Question

suppose that the probability that a passanger will miss a flight is 0.0963. airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passangers must be “bumped” from the flight. suppose that an airline has a seating capacity of 50 passengers.

A) if 52 tickets are sold, what is the probability that 51 or 52 passengers show up for the flight resulting in an overbooked flight?

B) suppose that 56 tickets are sold. What is the probability that a passenger will have to be bumped?

C) for a plane with seating capacity of 53 passengers, how many tickets may be sold to keep the probability of a passenger being “bumped” below 5%?

Answer 1

suppose that the probability that a passanger will miss a flight is 0.0963

Probability of miss a flight =0.0963

A) if 52 tickets are sold i.e. n=52

p= 1-0.0963 =0.9037

So, this follows a binomial distribution.

the probability that 51 or 52 passengers show up for the flight resulting in an overbooked flight is

B) suppose that 56 tickets are sold i.e. n =56

P(bumped) is that the capacity of airline is 50 but we sold 6 ticket more. So, probability that a passenger will have to be bumped is indirectly a airline sold more than 50 tickets.

C) For a plane with seating capacity of 53 passengers and the probability of a passenger being “bumped” below 5%

P(bumped) <0.05

that is more than 53 tickets are sold and what how many tickets is

For this we chosse n may be grater than 53 i.e. it may be 54,55,56,57,....under the condition that P(bumped) <0.05 so.

n	P(X<=53)	P(X>53)
54	0.99578	0.00422
55	0.973835	0.026165
56	0.91572	0.08428
57	0.811252	0.188748

From the table we see that 55 tickets may be sold to keep the probability of a passenger being “bumped” below 5%