# 2008 AMC 10B Problems/Problem 20

## Problem

The faces of a cubical die are marked with the numbers $1$, $2$, $2$, $3$, $3$, and $4$. The faces of another die are marked with the numbers $1$, $3$, $4$, $5$, $6$, and $8$. What is the probability that the sum of the top two numbers will be $5$, $7$, or $9$? $\mathrm{(A)}\ 5/18\qquad\mathrm{(B)}\ 7/18\qquad\mathrm{(C)}\ 11/18\qquad\mathrm{(D)}\ 3/4\qquad\mathrm{(E)}\ 8/9$

## Solution 1

The easiest way is to write a table of all $36$ possible outcomes, do the sums, and count good outcomes.

     1  3  4  5  6  8
------------------
1 |  2  4  5  6  7  9
2 |  3  5  6  7  8 10
2 |  3  5  6  7  8 10
3 |  4  6  7  8  9 11
3 |  4  6  7  8  9 11
4 |  5  7  8  9 10 12


We see that out of $36$ possible outcomes $4$ give the sum of $5$, $6$ the sum of $7$, and $4$ the sum of $9$, hence the resulting probability is $\frac{4+6+4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}$.

## Solution 2

Each die is equally likely to roll odd or even, so the probability of an odd sum is $\frac{1}{2}$.

So we can find the probability of rolling $3$ or $11$ instead and just subtract that from $\frac{1}{2}$, which seems easier. Without writing out a table, we can see that there are two ways to make $3$, and two ways to make $11$, for a probability of $\frac{4}{36}$. $\frac{1}{2}-\frac{4}{36}=\frac{14}{36}=\boxed{\frac{7}{18}}$.

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