Question

In: Statistics and Probability

Suppose that Teeny Weeny Airlines (TWA) operates a very small flight from Charleston to Bermuda with...

Suppose that Teeny Weeny Airlines (TWA) operates a very small flight from Charleston to Bermuda with only five seats available on the aircraft. They're interested in determining whether they should follow the lead of the larger carriers and oversell that flight to increase their revenue and minimize flights with empty seats. Assume that the event of a ticketed passenger not showing up for a flight is 0.10 and that the no-show events are independent.

A. Suppose TWA decides to sell up to 6 tickets per flight. Assuming that the distribution of passenger no-shows follows a binomial distribution, create a table showing the probability of 0, 1, 2, 3, 4, 5, and 6 no-shows on a flight when there are 6 tickets sold.

B. What is the probability that exactly 1 ticketed passenger out of 6 will not show up for a flight?

C. What is the probability that at least 1 ticketed passenger out of 6 will not show up for a flight?

D. Suppose TWA sells non-refundable tickets for $200 apiece. Any passenger who is bumped due to overselling will be guaranteed the next available seat (which it costs on average $250) at no additional cost and given vouchers for two free flights in the future (which it costs on average $400). Would you recommend that TWA pursue the overselling strategy? Why or why not?

Expert Solution

Hello Sir/ Mam

Let X be the random variable representing number of passengers not showing up for flight.

As the distribution of passenger no-shows follows binomial distribution,

(a) Hence,the required probability table is:

X	0	1	2	3	4	5	6
P(X)	0.531441	0.354294	0.098415	0.014580	0.001215	0.000054	0.000001

(b) The probability that exactly one ticketed passenger out of 6 will not show up = 35.43%

(c) The probability that atleast one ticketed passenger out of 6 will not show up = 46.86%

(d)
Benefit to TWA if atleast one of the passenger don't show up = $200

Additional cost to TWA if all of the passengers shows up = ($250-$200) + $400 = $450

	Probability	Cost/ Benefit	Expected Value
X = 0	0.531441	-$450.00	-$239.15
X > 0	0.468559	$200.00	$93.71
Total			-$145.44

The expected cost TWA will have to incur amounts to $145.44 and hence, TWA should not pursue the overselling strategy.

I hope this solves your doubt.

Do give a thumbs up if you find this helpful.

orchestra answered 1 year ago

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