AMC12 2015 A

AMC12 2015 A · Q17

AMC12 2015 A · Q17. It mainly tests Basic counting (rules of product/sum), Probability (basic).

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?

八个人围坐在一张圆桌旁，每人拿着一枚均匀硬币。八个人同时掷硬币，掷出正面的人站起来，掷出反面的人仍然坐着。问：没有任何两位相邻的人同时站起来的概率是多少？

(A) $\frac{47}{256}$ $\frac{47}{256}$

(B) $\frac{3}{16}$ $\frac{3}{16}$

(D) $\frac{25}{128}$ $\frac{25}{128}$

(E) $\frac{51}{256}$ $\frac{51}{256}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): There are $2^8=256$ equally likely outcomes of the coin tosses. Classify the possible arrangements around the table according to the number of heads flipped. There is 1 possibility with no heads, and there are 8 possibilities with exactly one head. There are $\binom{8}{2}=28$ possibilities with exactly two heads, 8 of which have two adjacent heads. There are $\binom{8}{3}=56$ possibilities with exactly three heads, of which 8 have three adjacent heads and $8\cdot4$ have exactly two adjacent heads (8 possibilities to place the two adjacent heads and 4 possibilities to place the third head). Finally, there are 2 possibilities using exactly four heads where no two of them are adjacent (heads and tails must alternate). There cannot be more than four heads without two of them being adjacent. Therefore there are $1+8+(28-8)+(56-8-32)+2=47$ possibilities with no adjacent heads, and the probability is $\frac{47}{256}$.

答案（A）：共有$2^8=256$种等可能的抛硬币结果。按出现正面（head）的个数对围桌排列的可能情况分类。没有正面的情况有1种；恰好1个正面的情况有8种。恰好2个正面的情况有$\binom{8}{2}=28$种，其中8种有两个相邻的正面。恰好3个正面的情况有$\binom{8}{3}=56$种，其中8种有三个相邻的正面，且有$8\cdot4$种恰好有两个相邻的正面（放置那对相邻正面有8种方法，放置第三个正面有4种方法）。最后，恰好4个正面且任意两个都不相邻的情况有2种（正反必须交替）。若正面超过4个，则必有两个正面相邻。因此，没有相邻正面的情况共有$1+8+(28-8)+(56-8-32)+2=47$种，概率为$\frac{47}{256}$。

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