Question

Suppose that the probability that a passenger will miss a flight is 0.0948. 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 59 passengers. ​

a) If 61 tickets are​ sold, what is the probability that 60 or 61 passengers show up for the flight resulting in an overbooked​ flight? ​

(b) Suppose that 65 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:

a)

Use binomial distribution with n=61 and p=1-0.0948=0.9052

P(x>=60)=P(x=60)+P(x=61)

=61C₆₀*0.9052^60*0.0948+0.9052^61

=0.0170

b)

Use binomial distribution with n=65 and p=1-0.0948=0.9052

P(x>=60)=1-P(x<=59)

=1-binomdist(59,65,0.9052,true)

=0.4109

c)

P(bumped)=P(x>=54)=1-P(x<=53)=1-binomdist(53,n,0.9052,true)

n	P(x>=54)
55	0.028242
56	0.089838
57	0.198837
58	0.346084