1995 AJHSME Problems/Problem 20

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 $2D+\frac{1}{6}=1$ $2D=\frac{5}{6}$ $D=\frac{5}{12} \Rightarrow \mathrm{(B)}$

1995 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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1995 AJHSME Problems/Problem 20

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