Question

Debra 反复抛一枚公平硬币，并记录到目前为止出现的正面数和反面数，直到出现连续两个正面或连续两个反面时停止。她出现连续两个正面，但在看到第二个正面之前先看到了第二个反面的概率是多少？

Debra flips a fair coin repeatedly, keeping track of how many heads and how many tails she has seen in total, until she gets either two heads in a row or two tails in a row, at which point she stops flipping. What is the probability that she gets two heads in a row but she sees a second tail before she sees a second head?

$\dfrac{1}{36}$

$\dfrac{1}{24}$

$\dfrac{1}{18}$

$\dfrac{1}{12}$

$\dfrac{1}{6}$

答案：B

知识点：基本概率等比数列

难度评级：1660

解答：

在最后一次重复结果出现之前，序列必须交替。若以 $H$ 开始，第二个正面必定早于第二个反面出现，所以成功序列必须以 $T$ 开始。为了在以 $HH$ 结束之前看到第二个反面，开头必须是 $THT$ 。

因此成功序列为 $THTHH,THTHTHH,\ldots$ ：对每个不小于 $5$ 的奇数长度，恰有一个序列。它们的总概率为 $\begin{gathered} \frac1{2^5}+\frac1{2^7}+\cdots\\ =\frac1{32}\cdot\frac1{1-\frac14}\\ =\frac1{24}. \end{gathered}$

所以答案是 B。

Before the final repeated flip, the sequence must alternate. If it starts with $H$ , the second head necessarily occurs before the second tail, so a successful sequence must start with $T$ . To see a second tail before ending with $HH$ , it must begin $THT$ .

Thus the successful sequences are $THTHH,THTHTHH,\ldots$ : exactly one sequence of each odd length at least $5$ . Their total probability is $\begin{gathered} \frac1{2^5}+\frac1{2^7}+\cdots\\ =\frac1{32}\cdot\frac1{1-\frac14}\\ =\frac1{24}. \end{gathered}$

Thus, the answer is B .

2019 AMC 10B 第 21 题

答案：B

难度评级：1660

解答：

其他年份的第 21 题