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Daily Airlines flies every day from Amsterdam to London. The price for a ticket on this popular route is \$75. The aircraft has a capacity of 150 passengers. Demand for tickets is greater than capacity, and tickets are sold out well in advance of flight departures. The airline company sells 160 tickets for each flight to protect itself against no-show passengers. The probability of a passenger being a no-show is q = 0.1. No-show passengers are refunded half the price of their tickets. Passengers that do show up and are not able to board the flight due to the overbooking are refunded the full amount of their tickets plus an extra \$425 compensation. What is the probability that more passengers will turn up for a flight than the aircraft has the seating capacity for? What are the expected value and standard deviation of the daily return for the airline?

My approach:

Let X equal the number of passengers who turn up for the flight. The probability that more passengers will turn up for a flight than the aircraft has the seating capacity for, will be a negative binomial distribution, meaning that P(X = k) = $ {{k-1}\choose{150}} {p^{151}} (1-p)^{(k-151)}$ for $k \geq 151$ and P = $0 $ for $k \leq 150$.

From here we find the probability P($X \geq 151$) by adding up the probabilities from the formula above, and the expected value and variance using their respective definitions.

Is this approach correct?

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  • $\begingroup$ Why a negative binomial? and not just a binomial? $\endgroup$ – vinnief Mar 16 '16 at 1:43
  • $\begingroup$ Your biggest hint that negative binomial isn't the appropriate model for this situation: what is the probability $P(X=161)$ with your formula? What's the actual probability given the problem statement? $\endgroup$ – Steven Stadnicki Mar 16 '16 at 1:44
  • $\begingroup$ @vinnief The question you linked doesn't include the part about how much the airline has to refund. $\endgroup$ – browngreen Mar 16 '16 at 1:52
  • $\begingroup$ Would the expected value be something like this: $\endgroup$ – Ben Thompson Mar 16 '16 at 2:03
  • $\begingroup$ @browngreen. true, new comment: the probabilities are calculated in this question $\endgroup$ – vinnief Mar 16 '16 at 2:28
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The number of passengers that show up has a binomial distribution with n=160, p=0.9.

To calculate the expected refund, you would add the expected refund from overbooking to the expected refund from no-shows.

The expected refund from overbooking would be: $$\sum_{k=151}^{160} P(X=k)*500(k-150) $$

The expected refund from no-shows would be $75*0.1*160=1200$

To calculate the standard deviation you would need to also calculate $E[Y^2]$ which is the sum of $E[Y^2]$ for overbooking and $E[Y^2]$ for no-shows (Y=amount refunded).

For the overbooking, $E[Y^2]$ would be: $$\sum_{k=151}^{160} P(X=k)*[500(k-150)]^2 $$ and for the no-shows, it would be $75^2*[(160*.1)(1-.1+160*.1)$

The standard deviation is $\sqrt{E[Y^2]-E[Y]^2}$

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