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Math Help - Help! Conditional Probability-Urn Problem

  1. #1
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    Help! Conditional Probability-Urn Problem

    The problem is stated as follows:

    You have an urn with 12 red balls, 20 green balls, and 13 yellow balls. Suppose 3 balls are drawn without replacement. What is the probability the third ball is yellow given the first ball is green.

    I am pretty sure I use Bayes formula, but Im not certain how to apply it in this case. Im really just baffled! Can anyone help me?
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  2. #2
    MHF Contributor matheagle's Avatar
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    It's just P(A|B).

     P(Y_3|G_1) ={P(Y_3 and G_1)\over P(G_1)}

     P(G_1) =20/45... the chance the first is green.

    NOW for the numerator you need order.... so hypergeometric formula isn't the best way to do it.
    first must be green, second can be anything left over and then the third must be yellow.
    Here you need to write down a few sample events.
    Last edited by matheagle; December 13th 2009 at 10:37 PM.
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  3. #3
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    I don't really understand how to use the formula you gave me but, I tried to do the problem using a probability tree, and this is what I got. Is it the same?
    Total Balls=45
    Green(G)=20
    Red(R)=12
    Yellow(Y)

    P(G,R,Y)=(20/45) X (12/44) X (13/43)=(52/1419)
    P(G,G,Y)=(20/45) X (19/44) X (13/43)=(247/4257)
    P(G,Y,Y)=(20/45) X (13/44) X (12/43)=(52/1419)
    P(1st G, 2nd (R,G,Y), 3rd Y)= (52/1419) + (247/4257) + (52/1419) = (13/99)
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  4. #4
    MHF Contributor matheagle's Avatar
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    That seems to be the correct numerator.
    BUT that's just the numerator, you asked for the conditional probability.
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