AMC12 2025 B

AMC12 2025 B · Q11

AMC12 2025 B · Q11. It mainly tests Basic counting (rules of product/sum), Probability (basic).

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?

九名运动员，他们两人之间身高都不相同，参加篮球队选拔。他们依次从一个袋子里随机抽取腕带，不放回，袋子里有3条蓝色腕带、3条红色腕带和3条绿色腕带。他们被分成蓝色组、红色组和绿色组。每组中最高的成员被任命为小组长。三名小组长是这九名运动员中三名最高的运动员的概率是多少？

(A) \dfrac{2}{9} \dfrac{2}{9}

(B) \dfrac{2}{7} \dfrac{2}{7}

(D) \dfrac{1}{3} \dfrac{1}{3}

(E) \dfrac{3}{8} \dfrac{3}{8}

Answer

Correct choice: (C)

正确答案：（C）

Solution

We will arrange the nine athletes in a line, with the first three being in the blue group, the next three being in the red group, and the last three being in the green group. Suppose the three tallest athletes are named $A, B, C$, in some order. We have $3!$ ways to choose which group each of these athletes can be in. We then have $3 \cdot 3 \cdot 3$ ways to determine which of the three positions in that group they can take. From here, we must distribute the remaining six athletes in the remaining six spaces, which can be done in $6!$ ways. There are therefore $6 \cdot 27 \cdot 6!$ favorable outcomes. There are also $9!$ total ways to arrange the athletes with no restrictions. Our answer is $\frac{6 \cdot 27 \cdot 6!}{9!} = \frac{6 \cdot 27}{9 \cdot 8 \cdot 7} = \boxed{\textbf{(C) }\frac{9}{28}}.$

我们将九名运动员排成一排，前三名为蓝色组，接下来三名为红色组，最后三名为绿色组。假设三名最高的运动员分别命名为$A,B,C$，顺序任意。这些运动员分配到各组有$3!$种方式。然后，在所分配的组中，他们有$3\cdot3\cdot3$种方式选择三个位置中的位置。从这里开始，将剩余的六名运动员分配到剩余的六个位置，有$6!$种方式。因此，有$6\cdot27\cdot6!$种有利结果。总共有$9!$种无限制排列运动员的方式。答案是$\frac{6\cdot27\cdot6!}{9!}=\frac{6\cdot27}{9\cdot8\cdot7}=\boxed{\textbf{(C) }\frac{9}{28}}$。

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