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?

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