Question

Because not all airline passengers show up for their reserved seat, an airline sells 125 tickets for a flight that holds only 124 passengers. The probability that a passenger does not show up is 0.13, and the passengers behave independently. Round your answers to two decimal places (e.g. 98.76).(a) What is the probability that every passenger who shows up gets a seat

Answer 1

100% probability that every passenger who shows up gets a seat

Explanation:

Since the passengers behave independently, we use the binomial probability distribution to solve this question.

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.

125 tickets:

So
`n = 125`

The probability that a passenger does not show up is 0.13.

So the probability that a passenger shows up is
`p = 1 - 0.13 = 0.87`.

What is the probability that every passenger who shows up gets a seat

This is the probability that at least one people does not show up.

Either all passengers show up, or at least one does not. The sum of the probabilities of these events is decimal 1. So

`P(X = 125) + P(X < 125) = 1`

We want P(X < 125). So

`P(X < 125) = 1 - P(X = 125)`

In which

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

`P(X = 125) = C_(125,125).(0.13)^(0).(0.87)^(125) \cong 0`

`P(X < 125) = 1 - P(X = 125) = 1-0 = 1`

100% probability that every passenger who shows up gets a seat