1994 AMC 12 第 27 题

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：

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