Question

一袋爆米花玉米粒中，白色玉米粒占 $\frac{2}{3}$ ，黄色玉米粒占 $\frac{1}{3}$ 。白色玉米粒中只有 $\frac{1}{2}$ 会爆开，而黄色玉米粒中有 $\frac{2}{3}$ 会爆开。从袋中随机选取一粒玉米，放入爆米花机后它爆开了。这粒玉米原本是白色的概率是多少？

A bag of popping corn contains $\frac{2}{3}$ white kernels and $\frac{1}{3}$ yellow kernels. Only $\frac{1}{2}$ of the white kernels will pop, whereas $\frac{2}{3}$ of the yellow ones will pop. A kernel is selected at random from the bag, and pops when placed in the popper. What is the probability that the kernel selected was white?

$\frac12$

$\frac59$

$\frac47$

$\frac35$

$\frac23$

答案：D

知识点：条件概率 Bayes’ theorem

难度评级：1750

小提示：

分别计算选中每种颜色且玉米粒爆开的概率

Compute the probabilities of selecting-and-popping for each color

大提示：

用选中白色且爆开的概率除以玉米粒爆开的总概率

Condition the white-and-popped probability on the total probability of popping

解答：

选中一粒会爆开的玉米的两种概率为 $\begin{gathered} P(W\text{ 且爆开})=\frac23\cdot\frac12=\frac13, \\ P(Y\text{ 且爆开})=\frac13\cdot\frac23=\frac29 \end{gathered}\text{。}$ 因此 $P(\text{爆开})=\frac{5}{9}$ ，且 $P(W\mid\text{爆开})=\frac{\frac{1}{3}}{\frac{5}{9}}=\frac35\text{。}$ 所以正确答案是 D。

The probabilities of selecting a kernel that pops are $\begin{gathered} P(W\text{ and pop})=\frac23\cdot\frac12=\frac13, \\ P(Y\text{ and pop})=\frac13\cdot\frac23=\frac29. \end{gathered}$ Hence $P(\text{pop})=\frac{5}{9},$ and $P(W\mid\text{pop})=\frac{\frac{1}{3}}{\frac{5}{9}}=\frac35.$ Thus the correct answer is D.

1994 AMC 12 第 27 题

答案：D

难度评级：1750

小提示：

大提示：

解答：

其他年份的第 27 题