AMC 8 1989 Problem 25
AMC8 1989 年第 25 题
概率★★★☆☆
Every time these two wheels are spun, two numbers are selected by the pointers. What is the probability that the sum of the two selected numbers is even?
每次转动这两个转盘时，指针各选中一个数字。两个选中数字之和为偶数的概率是多少？
- A.
- B.
- C.
- D.
- E.
正确答案 · Correct Answer
C
解析 · Solution
Solution 1
For the sum to be even, both numbers must have the same parity (i.e. both are odd or both are even).
The first spinner has 2 odd and 2 even numbers, so regardless of the second spinner's outcome, the probability the first spinner's number has the same parity as the second spinner's number, and thus the probability of the sum being even, is .
For the sum to be even, either both numbers are odd or both numbers are even.
The probability that both numbers are even is , while the probability that both numbers are odd is . So the probability that the sum is even is .-18, edited by j314andrews
We can complement count this, so we are looking for odd results. This happens when they have different parity. We can find the probability of an odd sum by using casework.
Case 1: The left spinner is odd. That has , and the right has probability of for an even number, of .
Case 2: The left spinner is even. That has , and the right has probability of for an odd number, of .
The sum is .
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