Question

An airline knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their policy is to sell 52 tickets for a flight that can hold only 50 passengers. What is the probability that there will be a seat available for every passenger who shows up?

Answer 1

Answer:

the probability that there will be a seat available for every passenger who shows up is 0.74

Step-by-step explanation:

if the probability of choosing a passenger that will not show up is 5% , then the probability of choosing a passenger that will show up is 95%.

Denoting event X= x passengers will show up from the total of 52 that purchased the ticket

Then P(X) follows a binomial probability distribution, since each passenger is independent from others. Thus

P(X) = n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of passengers hat purchased the ticket= 52

p= probability of choosing a passenger that will show up

x = number of passengers that will show up

therefore the probability that the number does not exceed the limit of 50 passengers is:

P(X≤50)=  ∑P(X=i)  from i=0 to i=50   =F(50)

where F(x) is the cumulative binomial probability distribution, then from tables:

P(X≤50)= F(50) = 0.74

therefore the probability that there will be a seat available for every passenger who shows up is 0.74