2004 AMC 10B Problems/Problem 11
Contents
Problem
Two eight-sided dice each have faces numbered through . 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?
Solution 1
We have , hence if at least one of the numbers is , the sum is larger. There such possibilities.
We have .
For we already have , hence all other cases are good.
Out of the possible cases, we find that in the sum is greater than or equal to the product, hence in cases the sum is smaller, satisfying the condition. Therefore the answer is .
Solution 2
Let the two rolls be , and .
From the restriction:
Since and are non-negative integers between and , either , , or
if and only if or .
There are ordered pairs with , ordered pairs with , and ordered pair with and . So, there are ordered pairs such that .
if and only if and or equivalently and . This gives ordered pair .
So, there are a total of ordered pairs with .
Since there are a total of ordered pairs , there are ordered pairs with .
Thus, the desired probability is .
See also
2004 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AMC 10 Problems and Solutions |
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