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2004 AMC 10B Problem 11

Problem 11 of 25IntermediateAlgebraCounting & Probability

Two eight-sided dice each have faces numbered 11 through 8.8. 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?

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Solution

There are 8⋅8=648 \cdot 8 = 64 ordered pairs. The inequality mn>m+nmn \gt m + n is equivalent to (m−1)(n−1)>1.(m-1)(n-1) \gt 1. This fails only when m=1,m = 1, n=1,n = 1, or m=n=2,m = n = 2, which account for 8+8−1+1=168 + 8 - 1 + 1 = 16 pairs. The probability is 64−1664=4864=34.\dfrac{64 - 16}{64} = \dfrac{48}{64} = \dfrac{3}{4}. Thus, the correct answer is C.

More practice

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

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.