Question

一个瓮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 进行下面的操作四次：从瓮中随机抽一个球，然后从盒中取一个相同颜色的球，把这两个同色球一起放回瓮中。四次操作后，瓮中有六个球。瓮中红球和蓝球各三个的概率是多少？

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：条件概率组合

难度评级：1540

解答：

最终有三个红球和三个蓝球，当且仅当四次抽球颜色中恰好有两次红色和两次蓝色。这样的颜色顺序有 $\binom42=6$ 种。

对任意固定的两红两蓝抽取顺序，其概率为因为每种颜色第一、二次被抽到时，分子分别为 $1$ 和 $2$ ，而四次抽取前瓮中球的总数依次为 $2,3,4,5$ 。 $\frac{1\cdot 2\cdot 1\cdot 2}{2\cdot 3\cdot 4\cdot 5}=\frac{4}{120}=\frac{1}{30},$

因此所求概率为 $6\cdot\frac{1}{30}=\frac15$ 。

所以正确答案是 B。

The urn ends with three red and three blue balls exactly when the four draws contain two red draws and two blue draws. There are $\binom42=6$ possible color orders of this type.

For any fixed order with two red draws and two blue draws, the probability is $\frac{1\cdot 2\cdot 1\cdot 2}{2\cdot 3\cdot 4\cdot 5}=\frac{4}{120}=\frac{1}{30},$ because the first and second draws of each color have numerators $1$ and $2,$ while the total number of balls before the four draws is $2,3,4,5.$

Thus the desired probability is $6\cdot\frac{1}{30}=\frac15.$

Thus, the correct answer is B .

2020 AMC 10B 第 18 题

答案：B

难度评级：1540

解答：

其他年份的第 18 题