Question

Suppose that the probability that a passenger will miss a flight is 0.0917. 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 51 passengers.

​(a) If 53 tickets are​ sold, what is the probability that 52 or 53 passengers show up for the flight resulting in an overbooked​ flight?

​(b) Suppose that 57 tickets are sold. What is the probability that a passenger will have to be​ "bumped"?

​(c) For a plane with seating capacity of 61 passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below 5​%?

Answer 1

a)

Probability that a passenger shows up for flight,p=1-0.0917=0.9083

n = 53, p = 0.9083, P(y > 51) = ?;

Given data indicates that this is a Binomial distribution;

P(Y=y) = C(n, y)p^(y).q^(n-y)

P(y52 or 52) = P(y = 52) + P(y = 53)
                = 0.033 + 0.006
                = 0.039

b)

n=57,p=0.9083

P(overbooked or bumped)=P(y > 51) = P(y = 52) + P(y = 53) + P(y = 54) + P(y = 55) + P(y = 56) + P(y = 57)
               = 0.183 + 0.171 + 0.125 + 0.068 + 0.024 + 0.004
                 = 0.575

c)

c)P(overbooked or bumped)=1-binomcdf(n,0.9042,56)

n	1-binomcdf(n,0.9038,61)
48	0.002
49	0.06
50	0.013

So,49 tickets may be sold to keep the probability of a passenger being? "bumped" below 5%