Question

九名运动员参加篮球队选拔，且没有两人身高相同。他们依次从一个袋子中随机抽取腕带，不放回；袋中有 $3$ 条蓝色、 $3$ 条红色、 $3$ 条绿色腕带。他们被分成蓝组、红组和绿组。每组最高的成员被指定为该组队长。三名队长正好是最高的三名运动员的概率是多少？

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?

$\dfrac{2}{9}$

$\dfrac{2}{7}$

$\dfrac{9}{28}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

答案：C

知识点：基本概率对称性

难度评级：1590

解答：

每组有 $3$ 个位置。最高的三名运动员正好成为队长，当且仅当他们落在三个不同的组中。将三人依次放入 $9$ 个位置，第二人在剩下的 $8$ 个位置中有 $6$ 个可选，第三人在剩下的 $7$ 个位置中有 $3$ 个可选。因此概率为 $\dfrac{6}{8} \cdot \dfrac{3}{7} = \dfrac{9}{28}$ 。

所以正确答案是 C。

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

Thus, the correct answer is C.

2025 AMC 12B 第 11 题

答案：C

难度评级：1590

解答：

其他年份的第 11 题