AMC 10 · 2025 · #14
Grade 7 probabilityNine athletes, no two of whom are the same height, try out for the basketball team. One at a time, they draw a wristband at random, without replacement, from a bag containing 3 blue bands, 3 red bands, and 3 green bands. They are divided into a blue group, a red group, and a green group. The tallest member of each group is named the group captain. What is the probability that the group captains are the three tallest athletes?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Nine athletes of all different heights each draw a colored wristband at random from a bag holding 3 blue, 3 red, and 3 green bands, splitting them into three groups of three. The tallest person in each group is that group's captain. Find the probability that the three group captains turn out to be the three tallest athletes overall.
Givens: 9 athletes, all with different heights; Bands drawn at random without replacement: 3 blue, 3 red, 3 green, forming three groups of 3; Each group's captain is its tallest member; Answer choices: (A) $\dfrac{2}{9}$, (B) $\dfrac{2}{7}$, (C) $\dfrac{9}{28}$, (D) $\dfrac{1}{3}$, (E) $\dfrac{3}{8}$
Unknowns: The probability that the three captains are exactly the three tallest athletes
Understand
Restated: Nine athletes of all different heights each draw a colored wristband at random from a bag holding 3 blue, 3 red, and 3 green bands, splitting them into three groups of three. The tallest person in each group is that group's captain. Find the probability that the three group captains turn out to be the three tallest athletes overall.
Givens: 9 athletes, all with different heights; Bands drawn at random without replacement: 3 blue, 3 red, 3 green, forming three groups of 3; Each group's captain is its tallest member; Answer choices: (A) $\dfrac{2}{9}$, (B) $\dfrac{2}{7}$, (C) $\dfrac{9}{28}$, (D) $\dfrac{1}{3}$, (E) $\dfrac{3}{8}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
Chasing captains directly is messy. Tool #16 (Change Focus) rewrites the event into something equivalent but simple: the three tallest are captains exactly when no two of them share a group. Once the event is about only the top three athletes, Tool #9 (Solve an Easier Related Problem) throws away the other six people entirely — they never affect who captains whom. Tool #7 (Identify Subproblems) then handles the top three one draw at a time: the tallest lands anywhere, then ask whether the second and third each grab a fresh color.
Execute — Answer: C
7.SP.C.8 Step 1 Rewrite the winning condition
- Each of the three tallest athletes is taller than all six shorter people.
- So the only way one of them fails to be a captain is if a taller person sits in the same group — and the only people taller are the other top-three athletes.
- Therefore the captains are the three tallest exactly when the three tallest land in three different-colored groups.
💡 A top-three athlete can only be beaten by another top-three athlete, so the whole race is really about keeping those three apart.
7.SP.C.7 Step 2 Track only the three tallest
- The six shorter athletes can be arranged in any way and it changes nothing, so ignore them.
- Drawing bands at random is the same as assigning colors to people at random, so imagine the three tallest draw their bands first, one after another.
- The tallest draws some color; it does not matter which.
💡 Only the top three decide the outcome, so watch them and let the rest fade into the background.
7.SP.C.7 Step 3 Second tallest needs a new color
- After the tallest takes one band, 8 bands remain: 2 of the tallest's color and 6 of the other two colors.
- For the second-tallest to end up in a different group, its band must be one of those 6.
💡 Six of the eight leftover bands are 'good,' because any color except the one already taken keeps the two apart.
7.SP.C.7 Step 4 Third tallest needs the last color
- Now two colors are already used by the top two, so the third-tallest must draw the one remaining color to avoid both of them.
- Of the 7 bands left, all 3 of that third color are 'good.'
💡 With two colors spoken for, only the single untouched color works, and there are 3 of those bands among the 7 left.
5.NF.B.4 Step 5 Multiply the three chances
The three groupings happen in sequence, so multiply the probabilities of each success and simplify the fraction.
💡 Each new draw must succeed on top of the last, and stacking chances means multiplying them.
7.SP.C.8 Each of the three tallest athletes is taller than all six shorter people. So the 7.SP.C.7 The six shorter athletes can be arranged in any way and it changes nothing, so i 7.SP.C.7 After the tallest takes one band, 8 bands remain: 2 of the tallest's color and 6 7.SP.C.7 Now two colors are already used by the top two, so the third-tallest must draw t 5.NF.B.4 The three groupings happen in sequence, so multiply the probabilities of each su Review
Reasonableness: Cross-check by whole-line counting: lay all nine athletes in a row where the first three positions are one color, the next three another, the last three the third. There are $9!$ orderings. For a favorable one, place the three tallest in three different color-blocks: $3!$ ways to pick which block each takes, $3\times3\times3$ ways to choose their exact seat inside their block, and $6!$ ways to seat the rest. That gives $\dfrac{3!\cdot 27\cdot 6!}{9!} = \dfrac{6\cdot 27}{9\cdot 8\cdot 7} = \dfrac{162}{504} = \dfrac{9}{28}$, matching (C). The value is a little under $\tfrac13$, which is sensible: it is fairly but not overwhelmingly likely the three tallest scatter.
Alternative: Count groups instead of draws. The teams can be formed in $\dfrac{\binom{9}{3}\binom{6}{3}\binom{3}{3}}{3!}=280$ unlabeled ways. Putting the three tallest in separate groups can be done in exactly $1$ way, and the remaining six fill the three groups in $\binom{6}{2}\binom{4}{2}\binom{2}{2}=90$ ways, so the probability is $\dfrac{90}{280}=\dfrac{9}{28}$.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Restating the target event as 'the three tallest land in three different groups' and identifying which outcomes count as favorable.)7.SP.C.7Develop probability models and use them to find probabilities of events (Modeling the random band draw as sequential draws for the top three and finding each single-draw probability, $\tfrac{6}{8}$ then $\tfrac{3}{7}$.)5.NF.B.4Apply and extend previous understandings of multiplication to multiply fractions (Multiplying the three success probabilities and simplifying $1\times\tfrac{6}{8}\times\tfrac{3}{7}=\tfrac{9}{28}$.)
⭐ The three tallest are captains only when they land in three different colors, so just track those three and multiply the chances of each grabbing a fresh color.
⭐ The three tallest are captains only when they land in three different colors, so just track those three and multiply the chances of each grabbing a fresh color.
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