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2025 AMC 10B Problem 14

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2025 AMC 10B Problem 14AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 3 of 5 · about 4 min · position P11-15

Nine 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 33 blue bands, 33 red bands, and 33 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?

  1. A)

    29\frac29

  2. B)

    27\frac 27

  3. C)

    928\frac9{28}

  4. D)

    13\frac13

  5. E)

    38\frac38

Answer

C

Key insight

The captains are the three tallest exactly when the three tallest land in three different groups, so only the bands of the top three matter.

Solution

Rank the athletes by height and call the three tallest A1,A2,A3A_1, A_2, A_3.

The captains are exactly A1,A2,A3A_1, A_2, A_3 if and only if these three are in three different groups. Indeed, if they are in different groups then each is taller than everyone else in its group, so each is a captain; and if two of them share a group, the shorter of the two is not a captain, so the set of captains is not {A1,A2,A3}\{A_1, A_2, A_3\}.

So only the colours drawn by the top three athletes matter. The nine bands are 33 of each colour, drawn without replacement.

  • A1A_1 draws some band; whatever it is, 66 of the remaining 88 bands have a different colour.
  • Given that A2A_2 has a second colour, 77 bands remain and exactly 33 carry the third colour, which is what A3A_3 needs.

Hence
P=68⋅37=1856=928.P = \frac68 \cdot \frac37 = \frac{18}{56} = \frac{9}{28}.

The answer is (C) 928\boxed{\textbf{(C)}\ \frac9{28}}.

Why this works

Naming the condition in terms of the objects that actually decide it, here the three tallest athletes, collapses a problem about nine people into a problem about three. The sequential conditional-probability count then works because the bands are exchangeable: at each stage every remaining band is equally likely, so only the counts by colour matter, never the order in which the other six athletes draw. The same reduction handles any "top kk are the group leaders" question.

Alternative approach

Count directly. The number of ways to split nine labelled athletes into three labelled groups of three is
9!3! 3! 3!=1680.\frac{9!}{3!\,3!\,3!} = 1680.
For a favourable split, assign A1,A2,A3A_1, A_2, A_3 to the three colours in 3!=63! = 6 ways, then distribute the remaining six athletes two to each group in 6!2! 2! 2!=90\frac{6!}{2!\,2!\,2!} = 90 ways, giving 6⋅90=5406 \cdot 90 = 540. The probability is 5401680=928\frac{540}{1680} = \frac{9}{28}.

The trap

Treating each athlete as independently picking one of three colours, which gives 3!/3^3 = 2/9 at choice (A); the bands are drawn without replacement, so the conditional probabilities are 6/8 and 3/7.

Common mistakes

  • Treating each athlete as independently picking one of three colours, which gives 3!/3^3 = 2/9 at choice (A); the bands are drawn without replacement, so the conditional probabilities are 6/8 and 3/7.
  • Using 69⋅38\frac69 \cdot \frac38 by forgetting that A1A_1's own band has already left the bag.
  • Dividing the favourable count by 3!3! as though the colours were interchangeable; the groups are named, and the same division is applied to the total, so the factor cancels only if it is applied consistently.

Techniques

Set up the equation/formula and compute; no special trick needed · Exploit symmetry to reduce work or pair up objects