Question

g On a certain daily flight, Air Northeast has a policy of booking as many as 22 people on an airplane that can seat only 19. Past studies have revealed that only 89% of the booked passengers actually arrive for the flight. If the airline books 22 people on a flight, find the probability that there will not be enough seats available for all booked passengers. Show sufficient work to justify answer

Answer 1

Answer:

55.82% probability that there will not be enough seats available for all booked passengers.

Step-by-step explanation:

For each booked passenger, there are only two possible outcomes. Either they arrive for the flight, or they do not arrive. The probability of a booked passenger arriving is independent of other booked passengers. So we used 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.

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

And p is the probability of X happening.

The airline books 22 people on a flight

This means that $n = 22$

Past studies have revealed that only 89% of the booked passengers actually arrive for the flight.

This means that $p = 0.89$

Find the probability that there will not be enough seats available for all booked passengers.

The airplane seats 19, so this is the probability of more than 19 passengers arriving.

$P(X > 19) = P(X = 20) + P(X = 21) + P(X = 22)$

In which

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

$P(X = 20) = C_{22,20}.(0.89)^{20}.(0.11)^{2} = 0.2718$

$P(X = 21) = C_{22,21}.(0.89)^{21}.(0.11)^{1} = 0.2094$

$P(X = 22) = C_{22,22}.(0.89)^{22}.(0.11)^{0} = 0.0770$

$P(X > 19) = P(X = 20) + P(X = 21) + P(X = 22) = 0.2718 + 0.2094 + 0.0770 = 0.5582$

55.82% probability that there will not be enough seats available for all booked passengers.