Question

United Flight 15 from New York's JFK airport to San Francisco uses a Boeing 757-200 with 182 seats. Because some people with reservations don't show up, United can overbook by accepting more than 182 reservations. Assume that there is a .0995 probability that a passenger with a reservation will not show up for the flight. Also assume that the airline accepts 200 reservations for the 182 seats that are available.Find the probability that when 200 reservations are accepted for United Flight 15, there are more passengers showing up than there are seats available. Table A-1 can not be used and calculations with the binomial probability formula would be extremely time-consuming and tedious, so use Excel. Is the probability of overbooking small enough so that it does not happen very often, or does it seem to high so that changes must be made to make it lower?

Answer 1

Answer:

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.

Step-by-step explanation:

For each passenger, there are only two outcomes possible. Either they show up for the flight, or they do not show up. This means that we can solve this problem using binomial distribution probability concepts.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

A probability is said to be unusually low if it is lower than 5%.

For this problem, we have that:

There are 200 reservations, so $n = 200$.

A passenger consists in a passenger not showing up. There is a .0995 probability that a passenger with a reservation will not show up for the flight. So $\pi = 0.0995$.

Find the probability that when 200 reservations are accepted for United Flight 15, there are more passengers showing up than there are seats available.

X is the number of passengers that do not show up. It needs to be at least 18 for the flight not being overbooked. So we want to find $P(X < 18)$, with $\pi = 0.0995, n = 200$. We can use a binomial probability calculator, and we find that:

$P(X < 18) = 0.2927$.

There is a 29.27% probability that the flight is overbooked. This is not an unusually low probability. So it does seem too high so that changes must be made to make it lower.