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Expected Value of Discrete Random Variables

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  • Expected Value of Discrete Random Variables

    Question:

    A tour operator has a bus that can accommodate 20 tourists. The operator knows that tourists may not show up, so he sells 21 tickets. The probability that an individual tourist will not show up is 0.02, independent of all other tourists.
    Each ticket costs 50, and is non-refundable if a tourist fails to show up. If a tourist shows up and a seat is not available, the tour operator has to pay 100 (ticket cost + 50 penalty) to the tourist. What is the expected revenue to the tourist?

    My solution:

    The way I started off this question was letting X= number of people who don't show and R= revenue.

    R= 1050 if X=1 and 950 if X=0. So:
    E(R) = 1050*P(X=1) + 950*P(X=0)

    I thought P(X=1) and P(X=0) would be binomial and solved it that way. However the answer I came up with was different from the one in the back of the book. The answer is supposed to be 984.6. Does anybody know how to get this answer?

  • #2
    The solution is 1050- P(X=0)*100.

    You might wonder why your solution is wrong. The concept is right, but the way you calculate the probability is a bit inaccurate. It would be right if you do this:

    E(R) = 1050*P(X>0) + 950*P(X=0)
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    • #3
      Right, you need to include the possibility of more than 1 not showing up. So it's really 1050*(probability you don't have to refund anyone's ticket) + 950*(probability you do have to refund a ticket), or as sohpmalvin said, 1050*P(X>0) + 950*P(X=0).

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      • #4
        Ah, I see what I did wrong now. Thank you very much to both of you.

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        • #5
          expected value question

          Originally posted by gmalivuk View Post
          Right, you need to include the possibility of more than 1 not showing up. So it's really 1050*(probability you don't have to refund anyone's ticket) + 950*(probability you do have to refund a ticket), or as sohpmalvin said, 1050*P(X>0) + 950*P(X=0).
          Can you please explain how to find the P(X>0)?

          Thanks
          Meem

          Comment


          • #6
            It's 1 minus the probability that x=0.

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