Question

Suppose that the probability that a passenger will miss a flight is 0.0905. 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 5​%?

Answer 1

Ans:

Probability of showing up for flight=1-0.0905=0.9095

Let x be the number of passengers showing up for flight.

Then x has binomial distribution with n=56 and p=0.9095

a)

P(x>54)=P(x=55)+P(x=56)

=56C₅₅*0.9095⁵⁵*(1-0.9095)¹+56C₅₆*0.9095⁵⁶*(1-0.0905)⁰

=0.0275+0.0049

=0.0324

b)Now,n=60

P(bumped)=P(x>54)=1-P(x<=54)

=1-binomdist(54,60,0.9095,true)

=0.5375

c)

P(bumped)=1-binomdist(53,n,0.9095,true)

n	P(bumped)
54	0.0060
55	0.0351
56	0.1076
57	0.2301

For n=55,probbaility of being bumped up is below 0.05.