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Method of Indicators help

  1. Nov 26, 2013 #1
    1. The problem statement, all variables and given/known data
    There are 20 couples who attend a party where there is a raffle. 10 Winners are chosen in the raffle. What is the expected number of couples for which both members win.


    2. Relevant equations
    Assuming method of indicators is used to solve the problem, Ai={members of ith couple win}.


    3. The attempt at a solution
    The Expected value would be the sum of all P(Ai), or 20P(Ai), correct?
    I'm having a little trouble figuring out what P(Ai) would be though. If there are 20 couples, then there are 40 people. So the total number of outcomes for winners is 40nCr10, right? I'm not quite sure what the numerator would be in this instance however... Then I thought just going through straight numerical probability, the probability of one person winning would be 10/40, and their counterpart winning would then be 1/39, so P(Ai)=10/(40*39)? I still feel as though that is incorrect however.

    Any insight you all could provide would be very helpful! Thank you!
     
  2. jcsd
  3. Nov 27, 2013 #2

    Office_Shredder

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    If Ai is 1 if both members of the ith couple win, and 0 otherwise, then the question is asking you to compute the expected value of

    [tex] \sum_{i=1}^{20} A_i [/tex]

    The two questions then are:
    1.) Do you see why this is the case?
    2.) Do you see how to calculate the expected value of this thing?
     
  4. Nov 27, 2013 #3
    Yes, essentially that is what I was trying to write in my first sentence under "Attempt at a solution" but I don't know how to write summation notation on this site haha. My question really was how to calculate P(Ai).

    I thought it might be something divided by 40choose10, but am unsure how to find the numerator in that case. Otherwise it might be (10/40) for picking the first winner, then (1/39) for picking their counterpart in the couple, e.g. P(Ai)=10/(39*40). I'm very skeptical of that answer for several reasons, the most important of which being that the expected value would only end up being .13 couples or somewhere thereabouts.
     
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