Question

Suppose that the probability that a passenger will miss a flight is 0.0995. 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 56 passengers. ​

(a) If 58 tickets are​ sold, what is the probability that 57 or 58 passengers show up for the flight resulting in an overbooked​ flight? Round to 4 decimal spaces.

​(b) Suppose that 62 tickets are sold. What is the probability that a passenger will have to be​ "bumped"? ​(Round to four decimal places as​ needed.)

Answer 1

(a) The probability that a passenger will board the flight = 1 - 0.0995 = 0.9005.

Let X be the random variable denoting the number of passengers showed up out of 58 passengers.

Thus, X ~ Bin(58, 0.9005).

The probability that the flight be overbooked i.e. 57 or 58 passengers show up for the flight = P(X $\geq$ 57) = = 0.0170. (Ans).

58 58 Σ ( ''o ) (0.9005)"(1-0.9005)58-1 57

(b) Let Y be the random variable denoting the number of passengers turned up out of 62 passengers. A passenger will be bumped if the number of passengers showing up is more than 56.

Hence, the probability that a passenger will have to be bumped = P(Y $\geq$ 57) = = 0.4093. (Ans).

6262 (0.9005)1-0.9005)2-