Because not all airline passengers show up for their reserved seat on a flight,
ID: 2924783 • Letter: B
Question
Because not all airline passengers show up for their reserved seat on a flight, an
airline sells 125 tickets for a flight that holds only 120 passengers. The airline estimates that the
probability that an individual passenger does not show up is 0.10 (assuming that each individual
passenger behaves independently and this probability is the same for each passenger).
a) What is the probability that there is a seat for every passenger who shows up for the flight?
b) The probability that a passenger does not show up is now estimated to be 0.01 (under the
same assumptions as above). What would be the probability from part (a) under these conditions?
c) What would be your recommendations regarding this airline’s “overbooking” strategy for the
scenarios in parts (a) and (b)? Explain.
Explanation / Answer
Ans:
Let the random variable X denote the number of passengers with tickets who do not show up. (X=number of passengers that don’t show up out of 125.)
There are 0,1,2,3,… ,n=125 possible passengers
The probability that a passenger does not show is 0.1 (10% do not show), or p = 0.1
Notice we have a binomial distribution with parameters p = 0.1 and n = 125 X~Bin(n = 125, p = 0.1)
a)
With 120 seats available, we are looking for the probability that 5 or more passengers do not show to take the flight. That is,
P(X>=5)=1-P(X<=4)=1-BINOMDIST(4,125,0.1,TRUE)=1-0.0039=0.9961
b)now,p=0.01
P(X>=5)=1-BINOMDIST(4,125,0.01,TRUE)=1-0.9913=0.0087
c)if the probability of not showing up is very low i.e probability of showing up is high,then ovrbooking should not be done,as the probability that each passenger will get seat is very low i.e. 0.0087