If the cubes are die such that each face has numbers from 1 to 6. Rolling two of them, there are 6x6 or 36 ways in which they may be arranged. For their numbers to sum to 7, the numbers should be 1 and 6, 2 and 5, 2 and 6, 3 and 4, 3 and 5, 3 and 6, 4 and 3, 4 and 4, 4 and 5, 4 and 6, 5 and 2, 5 and 3, 5 and 4, 5 and 5, 5 and 6, 6 and 1, 6 and 2, 6 and 3, 6 and 4, 6 and 5, 6 and 6. All in all there are 21 out of 36 and the probability is equal to 7/12.
I believe it’s the second & fourth one
You add 25 and 12 i think
