Respuesta :
Answer:
There is a 34.09% probability that there are more passengers showing up for the flight than seats are available.
Step-by-step explanation:
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}.\pi^{x}.(1-\pi)^{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 [tex]\pi[/tex] is the probability of X happening.
For this problem,
There are 560 tickets, so [tex]n = 560[/tex]
There is a 1% probability that a passenger will not show up for the flight, so [tex]\pi = 0.01[/tex]
How likely is it that there are more passengers showing up for the flight than seats are available?
An airline has sold 560 tickets for a certain flight (with capacity 555 seats).
So, this question is: What is the probability that at most 4 passengers do not show up for the flight.
[tex]P = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
[tex]P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}[/tex]
[tex]P(X = 0) = C_{560,0}.(0.01)^{0}.(0.99)^{560} = 0.0036[/tex]
[tex]P(X = 1) = C_{560,1}.(0.01)^{1}.(0.99)^{559} = 0.0203[/tex]
[tex]P(X = 2) = C_{560,2}.(0.01)^{2}.(0.99)^{558} = 0.0574[/tex]
[tex]P(X = 3) = C_{560,3}.(0.01)^{3}.(0.99)^{557} = 0.1079[/tex]
[tex]P(X = 3) = C_{560,4}.(0.01)^{4}.(0.99)^{556} = 0.1517[/tex]
So
[tex]P = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0036 + 0.0203 + 0.0574 + 0.1079 + 0.1517 = 0.3409[/tex]
There is a 34.09% probability that there are more passengers showing up for the flight than seats are available.
