Question

一个罐子中有一个红球和一个蓝球，旁边有一盒额外的红球和蓝球。George 重复如下操作四次：从罐子中随机取出一个球，然后从盒中取一个相同颜色的球，把这两个同色球都放回罐子。四次操作后，罐子中有六个球。罐子中红球和蓝球各有三个的概率是多少？

An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?

$\dfrac16$

$\dfrac15$

$\dfrac14$

$\dfrac13$

$\dfrac12$

答案：B

知识点：基本概率组合

难度评级：1660

解答：

要最后两种颜色各有三个，四次新增的球中必须恰有两个红球。当罐中有 $k$ 个球，其中有 $c$ 个某色球时，下一次抽到该色的概率为 $\tfrac{c}{k}$ 。

每个恰含两次红球和两次蓝球的指定序列概率都为 $\tfrac{1\cdot 2\cdot 1\cdot 2}{2\cdot 3\cdot 4\cdot 5}$ 这样的序列有 $\binom42 = 6$ 个，所以总概率为 $\tfrac{6\cdot 4}{120} = \tfrac15$ 事实上，四次后红球数在 $\{1, 2, 3, 4, 5\}$ 中均匀分布，其中目标值为 $3$ ，概率也为 $\tfrac15$ 。

所以正确答案是 B。

To end with three of each color, exactly two of the four added balls must be red. Consider any sequence of draws. When the urn holds $k$ balls, drawing a particular color with count $c$ has probability $\tfrac{c}{k}.$

Any ordering with two red and two blue additions gives the same product $\tfrac{1\cdot 2\cdot 1\cdot 2}{2\cdot 3\cdot 4\cdot 5},$ and there are $\binom42 = 6$ such orderings, for probability $\tfrac{6\cdot 4}{120} = \tfrac15.$ (Equivalently, the number of red balls after four steps is uniform on $\{1, 2, 3, 4, 5\},$ so $3$ red occurs with probability $\tfrac15.$ )

Thus, the correct answer is B.

2020 AMC 12B 第 16 题

答案：B

难度评级：1660

解答：

其他年份的第 16 题