Question

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 $3$ 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?

$\dfrac{2}{9}$

$\dfrac{2}{7}$

$\dfrac{9}{28}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

Solution:

The captains are the three tallest exactly when those three land in three different groups, since each is then the tallest of its own group. Drop them into the $9$ slots one at a time ( $3$ per group). The second tallest misses the first's group with probability $\tfrac{6}{8},$ and the third misses both with probability $\tfrac{3}{7}.$ So the probability is $\tfrac{6}{8} \cdot \tfrac{3}{7} = \tfrac{9}{28}.$ Therefore, the answer is C.

2025 AMC 10B Problem 14

Solution:

Problem 14 in Other Years