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Math Help - Marbles in bag

  1. #1
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    Marbles in bag

    In a bag, there is either a black or white marble: equal probability.
    Without looking, you insert a white marble into this bag.
    Then you randomly pick a marble from this bag and it turns out to be white.
    What is the probability that the marble left in the bag is black?

    I'm stuck: either 1/3 or 1/4; what is it and why? Thanks.

    Re 1/4:
    half the time, there is a black in the bag
    half that time, a white will ne pulled
    so 1/4
    ?????????????
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Marbles in bag

    Not quite - your reasoning only goes half way. Yes, the probability of there being a black in the bag and you pulling a white is 1/4. But there is also a chance that there was white on the bag (1/2) and you pull white (1), for a total of 1/2. Hence 3/4 of the time you will pull a white marble: 1 out of 4 times when there was a black in the bag and 2 out of 4 times when there is a white. Hence having pulled a white, the probability that there was a white to start is (2/4)/(2/4+1/4) = 2/3.

    Alternatively you can use Bayes' Theorem. If we let P(W_1) = probability that the original ball in the bag is white, and P(W_2) = probability that the ball drawn from the bag is white, then the probability that the original ball is white given that you drew a white from the bag is:

    $ P(W_1|W_2) = \frac {P(W_2|W_1)P(W_1)}{P(W_2)} = \frac {(1)(1/2)}{(3/4)} = \frac 2 3 $
    Thanks from mash and Wilmer
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    Re: Marbles in bag

    odd, no thanks button appears for your post.
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  4. #4
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    Re: Marbles in bag

    Gotcha! Thanks.
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