2019 AMC 12B 第 13 题

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