# 1992 AJHSME Problems/Problem 23

## Problem

If two dice are tossed, the probability that the product of the numbers showing on the tops of the dice is greater than 10 is $\text{(A)}\ \frac{3}{7}\qquad\text{(B)}\ \frac{17}{36}\qquad\text{(C)}\ \frac{1}{2}\qquad\text{(D)}\ \frac{5}{8}\qquad\text{(E)}\ \frac{11}{12}$

## Solution

When the first dice is $1$, there are no possibilities for the second dice.

When the first dice is $2$, the second dice can be $6$, giving one possibility.

When the first dice is $3$, the second dice can be $4,5,6$, giving three possibilities.

When the first dice is $4$, the second dice can be $3,4,5,6$, giving four possibilities.

When the first dice is $5$, the second dice can be $3,4,5,6$, giving four possibilities.

When the first dice is $6$, the second dice can be $2,3,4,5,6$, giving five possibilities.

The total number of ways for the product to greater than 10 is $1+3+4+4+5=17$ and the total number of possibilities is $(6)(6)=36$, yielding a probability of $\boxed{\text{(B)}\ \frac{17}{36}}$.

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