Question

星期一， $6$ 名学生同时来到辅导中心，每人随机分配给当天值班的 $6$ 名导师之一。星期二，同样的 $6$ 名学生又来了，同样的 $6$ 名导师也在值班，学生再次随机分配给导师。恰有 $2$ 名学生两天都见到同一位导师的概率是多少？

On Monday, $6$ students went to the tutoring center at the same time, and each one was randomly assigned to one of the $6$ tutors on duty. On Tuesday, the same $6$ students showed up, the same $6$ tutors were on duty, and the students were again randomly assigned to the tutors. What is the probability that exactly $2$ students met with the same tutor both Monday and Tuesday?

$\dfrac{1}{16}$

$\dfrac{3}{16}$

$\dfrac{1}{4}$

$\dfrac{3}{8}$

$\dfrac{1}{2}$

答案：B

知识点：错位排列排列基本概率

难度评级：1590

解答：

每天的分配都是把 $6$ 名学生分给 $6$ 名导师的一个排列。比较两天时，保持同一导师的学生数就是 $\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}}$ 的不动点个数，而这个置换是 $6$ 个元素上的均匀随机排列。要恰有 $2$ 个不动点，先选出这 $2$ 个，有 $\binom{6}{2}$ 种；其余 $4$ 个要错排，有 $D_4 = 9$ 种。因此概率为 $\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}$ 。所以正确答案是 B。

Each day's assignment is a permutation of the $6$ students among the $6$ tutors. Comparing the two days, the number who keep the same tutor is the number of fixed points of $\tau = \pi_{\text{Tue}}^{-1}\pi_{\text{Mon}},$ itself a uniformly random permutation of $6$ elements. We want exactly $2$ fixed points, so choose those $2$ in $\binom{6}{2}$ ways and derange the other $4,$ where $D_4 = 9.$ The probability is $\dfrac{\binom{6}{2} D_4}{6!} = \dfrac{15 \cdot 9}{720} = \dfrac{3}{16}.$ Thus, B is the correct answer.

2025 AMC 10B 第 11 题

答案：B

难度评级：1590

解答：

其他年份的第 11 题