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An airline always overbooks if possible. A particular
Chapter 2, Problem 11E(choose chapter or problem)
Problem 11E
An airline always overbooks if possible. A particular plane has 95 seats on a flight in which a ticket sells for $300. The airline sells 100 such tickets for this flight.
(a) If the probability of an individual not showing up is 0.05, assuming independence, what is the probability that the airline can accommodate all the passengers who do show up?
(b) If the airline must return the $300 price plus a penalty of $400 to each passenger that cannot get on the flight, what is the expected payout (penalty plus ticket refund) that the airline will pay?
Questions & Answers
QUESTION:
Problem 11E
An airline always overbooks if possible. A particular plane has 95 seats on a flight in which a ticket sells for $300. The airline sells 100 such tickets for this flight.
(a) If the probability of an individual not showing up is 0.05, assuming independence, what is the probability that the airline can accommodate all the passengers who do show up?
(b) If the airline must return the $300 price plus a penalty of $400 to each passenger that cannot get on the flight, what is the expected payout (penalty plus ticket refund) that the airline will pay?
ANSWER:
Answer :
Step 1 of 2 :
Given, An airline always overbooks if possible. A particular plane has 95 seat on a flight in which a ticket sells for $300. The airline sells 100 such tickets for this flight.
Let X be the number of passengers do not show up
Let X follows binomial distribution
X~B(n,p)
P(x) = , x = 0, 1, 2, ……, n.
Where, n = 100 and p = 0.05
The claim is to find the probability that at least 5 passengers do not show up
P(X 5) = 1 - P(X
4)
= 1 - [ P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)]
= 1 - ( +
+
+
= 1 - 0. 436
= 0.564
the probability that at least 5 passengers do not show up is 0.564 (56.4%)