Because not all airline passengers show up for their reserved seat on a flight,
ID: 3328396 • 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
a)
The Probability(P) that a passenger shows up is 1 - 0.10 = 0.90
:
We use the binomial probability formula
:
P(k successes out of n trials) = nCk * p^k * (1-p)^(n-k), where n is 125 and nCk = n! / (k! * (n-k)!)
:
Note that since the airline sold 125 tickets, then 125 passengers could show up,
so n=125
:
a) P ( X < or = 120) = 1 - P( k = 121) - P( k = 122) - P(k = 123) - P(k = 124) - P(k = 125) = 1 - 0.004 = 0.996
b) Now The Probability(P) that a passenger shows up is 1 - 0.01 = 0.99
P ( X < or = 120) = 1 - P( k = 121) - P( k = 122) - P(k = 123) - P(k = 124) - P(k = 125) = 0.999
c) P that seats will be empty:
P ( X < 120 ) = 1 - P( X > or = 120) = 1 - 0.011 = 0.989
So, overbooking strategy is not good as there is a high chance of 98.9% that one or more seat will be empty