Question

Suppose that the probability that a passenger will miss a flight is 0.09070 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 54 passengers.

​(a) If 56 tickets are​ sold, what is the probability that 55 or 56 passengers show up for the flight resulting in an overbooked​ flight?

​(b) Suppose that 60 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 10​%?

Answer 1

Ans:

a)Let x be the number of passenger who show up for flight.

Then x has binomial distribution with n=56 and p=0.9093

P(overbooked)=P(x>54)

=P(x=55)+P(x=56)

=⁵⁶C₅₅*0.9093^55*0.0907+0.9093^56

=0.0321

b)Now,x has binomial distribution with n=60 and p=0.9093

P(bumped)=P(overbooked)=P(x>54)

=1-P(x<=54)

=1-binomdist(54,60,0.9093,true)

=0.5353

c)

P(bumped)=P(overbooked)=P(x>53)=1-P(x<=53)

=1-binomdist(53,n,0.9093,true)

n	P(bumped)
54	0.0059
55	0.0347
56	0.1067
57	0.2286
58	0.3860

So,maximum number of ticket sold to keep probability of being bumped will be 55.

So,n=55