1. **State the problem:** We roll a fair six-sided die twice and add the results. We want to find the probability of two events: - Event A: The sum is greater than 5. - Event B: The sum is an even number. 2. **Total possible outcomes:** Each die has 6 faces, so rolling twice gives $6 \times 6 = 36$ equally likely outcomes. 3. **Event A: Sum greater than 5** Possible sums range from 2 to 12. We count outcomes where sum $> 5$. Sums $\leq 5$ are 2, 3, 4, 5. - Sum = 2: (1,1) → 1 outcome - Sum = 3: (1,2), (2,1) → 2 outcomes - Sum = 4: (1,3), (2,2), (3,1) → 3 outcomes - Sum = 5: (1,4), (2,3), (3,2), (4,1) → 4 outcomes Total outcomes with sum $\leq 5$ = $1 + 2 + 3 + 4 = 10$ Therefore, outcomes with sum $> 5$ = $36 - 10 = 26$ Probability: $$P(A) = \frac{26}{36} = \frac{\cancel{26}}{\cancel{36}} = \frac{13}{18}$$ 4. **Event B: Sum is even** Possible sums: 2 to 12. Even sums are 2, 4, 6, 8, 10, 12. Count outcomes for each even sum: - Sum = 2: (1,1) → 1 - Sum = 4: (1,3), (2,2), (3,1) → 3 - Sum = 6: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 - Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 - Sum = 10: (4,6), (5,5), (6,4) → 3 - Sum = 12: (6,6) → 1 Total even sum outcomes = $1 + 3 + 5 + 5 + 3 + 1 = 18$ Probability: $$P(B) = \frac{18}{36} = \frac{\cancel{18}}{\cancel{36}} = \frac{1}{2}$$ **Final answers:** $$P(A) = \frac{13}{18}$$ $$P(B) = \frac{1}{2}$$