In a simultaneous throw of a pair of dice, what is the probability of getting a total more than 8?

Question: In a simultaneous throw of a pair of dice, what is the probability of getting a total more than 8?

Solution:
The total number of possible outcomes for the pair of dice is the product of the outcomes for each die is-
= 6 × 6 = 36

And,
The favorable outcomes are the combinations where the sum of the two dice is more than 8. These sums can be 9, 10, 11, or 12. The combinations for each sum are-
Sum of 9: (3, 6), (4, 5), (5, 4), (6, 3) - 4 outcomes
Sum of 10: (4, 6), (5, 5), (6, 4) - 3 outcomes
Sum of 11: (5, 6), (6, 5) - 2 outcomes
Sum of 12: (6, 6) - 1 outcome

∴ The total number of favorable outcomes is the sum of these outcomes = 4 + 3 + 2 + 1 = 10

∴ P(sum > 8) = Favorable Outcomes/Total Outcomes = 10/36 = 5/18