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Harold and Maude plan to take a cruise together, but they live in separate cities. The cruise departs from Miami, and they each book a flight to arrive in Miami an hour before they need to be on the ship. Their travel planner explains that Harold's flight has an 80% chance of making it on time for him to get o the ship and that Maude's flight has a 90% chance of making it on time. Assume that the event of Harold making it on time is independent of the event of Maude making it on time. What is the probability that at least one of Harold and Maude will make it to the cruise

User David Kay
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1 Answer

7 votes

Answer:

98% probability that at least one of Harold and Maude will make it to the cruise

Explanation:

Independent probabilities:

When two events are independent, the probability of the two events happening simultaneously is the multiplication of each probability.

Probability that none makes it to the cruise:

Harold's flight has an 80% chance of making it, so 100 - 80 = 20% probability of missing.

Maude's flight has a 90% chance of making it on time, so 100 - 90 = 10% probability of missing.

Both missing: 0.2*0.1 = 0.02.

2% probability of both missing.

Probability that at least one makes it to the cruise:

Either both miss, or at least one makes it. The sum of the probabilities of these events is 100%. So

2 + p = 100

p = 98%

98% probability that at least one of Harold and Maude will make it to the cruise

User Ghayas
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