Answer:
There is a 8.74% probability that if Air America books 15 passengers and not enough seats will be available.
Step-by-step explanation:
For each passenger, there are only two possible outcomes. Either they arrive on time for their flight, or they do not. This means that we can solve this problem using the binomial probability distribution.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinatios of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
In this problem we have that:
[tex]p = 0.85, n = 15[/tex].
What is the probability that if Air America books 15 passengers and not enough seats will be available?
This is P(X = 15). So:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 15) = C_{15,15}.(0.85)^{15}.(0.15)^{0} = 0.0874[/tex]
There is a 8.74% probability that if Air America books 15 passengers and not enough seats will be available.
