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Expected tries to to remove ball from urn

  1. Mar 27, 2013 #1
    I have a problem dealing with expected values. I'll jump right into the problem:

    There are 3 red and 1 black balls in an urn. What is the expected number of times a ball must be removed before a black ball is removed.

    I originally thought of this using

    E[X]=1*P(X=1) + 2*P(X=2) + ...

    I filled in P's using hyper geometric distribution. But didn't get the correct result.

    Trying a uniform for each trial I did this
    1*1/4 + 2*1/3 + 3*1/2 + 4*1/1

    I found the following as a solution from a LONG derivation of formulas:
    k(r+b+1)/(b+1) where k=1 in this case.
    From a problem setup the same but using negative hyper geometric.

    This makes no sense to me. Is there some more simple way of thinking of this?
     
    Last edited: Mar 27, 2013
  2. jcsd
  3. Mar 27, 2013 #2

    mfb

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    The probability that the ball is picked in the nth attempt is just 1/4, as all balls are equal. This allows to calculate the expected value easily.

    You would have to consider the probability that the ball was not drawn before in each summand.
     
  4. Mar 27, 2013 #3
    Sorry, but I forgot to mention this is without replacement.

    So to start the black ball probability is 1/4, but after the 1st attempt the black ball is 1 of 3 remaining, etc.
     
  5. Mar 27, 2013 #4

    mfb

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    I know.
    Sure, but the probability that you draw a second ball is just 3/4. If you multiply 1/3 by 3/4, you get 1/4 again.
     
  6. Mar 27, 2013 #5
    Maybe I see what you're saying.

    P(X=1) = 1/4
    P(X=2) implies P(X=2) AND P(X!=1) = 1/3 * (1- P(X=1))

    ?
     
  7. Mar 27, 2013 #6

    mfb

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    Right.
    It is possible to extend this for P(X=3) and P(X=4), but that is unhandy. You can get the probabilities in a direct way as well.
     
  8. Mar 29, 2013 #7

    ssd

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    b, rb, rrb and rrrb are the 4 possible cases corresponding to your X=0,1,2,3. Find the later three probabilities and hence the expectation.
     
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