2004 AMC 10B Problem 11

Below is the professionally curated solution for Problem 11 of the 2004 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10B solutions, or check the answer key.

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

Difficulty rating: 1430

11.

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?

$\dfrac{1}{2}$

$\dfrac{47}{64}$

$\dfrac{3}{4}$

$\dfrac{55}{64}$

$\dfrac{7}{8}$

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.

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