AMC12 2020 B

AMC12 2020 B · Q16

AMC12 2020 B · Q16. 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。旁边有一个装有额外红蓝球的盒子。乔治进行以下操作四次：他从瓮中随机抽取一个球，然后从盒子中取出一个同色球，并将这两个同色球放回瓮中。四次操作后，瓮中含有六个球。瓮中含有每种颜色各三个球的概率是多少？

(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

Answer (B): 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 $$ \frac{2}{3}\cdot\frac{1}{4}\cdot\frac{2}{5}+\frac{1}{3}\cdot\frac{2}{5}=\frac{1}{5}. $$

答案（B）：由于对称性，不失一般性可设乔治第一次抽到的是红球，因此在第二次抽取前，罐中有两个红球和一个蓝球。以概率$\frac{2}{3}$，乔治接下来会抽到红球。在这种情况下，罐中最终要达到每种颜色各三个球，唯一的方法是接下来的两次抽取都为蓝球，其发生概率为$\frac{1}{4}\cdot\frac{2}{5}=\frac{1}{10}$。另一方面，乔治第二次抽到蓝球的概率为$\frac{1}{3}$，此时罐中两种颜色各有两个球。在这种情况下，无论第三次抽到什么颜色，为了使最终每种颜色各有三个球，第四次抽取必须与第三次不同，其发生概率为$\frac{2}{5}$。因此，罐中最终每种颜色各三个球的概率为 $$ \frac{2}{3}\cdot\frac{1}{4}\cdot\frac{2}{5}+\frac{1}{3}\cdot\frac{2}{5}=\frac{1}{5}. $$

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