Question

从满足 $10 \le x \le 99$ 的整数中选一个整数 $x$ 。若所有选择等可能， $x$ 的至少一个数字是 $7$ 的概率是多少？

An integer $x,$ with $10 \le x \le 99,$ is to be chosen. If all choices are equally likely, what is the probability that at least one digit of $x$ is a $7?$

$\dfrac{1}{9}$

$\dfrac{1}{5}$

$\dfrac{19}{90}$

$\dfrac{2}{9}$

$\dfrac{1}{3}$

答案：B

知识点：容斥原理数字基本概率

难度评级：1100

解答：

从 $10$ 到 $99$ 共有 $90$ 个整数。个位是 $7$ 的有十个，十位是 $7$ 的有九个。由于 $77$ 被重复计算，所以至少有一个 $7$ 的数共有 $10 + 9 - 1 = 18$ 个。概率为 $\dfrac{18}{90} = \dfrac{1}{5}$ 。

因此正确答案是 B。

There are $90$ integers from $10$ to $99.$ Ten have a units digit $7,$ and nine have a tens digit $7.$ Since $77$ is counted twice, there are $10 + 9 - 1 = 18$ with at least one $7.$ The probability is $\dfrac{18}{90} = \dfrac{1}{5}.$

Thus, the correct answer is B.

2004 AMC 12B 第 4 题

答案：B

难度评级：1100

解答：

其他年份的第 4 题