AMC 8 1997 Problem 20

AMC8 1997 年第 20 题

概率★★★☆☆

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

一对 8 面骰子，各面分别标有 1 到 8。每个面朝上的概率相同。两数乘积超过 36 的概率是

A. $\dfrac{5}{32}$
B. $\dfrac{11}{64}$
C. $\dfrac{3}{16}$
D. $\dfrac{1}{4}$
E. $\dfrac{1}{2}$

正确答案 · Correct Answer

解析 · Solution

Solution 1

There are 8×8 = 64 combinations to examine.

If one die is 1, then even with an 8 on the other die, no combinations will work.

If one die is 2, then even with an 8 on the other die, no combinations will work.

If one die is 3, then even with an 8 on the other die, no combinations will work.

If one die is 4, then even with an 8 on the other die, no combinations will work.

If one die is 5, then the other die must be an 8 to have a product over 36. Thus, (5,8) works.

If one die is 6, then the other die must be either 7 or 8 to have a product over 36. Thus, (6, 7) and (6, 8) both work.

If one die is 7, then the other die can be 6, 7, or 8 to have a product over 36. Thus, (7, 6), (7, 7), and (7, 8) all work.

If one die is 8, then the other die can be 5, 6, 7, or 8 to have a product over 36. Thus, (8, 5), (8,6), (8,7), and (8,8) work.

There are a total of 1 + 2 + 3 + 4 = 10 combinations that work out of a total of 64 possibilities.

Thus, the answer is

\displaystyle \frac{10}{64}

\displaystyle \frac{5}{32}

, and the answer is

A

But this may be slightly complicated, so you can also do solution 2 (scroll to bottom).

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