Question

一个红球和一个绿球随机且独立地被投入编号为正整数的箱子中。对每个球来说，投进第 $k$ 号箱子的概率为 $2^{-k}$ ，其中 $k=1$ ， $2$ ， $3$ ， $\ldots$ 。红球被投入编号比绿球更大的箱子的概率是多少？

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k=1,$ $2,$ $3,$ $\ldots$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

$\dfrac{1}{4}$

$\dfrac{2}{7}$

$\dfrac{1}{3}$

$\dfrac{3}{8}$

$\dfrac{3}{7}$

答案：C

知识点：基本概率对称性对立事件概率

难度评级：1440

小提示：

由对称性，红球编号更大和绿球编号更大的概率相等

By symmetry, red-higher and green-higher are equally likely

大提示：

先求平局概率： $\displaystyle\sum_{k\ge1}\left(2^{-k}\right)^2$

Find the probability of a tie, $\displaystyle\sum_{k\ge1}\left(2^{-k}\right)^2$

解答：

两球落入同一箱子的概率为 $\begin{gathered} \sum_{k=1}^\infty \left(2^{-k}\right)^2=\sum_{k=1}^\infty 4^{-k} \\ =\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac13 \end{gathered}\text{。}$

由对称性，红球编号更大和绿球编号更大的概率相等，所以各自概率为 $\dfrac{1-\tfrac13}{2}=\dfrac13\text{。}$

所以正确答案是 C。

The probability the balls land in the same bin is $\begin{gathered} \sum_{k=1}^\infty \left(2^{-k}\right)^2=\sum_{k=1}^\infty 4^{-k} \\ =\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac13. \end{gathered}$

By symmetry, the red ball being higher and the green ball being higher are equally likely, so each has probability $\dfrac{1-\tfrac13}{2}=\dfrac13.$

Thus, C is the correct answer.

2019 AMC 12B 第 13 题

答案：C

难度评级：1440

小提示：

大提示：

解答：

其他年份的第 13 题