# Difference between revisions of "1998 AJHSME Problems/Problem 19"

## Problem

Tamika selects two different numbers at random from the set $\{ 8,9,10 \}$ and adds them. Carlos takes two different numbers at random from the set $\{3, 5, 6\}$ and multiplies them. What is the probability that Tamika's result is greater than Carlos' result? $\text{(A)}\ \dfrac{4}{9} \qquad \text{(B)}\ \dfrac{5}{9} \qquad \text{(C)}\ \dfrac{1}{2} \qquad \text{(D)}\ \dfrac{1}{3} \qquad \text{(E)}\ \dfrac{2}{3}$

## Solution

The different possible (and equally likely) values Papa gets are: $8+9=17$ $8+10=18$ $9+10=19$

The different possible (and equally likely) values Carlos gets are: $3\times5=15$ $3\times6=18$ $5\times6=30$

The probability that if Tamika had the sum $17$ her sum would be greater than Carlos's set is $\frac{1}{3}$, because $17$ is only greater than $15$

The probability that if Tamika had the sum $18$ her sum would be greater than Carlos's set is $\frac{1}{3}$, because $18$ is only greater than $15$

The probability that if Tamika had the sum $19$ her sum would be greater than Carlos's set is $\frac{2}{3}$, because $19$ is greater than both $15$ and $18$

Each sum has a $\frac{1}{3}$ possibility of being chosen, so we have $\frac{1}{3}\left(\frac{1}{3}+\frac{1}{3}+\frac{2}{3}\right)=\frac{1}{3}\left(\frac{4}{3}\right)=\frac{4}{9}=\boxed{A}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 