1. **Problem Statement:** Two fair dice are rolled simultaneously. Find the probability that: 1. The sum of the numbers is an even number greater than 7. 2. At least one die shows a 5. 3. Compute the final probability of either event 1 or event 2 occurring. 2. **Step 1: Total possible outcomes** Since each die has 6 faces, total outcomes = $6 \times 6 = 36$. 3. **Step 2: Event 1 - Sum is an even number greater than 7** Possible sums greater than 7 are 8, 9, 10, 11, 12. Among these, even sums are 8, 10, 12. List favorable outcomes for each sum: - Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2) - Sum = 10: (4,6), (5,5), (6,4) - Sum = 12: (6,6) Total favorable outcomes for event 1 = $5 + 3 + 1 = 9$. 4. **Step 3: Event 2 - At least one die shows a 5** Calculate total outcomes with at least one 5: - First die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6 outcomes - Second die is 5: (1,5), (2,5), (3,5), (4,5), (6,5) → 5 outcomes (excluding (5,5) counted before) Total favorable outcomes for event 2 = $6 + 5 = 11$. 5. **Step 4: Compute probability of event 1 or event 2 (union)** Use formula: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Find intersection outcomes (both events): sum even >7 and at least one 5. From event 1 outcomes, those with at least one 5: - Sum 8: (3,5), (5,3) - Sum 10: (5,5) Total intersection outcomes = 3. 6. **Step 5: Calculate probabilities** $$P(\text{event 1}) = \frac{9}{36} = \frac{1}{4}$$ $$P(\text{event 2}) = \frac{11}{36}$$ $$P(\text{event 1} \cap \text{event 2}) = \frac{3}{36} = \frac{1}{12}$$ Therefore, $$P(\text{event 1} \cup \text{event 2}) = \frac{1}{4} + \frac{11}{36} - \frac{1}{12} = \frac{9}{36} + \frac{11}{36} - \frac{3}{36} = \frac{17}{36}$$ 7. **Final answers:** 1. Probability sum is even and >7: $\frac{1}{4}$ 2. Probability at least one die shows 5: $\frac{11}{36}$ 3. Probability either event occurs: $\frac{17}{36}$