I'm not sure which method is correct (statistics)

[defetey]

Homework Statement

Pythag-Air-US Airlines has determined that
5% of its customers do not show up for
their flights. If a passenger is bumped off a
flight because of overbooking, the airline
pays the customer $200. What is the
expected payout by the airline, if it
overbooks a 240-seat airplane by 5%?

The Attempt at a Solution

The textbook says the answer is 0. I think they get that by saying that 5% of 240 is 12. So the plane is overbooked by 12. So there are 252 people booked on the plane. Then the binomial expectation is (252*0.95)=239.4. Which means they payout nothing since less than 240 people are expected to show up.

However, I don't really think that's logical...the way I would do it is:

expectation=(200)(probability that 1 is bumped off)+(400)(probability that 2 are bumped off)+...etc, up to 12. Because how could they only be expected to pay out 0 as the textbook says, when sometimes they WILL have to pay out something (there is always a chance someone will get bumped off), which makes it >0.

Just to add, for example, if they were paying a trillion dollars for every person bumped off, the expected value wouldn't be 0 intuitively...just because they have a low chance of paying that, doesn't mean it can't happen.

This is all the solutions manual says:

Since the number of passengers overbooked is equal to the expected number of passengers who
will not show up for their flights, the expected payout is $0.

 

[Mark44]

Homework Statement

The Attempt at a Solution

The textbook says the answer is 0. I think they get that by saying that 5% of 240 is 12. So the plane is overbooked by 12. So there are 252 people booked on the plane. Then the binomial expectation is (252*0.95)=239.4. Which means they payout nothing since less than 240 people are expected to show up.

This sounds eminently reasonable to me.

However, I don't really think that's logical...the way I would do it is:

expectation=(200)(probability that 1 is bumped off)+(400)(probability that 2 are bumped off)+...etc, up to 12. Because how could they only be expected to pay out 0 as the textbook says, when sometimes they WILL have to pay out something (there is always a chance someone will get bumped off), which makes it >0.

But you aren't given any information about these probabilities. All you are given is that, on average, 5% of the booked passengers don't show up.

Just to add, for example, if they were paying a trillion dollars for every person bumped off, the expected value wouldn't be 0 intuitively...just because they have a low chance of paying that, doesn't mean it can't happen.

This is all the solutions manual says:

 

[defetey]

This sounds eminently reasonable to me.
But you aren't given any information about these probabilities. All you are given is that, on average, 5% of the booked passengers don't show up.

But that's enough to find the probabilities using a binomial distribution, isn't it?

Ex, to find the probability that one passenger is overbooked:

=252C241(.95)^241(.05)^11

 

[Mark44]

I guess this will work, but it seems the long way around.

 

[defetey]

I guess this will work, but it seems the long way around.

Yea it will take a lot longer, but the answer won't be 0 for sure though. Like I said, how could the expected payout be zero when there IS a chance they will pay SOMETHING out. That's what I've done for every other question, for http://en.wikipedia.org/wiki/Expected_value#Examples"