Question

一个红球和一个绿球独立随机地被投入按正整数编号的箱子中。对每个球，投入第 $k$ 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：几何分布对立事件概率对称性

难度评级：1460

解答：

已知两个球落在不同箱子中时，较大编号箱子中的球为红球的概率为 $\frac 12$ 。因此只需用补集求 $\frac{P(\text{Balls in different bins})}2$ $= \frac{1-P(\text{Balls in same bins})}2$

两个球都落在第 $k$ 号箱子的概率为 $2^{-k} \cdot 2^{-k} = 4^{-k}.$

所以它们落在同一个箱子的概率为用等比数列求和，得到 $\sum_{k=1}^\infty 4^{-k}.$ $\frac 14 \cdot \dfrac{1}{1-\frac 14} = \frac 13.$

故所求概率为 $\frac{1-\frac 13}2 = \frac 13 .$

所以答案是 C。

Given that the two balls were tossed into separate bins, the probability that the ball in the higher-numbered bin is red is $\frac 12.$ Thus we must find $\frac{P(\text{Balls in different bins})}2$ $= \frac{1-P(\text{Balls in same bins})}2$ by complementary counting.

The probability that both balls are in bin $k$ is $2^{-k} \cdot 2^{-k} = 4^{-k}.$

The probability that they are both in the same bin is therefore $\sum_{k=1}^\infty 4^{-k}.$ Using the geometric sequence formula, we get this to be $\frac 14 \cdot \dfrac{1}{1-\frac 14} = \frac 13.$

Therefore, our answer is $\frac{1-\frac 13}2 = \frac 13 .$

Thus, the answer is C .

2019 AMC 10B 第 17 题

答案：C

难度评级：1460

解答：

其他年份的第 17 题