1. **Problem Statement:** Two fair dice are rolled simultaneously. We need to find: - The probability that the sum of the numbers is an even number greater than 7. - The probability that at least one die shows a 5. - The probability that either event 1 or event 2 occurs. 2. **Total possible outcomes:** Each die has 6 faces, so total outcomes = $6 \times 6 = 36$. 3. **Event 1: Sum is an even number greater than 7.** - Even sums greater than 7 are 8, 10, and 12. - List 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 = $5 + 3 + 1 = 9$. - Probability(event 1) = $\frac{9}{36} = \frac{1}{4}$. 4. **Event 2: At least one die shows a 5.** - Outcomes where first die is 5: (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) → 6 outcomes - Outcomes where 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 = $6 + 5 = 11$. - Probability(event 2) = $\frac{11}{36}$. 5. **Event 1 and Event 2 intersection:** Outcomes where sum is even >7 and at least one die is 5. - From event 1 outcomes, those with a 5: - Sum=8: (3,5), (5,3) - Sum=10: (5,5) - Total intersection outcomes = 3. - Probability(event 1 and event 2) = $\frac{3}{36} = \frac{1}{12}$. 6. **Probability of either event 1 or event 2 occurring:** - Use formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ - $= \frac{1}{4} + \frac{11}{36} - \frac{1}{12} = \frac{9}{36} + \frac{11}{36} - \frac{3}{36} = \frac{17}{36}$. **Final answers:** - Probability(sum even >7) = $\frac{1}{4}$ - Probability(at least one 5) = $\frac{11}{36}$ - Probability(either event) = $\frac{17}{36}$