2011 AMC 8 Problems/Problem 18

Problem

A fair $6$ sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?

$\textbf{(A) }\dfrac16\qquad\textbf{(B) }\dfrac5{12}\qquad\textbf{(C) }\dfrac12\qquad\textbf{(D) }\dfrac7{12}\qquad\textbf{(E) }\dfrac56$

Solution

There are $6\cdot6=36$ ways to roll the two dice, and 6 of them result in two of the same number. Out of the remaining $36-6=30$ ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are $30/2=15$ ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is

$\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}$

2011 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AJHSME/AMC 8 Problems and Solutions

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2011 AMC 8 Problems/Problem 18

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