Suppose that the probability that a passenger will miss a flight is 0.0943. Airl

Question

Suppose that the probability that a passenger will miss a flight is

0.0943. 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 55 passengers. (a) If 57 tickets are sold, what is the probability that

56 or 57 passengers show up for the flight resulting in an overbooked flight? (b) Suppose that 61 tickets are sold. What is the probability that a passenger will have to be "bumped"? (c) For a plane with seating capacity of 58 passengers, how many tickets may be sold to keep the probability of a passenger being "bumped" below 55%?

Explanation / Answer

Solution:-

a) If 57 tickets are sold, what is the probability that 56 or 57 passengers show up for the flight resulting in an overbooked flight is 0.0245.

P(Not missing) = 1 - 0.0943 = 0.9057

n = 57

x = 56

By applying binomial distributiion:-

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

P(x > 56) = 0.0245

b) The probability that a passenger will have to be "bumped" if 61 are booked is 0.48.

P(Not missing) = 1 - 0.0943 = 0.9057

n = 61

x = 56

By applying binomial distributiion:-

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

P(x > 56) = 0.48

c) We can book a maximum of 64 tickets.

p = 0.9057

x = 58

P(x > 58) < 0.55

n = 64

P(x > 58, n = 64) = 0.4318

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