Question

八个人围坐在圆桌旁，每人手中有一枚公平硬币。八个人都抛硬币，抛出正面的人站起来，抛出反面的人仍坐着。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：环形排列基本概率分类讨论

难度评级：1910

小提示：

共有 $2^8 = 256$ 个等可能结果；数出没有相邻正面的结果

There are $2^8 = 256$ equally likely outcomes; count those with no two adjacent heads

大提示：

按站起来的人数 $(0,1,2,3,4)$ 分类；超过 $4$ 人必然相邻

Group by the number of people standing $(0,1,2,3,4);$ more than $4$ forces an adjacency

解答：

共有 $2^8 = 256$ 个等可能结果。按站起来的人数分类，数出圆周上 $8$ 个座位中没有两个相邻站立者（正面）的安排数。

从 $n$ 个圆周座位中选 $k$ 个不相邻座位的方法数为 $\dfrac{n}{n-k}\dbinom{n-k}{k}$ 。当 $n = 8$ 时，对 $k = 0,1,2,3,4$ 分别得到 $1,\; 8,\; 20,\; 16,\; 2$ 种；超过 $4$ 个站立者则不可能避免相邻。

总数为 $1 + 8 + 20 + 16 + 2 = 47$ ，所以概率是 $\dfrac{47}{256}$ 。

因此，正确答案是 A。

There are $2^8 = 256$ equally likely outcomes. Count the arrangements of standers (heads) with no two adjacent around the circle of $8$ seats, grouped by how many people stand.

The number of ways to choose $k$ non-adjacent seats from a circle of $n$ is $\dfrac{n}{n-k}\dbinom{n-k}{k}.$ For $n = 8$ this gives $1,\; 8,\; 20,\; 16,\; 2$ for $k = 0,1,2,3,4,$ and more than $4$ standers is impossible without an adjacency.

The total is $1 + 8 + 20 + 16 + 2 = 47,$ so the probability is $\dfrac{47}{256}.$

Thus, the correct answer is A.

2015 AMC 12A 第 17 题

答案：A

难度评级：1910

小提示：

大提示：

解答：

其他年份的第 17 题