AMC12 2016 B

AMC12 2016 B · Q19

AMC12 2016 B · Q19. It mainly tests Probability (basic), Counting & probability misc.

Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times?

汤姆、迪克和哈里在玩一个游戏。他们同时开始，每个人反复掷一枚公平硬币，直到第一次掷出正面为止，然后停止。问三个人掷硬币的次数都相同的概率是多少？

(A) \frac{1}{8} \frac{1}{8}

(B) \frac{1}{7} \frac{1}{7}

(D) \frac{1}{4} \frac{1}{4}

(E) \frac{1}{3} \frac{1}{3}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): The probability that a flipper obtains his first head on the $n^{\text{th}}$ flip is $\left(\frac{1}{2}\right)^n$, because the sequence of outcomes must be exactly $TT\cdots TH$, with $n-1$ Ts. Therefore the probability that all of them obtain their first heads on the $n^{\text{th}}$ flip is $\left(\left(\frac{1}{2}\right)^n\right)^3=\left(\frac{1}{8}\right)^n$. The probability that all three flip their coins the same number of times is computed by summing an infinite geometric series: \[ \left(\frac{1}{8}\right)^1+\left(\frac{1}{8}\right)^2+\left(\frac{1}{8}\right)^3+\cdots=\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac{1}{7}. \]

答案（B）：掷硬币者在第 $n^{\text{th}}$ 次掷出第一次正面的概率是 $\left(\frac{1}{2}\right)^n$，因为结果序列必须恰好为 $TT\cdots TH$，其中有 $n-1$ 个 $T$。因此，他们三人都在第 $n^{\text{th}}$ 次掷出第一次正面的概率为 $\left(\left(\frac{1}{2}\right)^n\right)^3=\left(\frac{1}{8}\right)^n$。三个人掷硬币次数相同的概率可通过对无穷等比级数求和得到： \[ \left(\frac{1}{8}\right)^1+\left(\frac{1}{8}\right)^2+\left(\frac{1}{8}\right)^3+\cdots=\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac{1}{7}. \]

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