AMC10 2020 B

AMC10 2020 B · Q18

AMC10 2020 B · Q18. It mainly tests Probability (basic), Conditional probability (basic).

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?

一个瓮中有一个红球和一个蓝球。旁边有一个额外红蓝球的盒子。George 执行以下操作 4 次：他随机从瓮中抽一个球，然后从盒子取一个同色球，并将这两个同色球放回瓮中。4 次迭代后，瓮中有 6 个球。瓮中包含每种颜色 3 个球的概率是多少？

(A) \frac{1}{6} \frac{1}{6}

(B) \frac{1}{5} \frac{1}{5}

(D) \frac{1}{3} \frac{1}{3}

(E) \frac{1}{2} \frac{1}{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

By symmetry it may be assumed without loss of generality that George's first draw is red and therefore the urn contains two red balls and one blue ball before the second draw. With probability $\frac{2}{3}$, George will draw a red ball next. In that case the only way for the urn to end up with three balls of each color is for the next two draws to be blue, which will happen with probability $\frac{1}{4} \cdot \frac{2}{5} = \frac{1}{10}$. On the other hand, George will get a blue ball on his second draw with probability $\frac{1}{3}$, and the urn will then have two balls of each color. In that case no matter what color he gets on his third draw, for the urn to end up with three balls of each color the fourth draw must be different from the third draw, which will happen with probability $\frac{2}{5}$. Therefore, the probability that the urn will end up with three balls of each color is $$\frac{2}{3} \cdot \frac{1}{10} + \frac{1}{3} \cdot \frac{2}{5} = \frac{1}{5}.$$

由对称性，可以不失一般性地假设 George 的第一次抽取是红球，因此第二次抽取前瓮中有两个红球和一个蓝球。以概率 $\frac{2}{3}$，George 下次抽红球。那时，瓮最终有每种颜色 3 个球的唯一方法是接下来两次抽蓝球，概率为 $\frac{1}{4} \cdot \frac{2}{5} = \frac{1}{10}$。另一方面，George 以概率 $\frac{1}{3}$ 在第二次抽蓝球，那时瓮中有每种颜色两个球。那时，无论第三次抽什么颜色，为使瓮最终有每种 3 个球，第四次抽取必须与第三次不同，概率为 $\frac{2}{5}$。因此，瓮最终有每种颜色 3 个球的概率为 $$\frac{2}{3} \cdot \frac{1}{10} + \frac{1}{3} \cdot \frac{2}{5} = \frac{1}{5}.$$

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