This problem uses PIE (Principle of Inclusion and Exclusion). In other words, we have to add the two probabilities, and then subtract the probability that a roll fits both conditions (because it would be counted twice and we only want to count it once).
The probability that one dice shows a 6 is 11/36 (Notice that it is not 12/36 because you can only roll two 6s one way).
To find the probability that the sum of both dice is a prime number, we first notice what prime numbers could be rolled. The numbers that could be rolled are: 2,3,5,7, or 11. The number 2 can be rolled in 1 way, 3 can be rolled in 2 ways, 5 can be rolled in 4 ways, 7 can be rolled in 6 ways, and 11 can be rolled in 2 ways. Adding up the number of ways gives us 15, so the probability is 15/36.
There are 4 rolls that fit both conditions: (1,6), (5,6), (6,1), and (6,5). Therefore, the probability of rolling something that fits both conditions is 4/36.
The final probability is therefore: 11/36 + 15/36 - 4/36 = 32/36 = 8/9.
