2016 AMC 12B 第 19 题

Question

Tom、Dick 和 Harry 正在玩一个游戏。他们同时开始，各自不断投掷一枚公平硬币，直到第一次出现正面时停止。三人投掷次数都相同的概率是多少？

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?

\dfrac18

\dfrac17

\dfrac16

\dfrac14

\dfrac13

答案：B

知识点：几何分布独立事件

难度评级：1910

小提示：

单个玩家在第

n

次投掷首次出现正面的概率为

\left(\tfrac12\right)^n

。

A single player gets his first head on flip

n

with probability

\left(\tfrac12\right)^n

大提示：

三人都在第

n

次停止的概率为

\left(\tfrac18\right)^n

；对

n\ge1

求这个等比级数。

All three stop on flip

n

with probability

\left(\tfrac18\right)^n;

sum this geometric series over

n\ge1

解答：

单个玩家第一次正面出现在第

n

次的概率为

\left(\tfrac12\right)^n

。三人都在第

n

次停止的概率为

\left(\left(\tfrac12\right)^n\right)^3=\left(\tfrac18\right)^n

。对

n\ge1

求和，得

\displaystyle\sum_{n=1}^\infty\left(\tfrac18\right)^n =\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac17

。

所以正确答案是 B。

A player’s first head comes on flip

n

with probability

\left(\tfrac12\right)^n.

All three stopping on the same flip

n

has probability

\left(\left(\tfrac12\right)^n\right)^3=\left(\tfrac18\right)^n.

Summing over

n\ge1,

\displaystyle\sum_{n=1}^\infty\left(\tfrac18\right)^n =\frac{\frac{1}{8}}{1-\frac{1}{8}}=\frac17.

Thus, the correct answer is B.

2016 AMC 12B 第 19 题

答案：B

难度评级：1910

小提示：

大提示：

解答：

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