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# Difference between revisions of "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 1

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

## Solution 2

We can use simple casework to solve this problem too. There are six cases based on Apollo's Roll. Apollo Rolls a 1: Diana could roll a $2$, $3$, $4$, $5$, or $6$. Apollo Rolls a 2: Diana could roll a $3$, $4$, $5$, or $6$. Apollo Rolls a 3: Diana could roll a $4$, $5$, or $6$. Apollo Rolls a 4: Diana could roll a $5$ or $6$. Apollo Rolls a 5: Diana could roll a $6$. Apollo Rolls a 6: There are no successful outcomes. The total amount of successful cases is $5+4+3+2+1 = 15$. The total amount of possible cases is $6(6) = 36$. Therefore, the probability of Diana rolling a bigger number is $\frac{15}{36} = \frac{5}{12} \Rightarrow \mathrm{(B)}$