Question

Suppose that the probability that a passenger will miss a flight is 0.0911. 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 58 passengers. ​(a) If 60 tickets are​ sold, what is the probability that 59 or 60 passengers show up for the flight resulting in an overbooked​ flight? ​(b) Suppose that 64 tickets are sold. What is the probability that a passenger will have to be​ "bumped"? ​(c) For a plane with seating capacity of 51 ​passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below 5​%?

Answer 1

Solution:-

a) The probability that 59 or 60 passengers show up for the flight resulting in an overbooked​ flight is 0.02275

P(Showing up) = 1 - 0.0911 = 0.9089, n = 60

x = 59

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x > 59) = P(x = 59) + P(x = 60)

P(x > 59) = 0.02275

b) The probability that a passenger will have to be​ "bumped" is 0.4667.

P(Showing up) = 1 - 0.0911 = 0.9089, n = 64

x = 58

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

P(x > 58) = 0.4667

c) Number of tickets to be booked for plane with seating capacity of 51 ​passengers, so that the probability of a passenger being​ "bumped" below 5​% is 53 or less.

P(Showing up) = 1 - 0.0911 = 0.9089

x = 51, n = ?

P(x > 51) < 0.05

By applying binomial distribution:-

P(x,n) = ⁿC_x*p^x*(1-p)^(n-x)

n = 53