[Combinatorics]
Problem: Tanner has two identical dice. Each die has six faces which are numbered 2, 3, 5, $7,11,13$. When Tanner rolls the two dice, what is the probability that the sum of the numbers on the top faces is a prime number?

Solution: We make a table of the 36 possible combinations of rolls and the resulting sums:

|  | 2 | 3 | 5 | 7 | 11 | 13 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 2 | 4 | 5 | 7 | 9 | 13 | 15 |
| 3 | 5 | 6 | 8 | 10 | 14 | 16 |
| 5 | 7 | 8 | 10 | 12 | 16 | 18 |
| 7 | 9 | 10 | 12 | 14 | 18 | 20 |
| 11 | 13 | 14 | 16 | 18 | 22 | 24 |
| 13 | 15 | 16 | 18 | 20 | 24 | 26 |

Of the 36 entries in the table, 6 are prime numbers (two entries each of 5, 7 and 13).

Therefore, the probability that the sum is a prime number is $\frac{6}{36}$ or $\frac{1}{6}$.

(Note that each sum is at least 4 and so must be odd to be prime. Since odd plus odd equals even, then the only possibilities that really need to be checked are even plus odd and odd plus even (that is, the first row and first column of the table).)

Final answer: $\frac{1}{6}$