Question

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?

$\dfrac{1}{6}$

$\dfrac{3}{7}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{5}{7}$

答案：C

知识点：基本概率奇偶性

难度评级：920

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 $2$ 个偶数、 $2$ 个奇数；第二个转盘有 $1$ 个偶数、 $2$ 个奇数

The first wheel has $2$ even and $2$ odd numbers; the second has $1$ even and $2$ odd

解答：

和为偶数需要两个数同奇偶。第一个转盘的偶数为 $\{4, 8\}$ ，奇数为 $\{3, 5\}$ ，所以两种奇偶性的概率都是 $\frac{2}{4} = \frac{1}{2}$ 。第二个转盘的偶数为 $\{6\}$ ，概率为 $\frac{1}{3}$ ；奇数为 $\{7, 9\}$ ，概率为 $\frac{2}{3}$ 。

因此和为偶数的概率是 $\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}$ 。

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 $\{4, 8\}$ and odds $\{3, 5\},$ each with probability $\frac{2}{4} = \frac{1}{2}.$ The second wheel has even $\{6\}$ with probability $\frac{1}{3}$ and odds $\{7, 9\}$ with probability $\frac{2}{3}.$

So the probability of an even sum is $\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{6} + \frac{2}{6} = \frac{1}{2}.$

Thus, the correct answer is C .

1989 AMC 8 第 25 题

答案：C

难度评级：920

小提示：

大提示：

解答：

其他年份的第 25 题