AMC10 2015 A

AMC10 2015 A · Q22

AMC10 2015 A · Q22. 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，每人拿着一枚公平的硬币。八个人同时掷硬币，掷出正面的人站起来，掷出反面的人继续坐着。问：没有任意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 $8 \cdot 4$ 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）：抛硬币共有 $2^8 = 256$ 种等可能结果。按出现正面（heads）的个数对圆桌上的可能排列分类。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，其中有 8 种是两个正面相邻。恰好 3 个正面的情况有 $\binom{8}{3} = 56$ 种，其中有 8 种是三个正面都相邻，且有 $8 \cdot 4$ 种是恰有一对相邻正面（相邻那一对有 8 种放法，第三个正面有 4 种放法）。最后，恰好 4 个正面且两两不相邻的情况有 2 种（正反必须交替）。若正面多于 4 个，则必有两个相邻。因此，没有相邻正面的排列数为 $1 + 8 + (28 - 8) + (56 - 8 - 32) + 2 = 47$，其概率为 $\frac{47}{256}$。

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