2004 AMC 10B Problem 11

Question

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}

Answer: C

Concepts:dice (probability)Simon’s Favorite Factoring Trick complementary counting

Difficulty rating: 1430

Small hint:

The condition

mn \gt m + n

rearranges to

(m-1)(n-1) \gt 1

Big hint:

Count the ordered pairs where

(m-1)(n-1) \le 1

instead

Solution:

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.