Respuesta :
Answer:
a) 0.2135 = 21.35% probability that overbooking occurs.
b) 0.4625 = 46.25% probability that the flight has empty seats.
Step-by-step explanation:
For each booked passengers, there are only two possible outcomes. Either they arrive, or they do not. The probability of a passenger arriving 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.
We are interest in the 19-14 = 5 remaining reservations, each of whom arrive with 52% probability
This means that [tex]n = 5, p = 0.52[/tex]
a. Find the probability that overbooking occurs.
There are 17-14 = 3 seats remaining. So overbooking occurs if more than 3 arrive.
[tex]P(X > 3) = P(X = 4) + P(X = 5)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 4) = C_{5,4}.(0.52)^{4}.(0.48)^{1} = 0.1755[/tex]
[tex]P(X = 5) = C_{5,5}.(0.52)^{5}.(0.48)^{0} = 0.0380[/tex]
[tex]P(X > 3) = P(X = 4) + P(X = 5) = 0.1755 + 0.0380 = 0.2135[/tex]
21.35% probability that overbooking occurs.
b. Find the probability that the flight has empty seats.
3 seats remaining, so this is the probability that less than 3 passengers arrive.
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = 0) = C_{5,0}.(0.52)^{0}.(0.48)^{5} = 0.0255[/tex]
[tex]P(X = 1) = C_{5,1}.(0.52)^{1}.(0.48)^{4} = 0.1380[/tex]
[tex]P(X = 2) = C_{5,2}.(0.52)^{2}.(0.48)^{3} = 0.2990[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0255 + 0.1380 + 0.2990 = 0.4625[/tex]
0.4625 = 46.25% probability that the flight has empty seats.
