Question

Suppose that the probability that a passenger will miss a flight is 0.0959. 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 5353 passengers.

​(a) If 55 tickets are​ sold, what is the probability that 54 or 55 passengers show up for the flight resulting in an overbooked​ flight?

​(b) Suppose that 59 tickets are sold. What is the probability that a passenger will have to be​ "bumped"?

​(c) For a plane with seating capacity of 55 ​passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below 55​%?

Answer 1

Ans:

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

Then x has binomial distribution with n=55 and p=1-0.0959=0.9041

P(overbooked)=P(x>53)=P(x=54)+P(x=55)

=⁵⁵C₅₄*0.9041^54*0.0959+0.9041^55

=0.0267

b)Then x has binomial distribution with n=59 and p=0.9041

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

=1-BINOMDIST(53,59,0.9041,TRUE)

=0.4964

c)

P(bumped)=P(x>55)=1-P(x<=55)

  =1-BINOMDIST(55,n,0.9041,TRUE)

n	1-BINOMDIST(55,n,0.9041,TRUE)
56	0.0035
57	0.0225
58	0.0744
59	0.1705

Maximum number of tickets to be sold,n=57 to keep the probability of a passenger being​ "bumped" below 5%.