The sum of the faces is going to fall in the range {2, 3, 4, …, 12}, which contains five prime numbers {2, 3, 5, 7, 11}. You can only have one sum at a time, so these outcomes are mutually exclusive. Then
Pr[sum is prime] = Pr[sum = 2] + Pr[sum = 3] + … + Pr[sum = 11]
• sum = 2 can only happen in 1 way (1 + 1)
• sum = 3 can happen in 2 ways (1 + 2, 2 + 1)
• sum = 5 can happen in 4 ways (1 + 4, 2 + 3, 3 + 2, 4 + 1)
• sum = 7 can happen in 6 ways (1 + 6, 2 + 5, …, 6 + 1)
• sum = 11 can happen in 2 ways (5 + 6, 6 + 5)
There are a total of 6² = 36 possible sums that can be rolled, each with probability 1/36 of occurring, so the probability of getting a prime sum is
(1 + 2 + 4 + 6 + 2)/36 = 15/36 = 5/12