Conditional Probability Dice 21872A

1. **State the problem:** We want to find the conditional probability that neither die shows a five given that the sum of the two dice is greater than 7. 2. **Define events:** - Let $A$ be the event that neither die is a five. - Let $B$ be the event that the sum of the dice is greater than 7. 3. **Formula for conditional probability:** $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ 4. **Calculate $P(B)$:** - Total possible outcomes when rolling two dice: $6 \times 6 = 36$. - Outcomes where sum $> 7$ are sums 8, 9, 10, 11, 12. - Count outcomes for each sum: - Sum 8: $(2,6),(3,5),(4,4),(5,3),(6,2)$ → 5 outcomes - 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 $>7$: $$5 + 4 + 3 + 2 + 1 = 15$$ - So, $$P(B) = \frac{15}{36} = \frac{5}{12}$$ 5. **Calculate $P(A \cap B)$:** - $A \cap B$ means sum $>7$ and neither die is 5. - Remove outcomes with any 5 from the 15 outcomes: - Sum 8 without 5: $(2,6),(4,4),(6,2)$ → 3 outcomes - Sum 9 without 5: $(3,6),(6,3)$ → 2 outcomes - Sum 10 without 5: $(4,6),(6,4)$ → 2 outcomes - Sum 11 without 5: none (both have 5) - Sum 12 without 5: $(6,6)$ → 1 outcome - Total outcomes in $A \cap B$: $$3 + 2 + 2 + 0 + 1 = 8$$ - So, $$P(A \cap B) = \frac{8}{36} = \frac{2}{9}$$ 6. **Calculate conditional probability:** $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{2}{9}}{\frac{5}{12}} = \frac{2}{9} \times \frac{12}{5} = \frac{24}{45}$$ 7. **Simplify fraction:** $$\frac{24}{45} = \frac{8}{15}$$ **Final answer:** $$\boxed{\frac{8}{15}}$$