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"A airline charges $560 for a ticket to a popular destination. There are 80 seats on the plane, and the probability that any particular customer will miss their flight is 20%.

The airline overbooks the flight by 5 seats (selling a total of 85 tickets) If a customer shows up and there are no seats available, the airline must refund the full price of their ticket plus 40%

Find the expected amount the airline will pay."

I just want to make sure I'm thinking about this right.

I think E(x) could have the values 0, 1, 2, 3, 4, 5

If it's $0$, the airline pays nothing, exactly the right amount of people shows up. The chance of this happening is $binompdf(85, 0.80, 80)$

If it's $1$, the airline pays $540 + 540 * (0.4) = 756.$ The chance of this happening is $binompdf(85, 0.80, 81)$

If it's $2$, the airline pays $756 + 756 = 1512.$ The chance of this happening is $binompdf(85, 0.80, 82)$

And so on.

This all seems pretty straightforward, but when I go to calculate E(x), I come up with ~$0.052, a number that seems way too small.

Am I on the right track? Or did I make some sort of weird mistake.

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You are on the right track. It's just that the probability that more than 80 people arrive is surprisingly small, leading to a very small expectation of payback.

Remark The low result isn't an indication that your answer is wrong, the question is meant to evoke the response, "Oh, so that's why they do that!"

You are looking at the extreme tail of a large-$n$ binomial distribution which has a very low probability for moderately high $p$. This is why airlines can overbook as a standard operating procedure.

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If $80$ or fewer people show up, the airline pays nothing. (Since we are multiplying by $0$, this does not change the expectation. But one should do the computation correctly.)

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