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Thread: Problem in Bayes' Theorem/ Conditional Probability

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    Problem in Bayes' Theorem/ Conditional Probability




    Hi everyone ! I'm new to this site and this is my first post.
    This is a problem I found in my Stats textbook under the Bayes Theorem section.
    Here's the problem:
    Three urns of the same appearance have the following proportion of balls.
    First urn:
    2 Black 1 White
    Second urn:
    1 Black 2 White
    Third Urn:
    2 Black 2 White
    One of the urns is selected and one ball is drawn. It turns out to be white. What is the probability of drawing a white ball again, the first one not having been returned ?

    Here's my attempt at the solution:
    Let A be the event of selecting a white ball.
    Let C be the event of selecting the second one.
    Let E_i be the event of selecting urn i = 1,2,3
    Now I suppose that that the white ball is already drawn from urn 1. So the probability of getting a second white ball given the first one is already drawn is:
    P(C | A \wedge E_1) = \frac{1}{3}\cdot0 + \frac{1}{3}\cdot\frac{2}{3} + \frac{1}{3}\cdot\frac{1}{2} = \frac{7}{18}

    Similarly selecting another white ball when a white ball was already selected from urn 2 is:
    P(C | A \wedge E_2) = \frac{1}{3}\cdot\frac{1}{3} + \frac{1}{3}\cdot\frac{1}{2} + \frac{1}{3}\cdot\frac{1}{2} = \frac{4}{9}

    Again, the prob of getting a white ball given a white was selected from third urn is:
    P(C| A \wedge E_3) = \frac{4}{9}

    I don't know how to proceed from here. I thought I would just add up the probabilities, but the answer I got was \frac{23}{18}.

    The answer given in the textbook is \frac{1}{3}.

    Can anyone please give me suggestions on how to approach/solve this problem ?
    Also how do I approach such problems in the future ?

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    Re: Problem in Bayes' Theorem/ Conditional Probability


    First thing to note is that in the 2nd draw, we will always use the same urn as the 1st draw, instead of independently select the urn again.

    Next by Law of Total Probability,
    P(C|A) = \sum_{i=1}^3 P(C|E_i \cap A)P(E_i|A)

    The crucial fact here is that P(E_i|A) maybe no longer equal to \frac {1} {3}; and it require you to apply the Baye's Theorem to calculate them.
    (That's why this question is put in this section.)

    You should be able to figure out the remaining part - you already know P(C|E_i \cap A)

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