AMC10 2019 B

AMC10 2019 B · Q21

AMC10 2019 B · Q21. It mainly tests Probability (basic), Recursion & DP style counting (basic).

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?

Debra 反复抛掷一枚公平的硬币，记录她总共看到的正面和反面的数量，直到她得到两个连续正面或两个连续反面，此时她停止抛掷。她得到两个连续正面但在看到第二个正面之前看到第二个反面的概率是多少？

(A) \frac{1}{36} \frac{1}{36}

(B) \frac{1}{24} \frac{1}{24}

(D) \frac{1}{12} \frac{1}{12}

(E) \frac{1}{6} \frac{1}{6}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): With probability 1, either HH or TT will occur after a finite number of flips. The desired event will occur if and only if the sequence of flips is THTHH or THTHTHH or THTHTHTHH or $\cdots$. The probabilities of these outcomes are $(\frac{1}{2})^5$, $(\frac{1}{2})^7$, $(\frac{1}{2})^9$, $\cdots$, a geometric sequence with common ratio $\frac{1}{4}$. The requested probability is the sum of these probabilities, $$ \frac{(\frac{1}{2})^5}{1-\frac{1}{4}}=\frac{1}{32}\cdot\frac{4}{3}=\frac{1}{24}. $$

答案（B）：以概率 1，经过有限次抛掷后，要么出现 HH，要么出现 TT。所求事件当且仅当抛掷序列为 THTHH 或 THTHTHH 或 THTHTHTHH 或 $\cdots$ 时发生。这些结果的概率分别为 $(\frac{1}{2})^5$、$(\frac{1}{2})^7$、$(\frac{1}{2})^9$、$\cdots$，构成一个公比为 $\frac{1}{4}$ 的等比数列。所求概率为这些概率之和： $$ \frac{(\frac{1}{2})^5}{1-\frac{1}{4}}=\frac{1}{32}\cdot\frac{4}{3}=\frac{1}{24}. $$

Topics