1995 AJHSME Problems/Problem 20

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Problem

Diana and Apollo each roll a standard die obtaining a number at random from $1$ to $6$. What is the probability that Diana's number is larger than Apollo's number?

$\text{(A)}\ \dfrac{1}{3} \qquad \text{(B)}\ \dfrac{5}{12} \qquad \text{(C)}\ \dfrac{4}{9} \qquad \text{(D)}\ \dfrac{17}{36} \qquad \text{(E)}\ \dfrac{1}{2}$

Solution

Note that the probability of Diana rolling a number larger than Apollo's is the same as the probability of Apollo's being more than Diana's. If we denote this common probability $D$, then $2D+P($Apollo=Diana$)=1$. Now all we need to do is find $P($Apollo=Diana$)$. There are $6(6)=36$ possibilities total, and 6 of those have Apollo=Diana, so $P($Apollo=Diana$)=\frac{6}{36}=\frac{1}{6}$. Going back to our first equation and solving for D, we get \[2D+\frac{1}{6}=1\] \[2D=\frac{5}{6}\] \[D=\frac{5}{12} \Rightarrow \mathrm{(B)}\]

See Also

1995 AJHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AJHSME/AMC 8 Problems and Solutions
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