1997 AMC 8 第 9 题

先试着解答 1997 AMC 8 第 9 题,然后核对你的答案与精心整理的解答,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 1997 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。

9.

三个姓名不同的学生排成一列。从前到后按字母顺序排列的概率是多少?
Three students, with different names, line up single file. What is the probability that they are in alphabetical order from front-to-back?
112\dfrac{1}{12}
19\dfrac{1}{9}
16\dfrac{1}{6}
13\dfrac{1}{3}
23\dfrac{2}{3}
答案:C
知识点:基本概率排列
小提示:
只有一种排列是按字母顺序的
Only one ordering is alphabetical
大提示:
数出三个学生所有可能的排列顺序
Count all possible orders of the three students
解答:
队首有 33 种选择,中间有 22 种选择。
最后剩 11 种选择,所以共有 3⋅2⋅1=63 \cdot 2 \cdot 1 = 6 种排列。
其中只有一种按字母顺序排列,所以概率为 16\dfrac{1}{6}。
所以正确答案是 C。
There are 33 options for the person in front. Then, there are 22 options for the person in the middle.
This leaves 11 choice for the person at the end. There are 3⋅2⋅1=63 \cdot 2 \cdot 1 = 6 ways for the people to line up.
Only one of these lines is correct, and the probability it occurs is 16.\dfrac{1}{6}.
Thus, C is the correct answer.
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