Question

两个转盘被转动，指针各指向一个数。第一个转盘有四个面积相等的区域，标着 $3$ 、 $4$ 、 $5$ 和 $8$ ；第二个转盘有三个面积相等的区域，标着 $6$ 、 $7$ 和 $9$ 。两个指到的数之和为偶数的概率是多少？

Two wheels are spun, and each wheel's pointer selects one number. The first wheel is divided into four equal regions numbered $3,$ $4,$ $5,$ and $8;$ the second wheel is divided into three equal regions numbered $6,$ $7,$ and $9.$ 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

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

解答：

其他年份的第 25 题