1. **State the problem:** We want to find the conditional probability that neither die shows a 2 given that the sum of the two dice is greater than 8. 2. **Recall the formula for conditional probability:** $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ where $A$ is the event "neither die is a 2" and $B$ is the event "sum > 8". 3. **Calculate $P(B)$, the probability that the sum is greater than 8:** Possible sums greater than 8 are 9, 10, 11, and 12. - Sum = 9: (3,6), (4,5), (5,4), (6,3) → 4 outcomes - Sum = 10: (4,6), (5,5), (6,4) → 3 outcomes - Sum = 11: (5,6), (6,5) → 2 outcomes - Sum = 12: (6,6) → 1 outcome Total outcomes with sum > 8 = 4 + 3 + 2 + 1 = 10 Total possible outcomes when rolling two dice = $6 \times 6 = 36$ So, $$P(B) = \frac{10}{36} = \frac{5}{18}$$ 4. **Calculate $P(A \cap B)$, the probability that neither die is 2 and the sum is greater than 8:** We list all outcomes with sum > 8 and exclude those with a 2: - Sum = 9: (3,6), (4,5), (5,4), (6,3) — none have 2 → 4 outcomes - Sum = 10: (4,6), (5,5), (6,4) — none have 2 → 3 outcomes - Sum = 11: (5,6), (6,5) — none have 2 → 2 outcomes - Sum = 12: (6,6) — no 2 → 1 outcome Total valid outcomes = 4 + 3 + 2 + 1 = 10 So, $$P(A \cap B) = \frac{10}{36} = \frac{5}{18}$$ 5. **Calculate the conditional probability:** $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{5}{18}}{\frac{5}{18}} = 1$$ **Final answer:** $$\boxed{1}$$ This means that if the sum is greater than 8, it is certain that neither die is a 2.