times \ldots
a is the number of choices for the first spot, b for the second, c for the third, and so on. Multiply them to get every possible way. It works because each choice for one spot can go with each choice for the next spot.

  1. Sort out the question

    You have five digits: 2, 3, 5, 7 and 8. A digit is one of the symbols 0 to 9. You build 5-digit numbers from them.

    “Without repeating” means each digit is used only once in a number. So 23578 is allowed, but 22578 is not.

    You need to find how many different numbers you can make.

  2. Count the choices for each spot

    Think of the number as five empty spots. Fill them from left to right.

    • First spot: any of the 5 digits, so 5 choices.
    • Second spot: one digit is used up, so 4 are left.
    • Third spot: 3 are left.
    • Fourth spot: 2 are left.
    • Last spot: only 1 is left.

  3. Multiply the choices together

    Use the counting principle. Each choice for one spot works with every choice for the next spot. So you multiply. You do not add.

    total ways=5×4×3×2×1=120\text{total ways} = 5 \times 4 \times 3 \times 2 \times 1 = 120

    There are 120 different numbers.

    Watch out: Adding 5 + 4 + 3 + 2 + 1 gives 15, which is far too small. Using 5 choices for every spot gives 3,125. That count lets digits repeat, which the question does not allow.

    So you would grid 120.

Check your answer

Try a smaller case you can list. With the digits 2, 3 and 5, the rule gives 3×2×1=63 \times 2 \times 1 = 6 numbers. The list is 235, 253, 325, 352, 523 and 532. That is 6, so the rule works.

More questions like this

The original DOE question
Question 58 as printed on page 153 of the DOE handbook.
As printed in the DOE’s Specialized High Schools Student Handbook 2018–2019, page 153, question 58. Open page 153 of the handbook (PDF)