1. **Stating the problem:** Given probabilities about piers and destinations on Coconut Island, we find various probabilities for one or two travelers, and then calculate the number of ferries needed. **Given:** - $P(A) = 0.6$, so $P(B) = 1 - 0.6 = 0.4$ - Destinations: $P(P) = 0.2$, $P(Q) = 0.35$, so $P(R) = 1 - (0.2 + 0.35) = 0.45$ --- **(a) Single traveler:** (i) Probability he chooses pier B: $$P(B) = 0.4$$ (ii) Probability island Q is *not* his destination: $$P(\text{not } Q) = 1 - P(Q) = 1 - 0.35 = 0.65$$ (iii) Probability he chooses pier A *and* destination is island R: Assuming independence of pier choice and destination: $$P(A \cap R) = P(A) \times P(R) = 0.6 \times 0.45 = 0.27$$ --- **(b) Two travelers:** For both travelers, pier choices and destinations are independent and identically distributed. (i) Probability their destinations are different: First, calculate probability both have the same destination: $$P(\text{same dest}) = P(P)^2 + P(Q)^2 + P(R)^2 = (0.2)^2 + (0.35)^2 + (0.45)^2 = 0.04 + 0.1225 + 0.2025 = 0.365$$ So, $$P(\text{different dest}) = 1 - 0.365 = 0.635$$ (ii) Probability they choose the same pier but different destinations: Probability same pier: $$P(\text{same pier}) = P(A)^2 + P(B)^2 = 0.6^2 + 0.4^2 = 0.36 + 0.16 = 0.52$$ Probability different destinations (from above) = $0.635$ Assuming independence between pier and destination, $$P(\text{same pier and different dest}) = P(\text{same pier}) \times P(\text{different dest}) = 0.52 \times 0.635 = 0.3302$$ (iii) Probability they choose different piers and none of their destinations is island Q: Probability different piers: $$P(\text{different pier}) = 1 - P(\text{same pier}) = 1 - 0.52 = 0.48$$ Probability destination is *not* Q: $0.65$ (from part a(ii)) Both travelers' destinations not Q: $0.65^2 = 0.4225$ Assuming independence, $$P(\text{different pier and both not Q}) = P(\text{different pier}) \times 0.4225 = 0.48 \times 0.4225 = 0.2028$$ --- **(c) Ferries needed from pier A to island R:** Expected number of passengers choosing pier A and traveling to R per day: $$5000 \times P(A \cap R) = 5000 \times 0.27 = 1350$$ Each ferry capacity: 60 passengers Minimum number of ferries required: $$\lceil \frac{1350}{60} \rceil = \lceil 22.5 \rceil = 23$$ **Final answers:** - (a)(i) $0.4$ - (a)(ii) $0.65$ - (a)(iii) $0.27$ - (b)(i) $0.635$ - (b)(ii) $0.3302$ - (b)(iii) $0.2028$ - (c) Minimum ferries needed from pier A to island R = $23$