Question

Find the indicated probability. Round to three decimal places. An airline estimates that 91% of people booked on their flights actually show up. If the airline books 76 people on a flight for which the maximum number is 74, what is the probability that the number of people who show up will exceed the capacity of the plane?

Answer 1

0.66% probability that the number of people who show up will exceed the capacity of the plane

Explanation:

For each person booked, there are only two possible outcomes. Either they show up, or they do not. The probability of a person not showing up is independent from other people. So we use the binomial probability distribution to solve this problem.

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).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

An airline estimates that 91% of people booked on their flights actually show up.

This means that
`p = 0.91`

If the airline books 76 people on a flight for which the maximum number is 74, what is the probability that the number of people who show up will exceed the capacity of the plane?

This is
`P(X > 74)` when
`n = 76`.

`P(X > 74) = P(X = 75) + P(X = 76)`

So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 75) = C_(76,75).(0.91)^(75).(0.09)^(1) = 0.0058`

`P(X = 76) = C_(76,76).(0.91)^(76).(0.09)^(0) = 0.0008`

`P(X > 74) = P(X = 75) + P(X = 76) = 0.0058 + 0.0008 = 0.0066`

0.66% probability that the number of people who show up will exceed the capacity of the plane