AMC8 1997

AMC8 1997 · Q20

AMC8 1997 · Q20. It mainly tests Basic counting (rules of product/sum), Probability (basic).

A pair of 8-sided dice have sides numbered 1 through 8. Each side has the same probability of landing face up. The probability that the product of the two numbers on the sides that land face-up exceeds 36 is

一对 8 面骰子，面编号 1 到 8。每面朝上的概率相同。两个朝上面数字的乘积超过 36 的概率是

(A) $\frac{5}{32}$ $\frac{5}{32}$

(B) $\frac{11}{64}$ $\frac{11}{64}$

(D) $\frac{1}{4}$ $\frac{1}{4}$

(E) $\frac{1}{2}$ $\frac{1}{2}$

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): There are 64 equally likely possibilities for the numbers on the two dice. Of these, only (5, 8), (6, 7), (6, 8), (7, 6), (7, 7), (7, 8), (8, 5), (8, 6), (8, 7), and (8, 8) give products exceeding 36, so the probability of this occurring is $\frac{10}{64}=\frac{5}{32}$.

答案（A）：两个骰子上的数字共有 64 种等可能的情况。其中，只有 (5, 8)、(6, 7)、(6, 8)、(7, 6)、(7, 7)、(7, 8)、(8, 5)、(8, 6)、(8, 7) 和 (8, 8) 的乘积大于 36，因此该事件发生的概率为 $\frac{10}{64}=\frac{5}{32}$。

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