Two unbiased dice are rolled. What is the probability that the sum of the numbers on the dice is greater than 7, but the numbers on the two dice are not the same?

Question

Accepted Answer

**Solution:**

1. Total possible outcomes when two dice are rolled = $6 	imes 6 = 36$.

2. We need to find the probability of the sum being greater than 7 and the numbers on the dice not being the same.

3. First, list all pairs where the sum is greater than 7:
   - Sum = 8: (2,6), (3,5), (4,4), (5,3), (6,2)
   - Sum = 9: (3,6), (4,5), (5,4), (6,3)
   - Sum = 10: (4,6), (5,5), (6,4)
   - Sum = 11: (5,6), (6,5)
   - Sum = 12: (6,6)

4. Now exclude pairs where the numbers are the same:
   - Remove (4,4), (5,5), and (6,6)

5. Remaining favorable outcomes:
   - Sum 8: (2,6), (3,5), (5,3), (6,2) → 4 outcomes
   - Sum 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
   - Sum 10: (4,6), (6,4) → 2 outcomes
   - Sum 11: (5,6), (6,5) → 2 outcomes

6. Total favorable outcomes = $4 + 4 + 2 + 2 = 12$.

7. Therefore, the required probability = $\frac{12}{36} = \frac{1}{3}$.

**Answer:** $\frac{1}{3}$

Answer

Answer

Answer

Answer