Question

八个人围坐在一张圆桌旁，每人手中有一枚公平硬币。所有人同时抛硬币，正面朝上的人站起来，反面朝上的人仍坐着。没有两个相邻的人都站起来的概率是多少？

Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?

$\dfrac{47}{256}$

$\dfrac{3}{16}$

$\dfrac{49}{256}$

$\dfrac{25}{128}$

$\dfrac{51}{256}$

答案：A

知识点：有限制的排列基本概率分类讨论

难度评级：1970

解答：

统计站起来的人构成的集合。站起 $0$ 人和 $1$ 人时，分别有 $1$ 种和 $8$ 种。

站起 $2$ 人时，从任意两人的组合中减去 $8$ 对相邻者，共有 $\binom82-8=20$ 种。

站起 $3$ 人时，先选一人；其余五个非相邻座位有 $10$ 对，其中 $4$ 对相邻，剩下 $6$ 对。每个最终集合被计数三次，所以共有 $8\cdot6/3=16$ 种。

站起 $4$ 人时，只有两种交替站立的情况。因此有利结果共有 $1+8+20+16+2=47$ 种。全部 $2^8=256$ 种结果等可能，所以概率为 $\frac{47}{256}$ 。

所以正确答案是 A。

Count the possible sets of people who stand. For $0$ and $1$ people standing, there are $1$ and $8$ possibilities.

For $2$ people standing, choose any pair and subtract the $8$ adjacent pairs: $\binom82-8=20$ .

For $3$ people standing, first choose one standing person. Among the remaining five non-neighbor seats, $10$ pairs are possible, but $4$ of those pairs are adjacent, leaving $6$ . This counts each final set three times, so there are $8\cdot6/3=16$ possibilities.

For $4$ people standing, the only possibilities are the two alternating sets. Thus the number of favorable coin-flip outcomes is $1+8+20+16+2=47$ . Since all $2^8=256$ outcomes are equally likely, the probability is $\frac{47}{256}$ .

Thus, A is the correct answer.

2015 AMC 10A 第 22 题

答案：A

难度评级：1970

解答：

其他年份的第 22 题