Respuesta :
Answer:
90.7% probability that the flight will depart with (at least one) empty seats
Step-by-step explanation:
For each passenger, there are only two possible outcomes. Either they show up for the flight, or they do not. The probability of a passenger showing up for the flight is independent of other passengers. So 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.
[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 combinations 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]n = 338, p =0.98[/tex]
Find the probability that the flight will depart with (at least one) empty seats?
This is the probability that there are at most 334 passengers showing up for the flight.
Either they are at most 334 passengers showing up for the flight, or there are at least 335. The sum of the probabilities of these events is decimal 1. So
[tex]P(X \leq 334) + P(X \geq 335) = 1[/tex]
We want [tex]P(X \leq 334)[/tex]. So
[tex]P(X \leq 334) = 1 - P(X \geq 335)[/tex]
In which
[tex]P(X \geq 335) = P(X = 335) + P(X = 336) + P(X = 337) + P(X = 338)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 335) = C_{338,335}.(0.98)^{335}.(0.02)^{3} = 0.0587[/tex]
[tex]P(X = 336) = C_{338,336}.(0.98)^{336}.(0.02)^{2} = 0.0257[/tex]
[tex]P(X = 337) = C_{338,337}.(0.98)^{337}.(0.02)^{1} = 0.0075[/tex]
[tex]P(X = 338) = C_{338,338}.(0.98)^{338}.(0.02)^{0} = 0.0011[/tex]
[tex]P(X \geq 335) = P(X = 335) + P(X = 336) + P(X = 337) + P(X = 338) = 0.0587 + 0.0257 + 0.0075 + 0.0011 = 0.093[/tex]
[tex]P(X \leq 334) = 1 - P(X \geq 335) = 1 - 0.093 = 0.907[/tex]
90.7% probability that the flight will depart with (at least one) empty seats
