1. **State the problem:** We have 71 guests with bookings involving breakfast (B), lunch (L), and supper (S). We need to find probabilities related to meal bookings based on the Venn diagram data. 2. **Identify given data from the Venn diagram:** - Only B: 24 - Only L: 9 - Only S: 16 - B and L only: 6 - B and S only: 7 - L and S only: 5 - All three (B, L, S): 3 - Total guests: 71 3. **Calculate total guests who booked all three meals:** 3 4. **Calculate guests who have not booked all three meals:** $$71 - 3 = 68$$ --- **Part a i:** Find the probability that a guest who has not booked all three meals has booked breakfast or supper. 5. Calculate number of guests who booked breakfast or supper but not all three: - Breakfast only: 24 - Supper only: 16 - B and L only: 6 (includes B) - B and S only: 7 (includes B and S) - L and S only: 5 (includes S) - Exclude all three: 3 Guests with breakfast or supper (excluding all three): $$24 + 16 + 6 + 7 + 5 = 58$$ 6. Probability: $$P = \frac{58}{68} = \frac{29}{34} \approx 0.8529$$ --- **Part a ii:** Find the probability that a guest has not booked supper given that they have booked lunch. 7. Calculate guests who booked lunch: - Only L: 9 - B and L only: 6 - L and S only: 5 - All three: 3 Total lunch bookings: $$9 + 6 + 5 + 3 = 23$$ 8. Calculate guests who booked lunch but not supper: - Only L: 9 - B and L only: 6 Total lunch but no supper: $$9 + 6 = 15$$ 9. Conditional probability: $$P(\text{no supper} | L) = \frac{15}{23} \approx 0.6522$$ --- **Part b:** Find the probability that two randomly selected guests both booked lunch given that both booked at least two meals. 10. Calculate guests who booked at least two meals: - B and L only: 6 - B and S only: 7 - L and S only: 5 - All three: 3 Total at least two meals: $$6 + 7 + 5 + 3 = 21$$ 11. Calculate guests who booked lunch and at least two meals: - B and L only: 6 - L and S only: 5 - All three: 3 Total lunch and at least two meals: $$6 + 5 + 3 = 14$$ 12. Probability a randomly selected guest booked lunch given at least two meals: $$P(L | \geq 2) = \frac{14}{21} = \frac{2}{3}$$ 13. Probability that two randomly selected guests both booked lunch given both booked at least two meals: Assuming independence: $$P = \left(\frac{2}{3}\right)^2 = \frac{4}{9} \approx 0.4444$$ --- **Final answers:** - a i: $\frac{29}{34}$ - a ii: $\frac{15}{23}$ - b: $\frac{4}{9}$