Two standard dice are rolled. What is the probability that the sum of the numbers on the top faces is a prime number?
(A) $\frac{5}{12}$
(B) $\frac{7}{12}$
(C) $\frac{1}{2}$
(D) $\frac{5}{6}$
(E) $\frac{1}{3}$

## Part C: Each correct answer is worth 8.
The table below shows the possible sums when two standard dice are rolled.

Each sum in bold is equal to a prime number.

Number on the First Die

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

Looking at the table above, the total number of possible outcomes is $6 \times 6=36$.

The total number of outcomes for which the sum is a prime number is 15 .

The probability that the sum of the numbers on the top faces is a prime number is $\frac{15}{36}=\frac{5}{12}$.