# Binomial probability expected value

"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.

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.
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.)