AMC8 2013

AMC8 2013 · Q14

AMC8 2013 · Q14. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Abe holds 1 green and 1 red jelly bean in his hand. Bea holds 1 green, 1 yellow, and 2 red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?

Abe手里拿着1颗绿色和1颗红色果冻豆。Bea手里拿着1颗绿色、1颗黄色和2颗红色果冻豆。他们各自随机挑选一颗果冻豆给对方看。颜色匹配的概率是多少？

(A) \frac{1}{4} \frac{1}{4}

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

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

(E) \frac{2}{3} \frac{2}{3}

Answer

Correct choice: (C)

正确答案：（C）

Solution

The favorable responses are either they both show a green bean or they both show a red bean. The probability that both show a green bean is $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$. The probability that both show a red bean is $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$. Therefore the probability is $\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}$

有利情况是他们都展示绿色果冻豆或都展示红色果冻豆。两者都展示绿色果冻豆的概率为 $\frac{1}{2}\cdot\frac{1}{4}=\frac{1}{8}$。两者都展示红色果冻豆的概率为 $\frac{1}{2}\cdot\frac{2}{4}=\frac{1}{4}$。因此，总概率为 $\frac{1}{4}+\frac{1}{8}=\boxed{\textbf{(C)}\ \frac{3}{8}}$

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