2013 AMC 8 第 14 题

先试着解答 2013 AMC 8 第 14 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2013 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

14.

艾布手里有 11 颗绿色和 11 颗红色软糖。比娅手里有 11 颗绿色、11 颗黄色和 22 颗红色软糖。两人各随机拿出一颗软糖给对方看。颜色相同的概率是多少?
Abe holds 11 green and 11 red jelly bean in his hand. Bea holds 11 green, 11 yellow, and 22 red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?
14\dfrac{1}{4}
13\dfrac{1}{3}
38\dfrac{3}{8}
12\frac{1}{2}
23\dfrac{2}{3}
答案:C
知识点:基本概率分类讨论
小提示:
共有 2⋅42\cdot4 种等可能的软糖配对
There are 2⋅42\cdot4 equally likely jelly-bean pairings.
大提示:
统计绿色匹配和两个红色匹配
Count the green match and the two red matches.
视频讲解:
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文字解答:
艾布和比娅选择软糖共有 2⋅4=82\cdot4=8 种等可能结果。颜色相同可能是两人都选绿色,有 11 种;也可能两人都选红色,有 1⋅2=21\cdot2=2 种。
因此 88 个结果中有 33 个颜色相同,概率为 38\dfrac{3}{8}。
所以正确答案是 C。
There are 2⋅4=82\cdot4=8 equally likely ways for Abe and Bea to choose their jelly beans. The colors match if they both choose green, which can happen in 11 way, or if they both choose red, which can happen in 1⋅2=21\cdot2=2 ways.
Thus 33 of the 88 outcomes match, for probability 38\dfrac{3}{8}.
Thus, C is the correct answer.
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