Question

10) The overbooking problem: A plane has a capacity of 150 passengers. The airline, which knows that the industry standard is that one person out of 12 is a no-show at the airport, sells 160 tickets. a) What is the probability that all the passengers that show up at departure will be accommodated? b) What is the maximum number of tickets that the airline should sell so that they should be able to accommodate all the passengers with probability at least equal to 90%?

Answer 1

a)

Let X is a random variable shows the number of people show up of n. Here X has binomial distribution with parameter n=160 and p= 1 -(1/12) = 11/12

Here we need to find the probability that number of passengers show up is less than equal to 150. The required probability is

(2)

Let X is a random variable shows the number of people show up of n. Here X has binomial distribution with parameter n and p= 11/12.

Here we need to find n such that

Following table shows the values of above expression for various n:

n	P(X<=150)
151	0.999998032
152	0.999973274
153	0.999816471
154	0.999150056
155	0.997011977
156	0.991488606
157	0.979521303
158	0.957153842
159	0.920340729
160	0.866143647
161	0.793880871

From above table, for n= 159 airline can accommodate all the passengers with probability at least equal to 90%.

Excel function used to find the probabilities: "=BINOMDIST(150,n,11/12,TRUE)"