Question

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 83% chance of making it on time for him to get to the ship and that Maude's flight has a 93% chance of making it on time.

a)Whether realistic or not, assuming that the probabilities are independent, what is the probability that one of the two will be cruising alone?

Answer 1

Solution:-

Given that

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 83% chance of making it on time for him to get to the ship and that Maude's flight has a 93% chance of making it on time.

a)Whether realistic or not, assuming that the probabilities are independent, what is the probability that one of the two will be cruising alone?

If we assume the two probabilities are independent than the probability that the two people both arrive on time can be calculated using the multiplicative rule:

= P(1st person on time) . P(2nd person on time)

= 0.83 (0.93)

= 0.772

The probability that will both arrive on time is 0.772

The probability that only one of the people will make it on time is calculated by using the addition rule then subtracting the probability that both people will make it on time, which is:

= (0.83 + 0.93 - 0.772) - 0.772

= 0.988 - 0.772

= 0.216

The probability that one of the two will be cruising alone is 0.216

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