AMC10 2019 B

AMC10 2019 B · Q17

AMC10 2019 B · Q17. It mainly tests Probability (basic), Casework.

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 $k = 1, 2, 3, \dots$. What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?

一个红球和一个绿球被随机且独立地扔入编号为正整数的箱子中，对于每个球，扔入箱子 $k$ 的概率为 $2^{-k}$，其中 $k = 1, 2, 3, \dots$。红球被扔入编号高于绿球的箱子的概率是多少？

(A) $\frac{1}{4}$ $\frac{1}{4}$

(B) $\frac{2}{7}$ $\frac{2}{7}$

(D) $\frac{3}{8}$ $\frac{3}{8}$

(E) $\frac{3}{7}$ $\frac{3}{7}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): The probability that the two balls are tossed into the same bin is $\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{8}\cdot\frac{1}{8}+\cdots=\sum_{n=1}^{\infty}\frac{1}{4^n}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}.$ Therefore the probability that the balls are tossed into different bins is $\frac{2}{3}$. By symmetry the probability that the red ball is tossed into a higher-numbered bin than the green ball is $\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$.

答案（C）：两个球被投进同一个箱子的概率为 $\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{8}\cdot\frac{1}{8}+\cdots=\sum_{n=1}^{\infty}\frac{1}{4^n}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{1}{3}.$ 因此，两个球被投进不同箱子的概率是 $\frac{2}{3}$。由对称性，红球被投进编号比绿球更大的箱子的概率为 $\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$。

Topics