2004 AMC 10B Problems/Problem 11

Problem

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?

$\mathrm{(A) \ } \frac{1}{2} \qquad \mathrm{(B) \ } \frac{47}{64} \qquad \mathrm{(C) \ } \frac{3}{4} \qquad \mathrm{(D) \ } \frac{55}{64} \qquad \mathrm{(E) \ } \frac{7}{8}$

Solution

We have $1\times n = n < 1 + n$ , hence if at least one of the numbers is $1$ , the sum is larger. There $15$ such possibilities.

We have $2\times 2 = 2+2$ .

For $n>2$ we already have $2\times n = n + n > 2 + n$ , hence all other cases are good.

Out of the $8\times 8$ possible cases, we found that in $15+1=16$ the sum is greater than or equal to the product, hence in $64-16=48$ it is smaller. Therefore the answer is $\frac{48}{64} = \boxed{\frac34}$ .

2004 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2004 AMC 10B Problems/Problem 11

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