Question

Suppose that the probability that a passenger will miss a flight is 0.0913. 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 52 passengers. ​(a) If 54 tickets are​ sold, what is the probability that 53 or 54 passengers show up for the flight resulting in an overbooked​ flight? ​(b) Suppose that 58 tickets are sold. What is the probability that a passenger will have to be​ "bumped"? ​(c) For a plane with seating capacity of 57 ​passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below 5​%?

Answer 1

Solutions:-

a) If 54 tickets are​ sold, what is the probability that 53 or 54 passengers show up for the flight resulting in an overbooked​ flight is 0.0253

P(Not missing) = 1 - 0.0913 = 0.9087

n = 54

x = 53

By applying binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x > 53) = 0.0253

b) The probability that a passenger will have to be​ "bumped" if 58 are booked is 0.48

P(Not missing) = 1 - 0.0913 = 0.9087

n = 58

x = 53

By applying binomial distributiion:-

P(x,n) = nCx*px*(1-p)(n-x)

P(x > 53) = 0.48

c) We can book a maximum of 61 tickets.

P = 0.9087

X= 57

P(x>57)0.05

n= 61

P(x>57,n=61) = 0.1862

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