2025 AMC 12B Problem 11

Below is the professionally curated solution for Problem 11 of the 2025 AMC 12B, 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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Concepts:basic probabilitysymmetry

Difficulty rating: 1590

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}

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 11 in Other Years