Question

$6$ 个球各自独立且等可能地被涂成黑色或白色。每个球都与其他 $5$ 个球中超过一半的球颜色不同的概率是多少？

Each of $6$ balls is randomly and independently painted either black or white with equal probability. What is the probability that every ball is different in color from more than half of the other $5$ balls?

$\dfrac{1}{64}$

$\dfrac{1}{6}$

$\dfrac{1}{4}$

$\dfrac{5}{16}$

$\dfrac{1}{2}$

答案：D

知识点：基本概率组合

难度评级：900

解答：

要使每个球都与超过一半的其他球颜色不同，任意一个球必须看到至少 $3$ 个相反颜色的球。

因此必须正好有三颗黑球和三颗白球。所有涂色共有 $2^6 = 64$ 种，其中选择哪三颗球为白色有 $\binom{6}{3} = 20$ 种。

所以所求概率为 $\dfrac{20}{64} = \dfrac{5}{16}$ 。

所以正确答案是 D。

Note that for this restriction to hold, there must be $3$ balls of each color.

There are $2^6 = 64$ ways to color the balls and $\binom{6}{3} = 20$ to choose which balls are white.

The desired probability is therefore $\dfrac{20}{64} = \dfrac{5}{16}.$

Thus, D is the correct answer.

2021 AMC 10A Fall 第 13 题

答案：D

难度评级：900

解答：

其他年份的第 13 题