AMC12 2019 B

AMC12 2019 B · Q13

AMC12 2019 B · Q13. It mainly tests Probability (basic), Counting & probability misc.

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随机独立地投放到编号为正整数的箱子中，对于每个球，投放到箱子 $k$ 的概率为 $2^{-k}$，$k = 1, 2, 3, \dots$。红球被投放到比绿球更高编号箱子的概率是多少？

(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）：两球被投到同一个箱子的概率是 $$ \frac{1}{2}\cdot\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{4}+\frac{1}{8}\cdot\frac{1}{8}+\cdots =\sum_{n=1}^{\infty}\frac{1}{4^n} =\frac{\frac{1}{4}}{1-\frac{1}{4}} =\frac{1}{3}. $$ 因此，两球被投到不同箱子的概率是 $\frac{2}{3}$。由对称性，红球被投到编号比绿球更大的箱子的概率是 $\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$。

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