2016 AMC 8 第 21 题

先试着解答 2016 AMC 8 第 21 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 2016 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。

21.

一个盒子里有 33 枚红色筹码和 22 枚绿色筹码。每次随机抽出一枚筹码且不放回,直到 33 枚红色筹码都被抽出,或直到两枚绿色筹码都被抽出。33 枚红色筹码都被抽出的概率是多少?
A box contains 33 red chips and 22 green chips. Chips are drawn randomly, one at a time without replacement, until all 33 of the reds are drawn or until both green chips are drawn. What is the probability that the 33 reds are drawn?
310\dfrac{3}{10}
25\dfrac{2}{5}
12\dfrac{1}{2}
35\dfrac{3}{5}
23\dfrac{2}{3}
答案:B
知识点:基本概率对称性
小提示:
停止时的结果由完整排列中哪种颜色最后出现决定
The stopping result is determined by which color appears last in the full ordering of the chips.
大提示:
在五个筹码位置中统计两枚绿色筹码的位置
Count equally likely positions for the two green chips among the five total chip positions.
视频讲解:
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文字解答:
33 枚红色筹码先于两枚绿色筹码都被抽出,当且仅当完整抽取顺序中的最后一枚筹码是绿色。选择两枚绿色筹码位置的等可能方式共有 1010 种,其中最后一个位置为绿色的有 44 种。
因此所求概率为 410=25。\dfrac{4}{10} = \dfrac{2}{5}\text{。}
所以正确答案是 B。
The 33 reds are drawn before both green chips exactly when a green chip is the last chip in the full ordering. There are 1010 equally likely ways to choose the two positions of the green chips, and 44 of them have a green chip in the last position.
Therefore, the desired probability is 410=25.\dfrac{4}{10} = \dfrac{2}{5}.
Thus, B is the correct answer.
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