Question

五个球围成一圈。Chris 随机选择两个相邻的球并交换它们。然后 Silva 也这样做，她选择的相邻球与 Chris 的选择独立。经过这两次相邻交换后，仍在原来位置上的球的期望个数是多少？

Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpositions?

$1.6$

$1.8$

$2.0$

$2.2$

$2.4$

答案：D

知识点：期望值分类讨论

难度评级：1420

解答：

Chris 选定一对相邻球后，Silva 有 $5$ 对相邻球可等可能选择。

如果 Silva 选择同一对，则所有 $5$ 个球都回到原位，概率为 $\frac15$ 。

如果 Silva 选择与 Chris 的那对恰好共享一个球的相邻对，则有 $2$ 个球在原位。这样的对有 $2$ 个，所以概率为 $\frac25$ 。

如果 Silva 选择与 Chris 的那对不相交的相邻对，则有 $1$ 个球在原位。概率同样为 $\frac25$ 。

期望为 $\begin{aligned} 5\cdot\frac15+2\cdot\frac25+1\cdot\frac25&=\frac{11}{5}\\ &=2.2. \end{aligned}$

所以答案是 D。

After Chris chooses an adjacent pair, Silva has $5$ equally likely adjacent pairs to choose.

If Silva chooses the same pair, all $5$ balls return to their original positions. This has probability $\frac15.$

If Silva chooses a pair sharing exactly one ball with Chris's pair, then $2$ balls are in their original positions. There are $2$ such pairs, so this has probability $\frac25.$

If Silva chooses a disjoint adjacent pair, then $1$ ball is in its original position. This also has probability $\frac25.$

The expected number is $\begin{aligned} 5\cdot\frac15+2\cdot\frac25+1\cdot\frac25&=\frac{11}{5}\\ &=2.2. \end{aligned}$

Thus, the answer is D .

2021 AMC 10B Fall 第 16 题

答案：D

难度评级：1420

解答：

其他年份的第 16 题