sTATS Find the probability that overbooking occurs. Find the probability that the flight has empty seats

Question

A small regional carrier accepted 15 reservations for a particular flight with 14 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 55% chance, independently of each other.

(Report answers accurate to 4 decimal places.)

Find the probability that overbooking occurs.   

Find the probability that the flight has empty seats

Answer 1

Hi,

This is an example of binomial probability.

a) Since 11 passengers are guaranteed to arrive, the only uncertainty comes from the remaining 4 passengers. There are only 3 seats available for these passengers, so overbooking occurs only if all 4 of them show up.

Each passenger has a 0.55 probability of arriving, independently of the others. For independent events, the probability that all of them occur is calculated by multiplying their individual probabilities:

P(all 4 arrive) = (0.55)(0.55)(0.55)(0.55) = (0.55)^4 = 0.09150625

Therefore, the probability of overbooking is 0.0915.

b) There are only three seats available for the last four passengers. Therefore, the flight will have at least one empty seat only if 0, 1, or 2 of those passengers arrive. Instead of calculating the probabilities of these three cases separately, it is easier to use the complement rule. We subtract the probability that exactly three passengers arrive (the flight is full) and the probability that all four passengers arrive (the flight is overbooked) from 1.

P(empty seats) = 1 − P(X = 3) − P(X = 4)

To calculate the probability that exactly three passengers arrive, we must consider that there is more than one possible outcome. Any one of the four passengers could be the one who does not show up, so there are four different combinations that satisfy this condition.

Each combination has the same probability: three passengers arrive (0.55 × 0.55 × 0.55) and one passenger does not (0.45). Since these four outcomes are mutually exclusive, we add their probabilities together by multiplying one of them by 4.

P(X = 3) = 4 × (0.55)^3 × (0.45) = 0.2995

In contrast, there is only one way for all four passengers to arrive, so no multiplication by 4 is needed:

P(X = 4) = (0.55)^4 = 0.0915

Then,

P(empty seats) = 1 − 0.2995 − 0.0915 = 0.6090

Note: Although the complement rule may seem unnecessary in this example, it is worth remembering because it becomes much more useful in more complex problems. As the number of possible outcomes increases, calculating the complement is often much easier than computing and adding many individual probabilities.