2025 AMC 12B 第 11 题

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

小提示：

队长是最高的三人，当且仅当这三人分到三个不同的组。

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

大提示：

依次放置前三高的人；每个人都必须避开已经用过的组，而剩余位置数也在减少。

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

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

小提示：

大提示：

解答：

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