In: Statistics and Probability
Suppose that the probability that a passenger will miss a flight is 0.0976. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be "bumped" from the flight. Suppose that an airplane has a seating capacity of 59 passengers.
(a) If 61 tickets are sold, what is the probability that 60 or 61 passengers show up for the flight resulting in an overbooked flight?
(b) Suppose that 65 tickets are sold. What is the probability that a passenger will have to be "bumped"?
(c) For a plane with seating capacity of 58 passengers, how many tickets may be sold to keep the probability of a passenger being "bumped" below 5%?
a) The number of passengers who would show up for the flight is modelled here as binomial distribution given as:
The required probability here is computed as:
P(X = 60) + P(X = 61)
This is computed in EXCEL as:
=binom.dist(60,61,0.9024,FALSE)+binom.dist(61,61,0.9024,FALSE)
0.0145 is the output here.
Therefore 0.0145 is the required probability here.
b) Probability that passenger will be bumped here is computed as:
P(X >= 60) = 1 - P(X <= 59)
This is computed for n = 65 in EXCEL here as:
=1-binom.dist(59,65,0.9024,TRUE)
0.3817 is the output here.
Therefore 0.3817 is the required probability here.
c) This is sold using hit and trial method here.
For n = 62, we have here:
P(X <= 58)
This is computed as:
=binom.dist(58,62,0.9024,TRUE)
the output here is 0.8666 but we can take it to 0.95 as we want only 0.05 probability of a passenger being bumped.
For n = 60
P(X <= 58)
This is computed as:
=binom.dist(58,60,0.9024,TRUE)
0.9842 is the output here.
For n = 61,
This is computed as:
=binom.dist(58,61,0.9024,TRUE)
The output here is 0.9448 < 0.95
Therefore n = 60 is the required number of tickets which should be sold here.