2025 AMC 12B Problem 11

Attempt Problem 11 of the 2025 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12B solutions, or check the answer key.

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11.

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?

29\dfrac{2}{9}

27\dfrac{2}{7}

928\dfrac{9}{28}

13\dfrac{1}{3}

38\dfrac{3}{8}

Answer: C
Concepts:basic probabilitysymmetry
Difficulty rating: 1590
Small Hint:

The captains are the three tallest exactly when the three tallest land in three different groups

Big Hint:

Place the tallest three one at a time; each must avoid the groups already used, whose slots shrink the pool

Solution:

Each group has 33 slots. The three tallest athletes are the captains precisely when they fall into three different groups. Placing them one at a time into the 99 slots, the second must avoid the first’s group (66 of the remaining 88 slots) and the third must avoid both used groups (33 of the remaining 77 slots). The probability is 6837=928.\dfrac{6}{8} \cdot \dfrac{3}{7} = \dfrac{9}{28}.

Thus, the correct answer is C.

Problem 10#10
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