1989 AMC 8 第 25 题

先试着解答 1989 AMC 8 第 25 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 1989 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。

25.

每次转动图中的两个转盘时,两个指针各选出一个数。所选两个数之和为偶数的概率是多少?
Every time the two wheels shown are spun, two numbers are selected by the pointers. What is the probability that the sum of the two selected numbers is even?
16\dfrac{1}{6}
37\dfrac{3}{7}
12\dfrac{1}{2}
23\dfrac{2}{3}
57\dfrac{5}{7}
答案:C
知识点:基本概率奇偶性
小提示:
两个数之和为偶数,当且仅当两个数同为偶数或同为奇数
A sum is even exactly when both numbers are even or both are odd
大提示:
第一个转盘有 22 个偶数、22 个奇数;第二个转盘有 11 个偶数、22 个奇数
The first wheel has 22 even and 22 odd numbers; the second has 11 even and 22 odd
解答:
和为偶数需要两个数同奇偶。第一个转盘的偶数为 {4,8}\{4, 8\},奇数为 {3,5}\{3, 5\},所以两种奇偶性的概率都是 24=12\frac{2}{4} = \frac{1}{2}。第二个转盘的偶数为 {6}\{6\},概率为 13\frac{1}{3};奇数为 {7,9}\{7, 9\},概率为 23\frac{2}{3}。
因此和为偶数的概率是 12⋅13+12⋅23=16+26=12\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}。
所以正确答案是 C。
The sum is even when both numbers are even or both are odd. The first wheel has evens {4,8}\{4, 8\} and odds {3,5},\{3, 5\}, each with probability 24=12.\frac{2}{4} = \frac{1}{2}. The second wheel has even {6}\{6\} with probability 13\frac{1}{3} and odds {7,9}\{7, 9\} with probability 23.\frac{2}{3}.
So the probability of an even sum is 12⋅13+12⋅23=16+26=12.\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}.
Thus, the correct answer is C .
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