Question

Suppose that the probability that a passenger will miss a flight is 0.09420. 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 a seating capacity of 53 ​passengers, how many tickets may be sold to keep the probability of a passenger being​ "bumped" below 5%?

Answer 1

(a)

Let X be the number of passenger miss a flight. Then X ~ Binomial(n = 60, p = 0.09420)

Probability that 59 or 60 passengers show up for the flight = P(X = 0) + P(X = 1)

= 60C0 * 0.094200 * (1 - 0.09420)60-0 + 60C1 * 0.094201 * (1 - 0.09420)60-1

=  0.002642038 + 0.019127792

= 0.02176983

(b)

Now, X ~ Binomial(n = 64, p = 0.09420)

Probability that a passenger will have to be​ "bumped" = Probability that less than 6 passengers miss the flight

= P(X < 6)

= P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

= 64C0 * 0.094200 * (1 - 0.09420)64-0 + 64C1 * 0.094201 * (1 - 0.09420)64-1 + 64C2 * 0.094202 * (1 - 0.09420)64-2 + 64C3 * 0.094203 * (1 - 0.09420)64-3  + 64C4 * 0.094204 * (1 - 0.09420)64-4 + 64C5 * 0.094205 * (1 - 0.09420)64-5

= 0.001778559 + 0.011837686 + 0.038778941 + 0.083346040 + 0.132182329 + 0.164957942

= 0.4328815

(c)

Let N be the number of tickets Sold. Let Y be the number of passengers showed up.

Probability that a passenger will not miss a flight is 1 - 0.09420 = 0.9058

Then, Y ~ Binomial(n = N, p = 0.9058)

Probability of a passenger being​ "bumped = P(Y > 53)

For N = 54, Y ~ Binomial(n = 54, p = 0.9058)

P(Y > 53) = P(Y = 54) =  54C54 * 0.905854 * (1 - 0.9058)54-54 =  0.004783492

For N = 55, Y ~ Binomial(n = 55, p = 0.9058)

P(Y > 53) = P(Y = 54) + P(Y = 55) =  55C54 * 0.905854 * (1 - 0.9058)55-54 + 55C55 * 0.905855 * (1 - 0.9058)55-55 =  0.02911616

For N = 56, Y ~ Binomial(n = 56, p = 0.9058)

P(Y > 53) = P(Y = 54) + P(Y = 55) + P(Y = 56) =  56C54 * 0.905854 * (1 - 0.9058)56-54 + 56C55 * 0.905855 * (1 - 0.9058)56-55 + 56C56 * 0.905856 * (1 - 0.9058)56-56 =  0.09214994 which is greater than 0.05

Thus, 55 tickets to be sold to keep the probability of a passenger being​ "bumped" below 5%.