2019 AMC 10B 第 17 题

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

小提示：

先求两个球落在同一个箱子的概率。

First compute the probability that both balls land in the same bin

大提示：

若它们落在不同箱子中，由对称性决定哪种颜色编号更大。

If they land in different bins, symmetry decides which color is higher

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

。因此只需求出

\frac{P(\text{两球落在不同箱子})}2

= \frac{1-P(\text{两球落在同一箱子})}2

这个值，这里用了补集计数。

两个球都落在第

k

号箱子的概率为

2^{-k} \cdot 2^{-k} = 4^{-k}\text{。}

所以两个球落在同一个箱子的概率为

\sum_{k=1}^\infty 4^{-k}\text{。}

用等比数列求和公式，得到它等于

\frac 14 \cdot \dfrac{1}{1-\frac 14} = \frac 13\text{。}

故所求概率为

\frac{1-\frac 13}2 = \frac 13\text{。}

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

小提示：

大提示：

解答：

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