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：基本概率条件概率

难度评级：1500

解答：

三名队长正好是最高的三人，当且仅当这三人进入三个不同的组；这样每人都是自己组中最高的。把这三人依次放入 $9$ 个位置（每组 $3$ 个）中。第二高的人避开第一高所在组的概率是 $\tfrac{6}{8}$ ，第三高的人避开前两人所在组的概率是 $\tfrac{3}{7}$ 。所以所求概率为 $\tfrac{6}{8} \cdot \tfrac{3}{7} = \tfrac{9}{28}$ 。因此正确答案是 C。

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 第 14 题

答案：C

难度评级：1500

解答：

其他年份的第 14 题