Question

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

$\dfrac{5}{32}$

$\dfrac{11}{64}$

$\dfrac{3}{16}$

$\dfrac{1}{4}$

$\dfrac{1}{2}$

Solution:

We can case on the value of the first die. If its value is $1 - 4,$ then it is impossible for the product to be greater than $36.$

If it is a $5,$ then the other die has to roll an $8,$ otherwise the product is less than $36.$

If the first die is a $6$ or $7,$ then other die has to be at least a $6,$ giving $2 \cdot 3 = 6$ possibilities.

Finally, if the first roll is an $8,$ the other die must roll at least a $5,$ giving us $3$ more possibilities.

The total number of working pairs is $1 + 6 + 3 = 10,$ and the total number of pairs is $8^2 = 64.$ The desired probability is then $\dfrac{10}{64} = \dfrac{5}{32}.$

Thus, A is the correct answer.

1997 AMC 8 Problem 20

Solution:

Problem 20 in Other Years