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An airline operates a flight having 50 seats. As they expect some passenger to not show up, they overbook the flight by selling 51 tickets. The probability that an individual passenger will not show up is 0.01, independent of all other tourists. Each ticket costs 10,000 and is non refundable if a tourist fails to show up. If a tourist show up and a set is not available, the airline has to pay a compensation of 100,000 to that passenger. What is the expected revenue of the airline?

My solution

Let $X$ be a random variable that represents the number of passengers that shows up to the flight and $r(X)$ the revenue of the company when a given number of passengers shows up. Then,

$$ X \sim Bin(51,0.99) $$

If the company didn't have to pay a compensation to the passenger in the case that he shows up and a seat is not available, then the company's expected revenue would be

$$E_X(r(X)) = 10,000 np = 504,900$$

But in the case that a passenger shows up and no seat is available, the company must pay 100,000 to that passenger. Note that this situation occurs only when all the 51 passengers shows, which occurs with probability $0.99^{51}=0.598956006$. Hence, the expected loss in this situation to the company would be

$$0.99^{51} (-100,000)=-59,895.6006$$

And the company's expected revenue accounting for overbooking is

$$504,900 - 59,895.6006 = 445,004.399$$

Book's answer: $450,104.4$

I don't understand where my logic is flawed. Can you help me with it?

Giiovanna
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    All $51$ passengers pay $10,000 whether or not they show up. – Xi'an Jul 11 '21 at 13:54
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    I'm not sure the "Right Answer" is actually correct--maybe a typo. \ If the nr. of arriving passengers is $0,1,2,\dots, 51,$ then the airline nets dllr = 10000*c(rep(50,51),40), so that expected gain is e.gain = sum(dllr*dbinom(0:51,51,.99)) or $$ 440,104.4$ per flight. (Revenue from reservations 41 through 50 pays for the penalty to rejected arrival 51. [Maybe $$450,104.4$ is a typo for $$ 440,104.4?]$ \ With full booking (no overbooking) the average gain would be 10000*50*.99 or $$495,000$ per flight. So the overbooking policy seems misguided. // – BruceET Jul 11 '21 at 21:32
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    Quibbles: (a) The problem has $0.001$ as chance of no show and OP attempted answer seems to use $0.99$ as chance of show. // I used the latter in my previous comment. (b) There seems to be an unstated assumption in the original Q and in OP's attempted Answer that all flights are booked at 51. // Hoping for clarifications. – BruceET Jul 11 '21 at 21:43
  • @BruceET 0.001 was a typo. The correct probability of not showing is 0.01. Regarding the book's answer, it is 450,104.4. I am not also sure that it is the right answer – Giiovanna Jul 12 '21 at 00:18
  • @Xi'an yes, I considered that in my answer. – Giiovanna Jul 12 '21 at 00:20
  • @BruceET About point (b), there is no assumption. I used the case when all the 51 passengers shows in accounting for losses because this is the only case that the company will have to pay a passenger 100,000 compensation (flight has 50 seats available, 51 passengers shows) Note that I split the problem in two: accounted for profits (E(r(X), which accounts for profits in all cases when 1,2,...,51 passengers shows) and losses (-100,000*(0.99)^51) in order to calculate the company's revenue. I tried to explain that in my attempt, but I hope it is clear now. Thanks for your time and patience. – Giiovanna Jul 12 '21 at 00:40
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    51 x 10,000 - 0.99⁵¹ x 100,000 = 450,104.4 – Xi'an Jul 12 '21 at 06:03
  • @Xi'an I see now what you tried to say on your previous comment. Thanks! – Giiovanna Jul 12 '21 at 10:30

1 Answers1

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Thanks to @Xi'an I was able to see what I was doing wrong.

The point is that all the 51 seats were sold, independent of the number of passengers showing to the flight. Hence the expected company's revenue is

51 × 10,000 - (0.99)^{51} × 100,000 = 450,104.4

Giiovanna
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