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Urn Conditional Probability Problem

  1. Dec 13, 2009 #1
    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?
    Last edited: Dec 13, 2009
  2. jcsd
  3. Dec 13, 2009 #2
    First step is to work out the probability of each combination of 3 balls. Might be easiest with the aid of a probability tree diagram.
  4. Dec 13, 2009 #3
    Im not certain I understand how to set up a probability tree for a problem like this. Im going to try to work it through. Give me a moment, I'll let you double check my results.
  5. Dec 14, 2009 #4
    Ok! I think I got it. This is what I concluded.
    Total Balls=45

    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)
    (52/1419) + (247/4257) + (52/1419) = (13/99)

    If this is incorrect please let me know! It is very important I get this question correct. Thank you for all your help.
  6. Dec 14, 2009 #5
    You're half-way there - what you've worked out is the probability that 3rd is yellow AND 1st is green. Next step uses Bayes formula.
  7. Dec 14, 2009 #6


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    Science Advisor

    I can't resist. Did you notice that the value for P(G,Y) is also 13/99?

    This is no accident. As long as you have no information about the second ball, the third ball and the second ball have exactly the same probability distributions. An easy way to get this first step would be to observe this equivalence and then just calculate P(G,Y) = (20/45)*(13/44)

    It all makes no difference to the next step, where you apply Bayes theorem to get the required conditional probability.

    Cheers -- sylas
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