Question

Suppose that the probability that a passenger will miss a flight is .0946. 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 56 passengers.

​(a) If 58 tickets are​ sold, what is the probability that 57 or 58 passengers show up for the flight resulting in an overbooked​ flight?

​(b) Suppose that 62 tickets are sold. What is the probability that a passenger will have to be​ "bumped"?

​(c) For a plane with seating capacity of 51 ​passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below 5​%?

Answer 1

a) Given that 58 tickets are sold, the number of people who show up is modelled here as:

as 0.0946 is the probability of missing the flight

The probability that 57 or 58 passengers show up for the flight is computed here as:

Therefore 0.0222 is the required probability here.

b) Given that 62 tickets are sold, probability that a passenger has to be bumped is computed here as the probability that more than 56 passengers show up.

The distribution here is given as:

The probability here is computed using EXCEL as:
=1-binom.dist(56,62,0.9054,TRUE)

0.4605 is the output here.

Therefore 0.4605 is the required probability here.

c) Let the number of tickets sold here be K.

For K= 56, we have
P(X > 51) = 1 - P(X <= 51)

This is computed in EXCEL as:
=1-BINOM.DIST(51,56,0.9054,TRUE)

0.38 is the output here but we need 0.05 only.

For K = 53, we have here:
P(X > 51) = 1 - P(X <= 51)

This is computed in EXCEL as
=1-BINOM.DIST(51,53,0.9054,TRUE)
0.0337 is the output here.

For K = 54, we have here:
P(X > 51) = 1 - P(X<= 51)

This is computed in EXCEL as:
=1-BINOM.DIST(51,54,0.9054,TRUE)

0.104 is the output here which is greater than 0.05.

Therefore 53 tickets should be sold here.