Question

两个八面骰子的面都标有 $1$ 到 $8$ 。掷骰子时，每个面朝上的概率相同。两个朝上数字的乘积大于它们的和的概率是多少？

Two eight-sided dice each have faces numbered $1$ through $8.$ When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?

$\dfrac{1}{2}$

$\dfrac{47}{64}$

$\dfrac{3}{4}$

$\dfrac{55}{64}$

$\dfrac{7}{8}$

答案：C

知识点：骰子（概率）西蒙最爱的因式分解技巧补集计数

难度评级：1430

解答：

共有 $8 \cdot 8 = 64$ 个有序结果。条件 $mn \gt m + n$ 等价于 $(m-1)(n-1) \gt 1$ 。

不满足条件的情况只有 $m = 1$ 、 $n = 1$ ，或 $m = n = 2$ 。这些情况共有 $8 + 8 - 1 + 1 = 16$ 个。

因此概率为 $\dfrac{64 - 16}{64} = \dfrac{48}{64} = \dfrac{3}{4}$ 。

所以正确答案是 C。

There are $8 \cdot 8 = 64$ ordered pairs. The inequality $mn \gt m + n$ is equivalent to $(m-1)(n-1) \gt 1.$

This fails only when $m = 1,$ $n = 1,$ or $m = n = 2,$ which account for $8 + 8 - 1 + 1 = 16$ pairs.

The probability is $\dfrac{64 - 16}{64} = \dfrac{48}{64} = \dfrac{3}{4}.$

Thus, the correct answer is C.

2004 AMC 10B 第 11 题

答案：C

难度评级：1430

解答：

其他年份的第 11 题