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Math Help - [SOLVED] Total probability and Baye's rule

  1. #1
    Senior Member Danneedshelp's Avatar
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    [SOLVED] Total probability and Baye's rule

    Q: Five identical bowls are labeled 1, 2, 3, 4, and 5. Bowl i contains i white balls and 5-i black balls, with i=1, 2,...,5. A bowl is randomly selected and two balls are randomly selected (without replacement) from the contents of the bowl.

    a) What is the probability that both balls selected are white?

    b) Given that both balls selected are white, what is the probability that bowl 3 was selected?


    EDIT: I got it, sorry for posting this.
    Last edited by Danneedshelp; September 24th 2009 at 02:02 PM.
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  2. #2
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    I'm new here but...

    Hey I'm new here, but I would like to hear the answer to this one.

    I'm sure I'm doing this wrong. Could someone correct me?

    All bowls have a 1/5 chance of being selected. In order to select two white balls bowl 2, 3, 4 or 5 must be selected (since bowl 1 only has 1 white ball).

    So the chances of getting 2 white balls should be:
    1/5 * (the chance of picking 2 white balls from bowl 2) + 1/5 * (the chance of picking 2 white balls from bowl 3 + .... + 1/5 * (the chance of picking 2 white balls from bowl 5)

    If bowl 2 is selected there is a 2/5 chance of getting a white ball and then a 1/4 chance of getting another one. So the chance of selecting bowl 2, and getting two white balls out of it is 1/5 * (2/5 * 1/4) = 1/50. Applying this same logic to the other bowls gives me:


    {1 \over 5} \left ({2 \over 5} * {1 \over 4} \right) + {1 \over 5} \left ({3 \over 5} * {2 \over 4} \right ) + {1 \over 5} \left ({4 \over 5} * {3 \over 4} \right ) + {1 \over 5} \left ({5 \over 5} * {4 \over 4} \right )

    = {1 \over 50} + {3 \over 50} + {3 \over 25} + {1 \over 5}
    = 2 \over 5

    So there is a 40% chance of selecting 2 white balls, right?

    So to answer B) I think it would be that if we selected 2 white balls, we must have selected either bowl 2, 3, 4 or 5. Now we only have 4 possible bowls that were selected. So I believe our equation would end up being

    {{1 \over 4} \left (P(2 white balls from bowl 3) \right )<br />
\over <br />
{1 \over 4} \left (P(2 white balls from bowl 2) \right) +... + {1 \over 4} \left(P(2 white balls from bowl 5) \right)}

    Therefore, we end up with:
    {{1 \over 4} \left({3 \over 5} * {2 \over 4} \right)<br />
\over<br />
{1 \over 4} \left({2 \over 5} * {1 \over 4} \right) + {1 \over 4} \left ({3 \over 5} * {2 \over 4} \right) + {1 \over 4} \left({4 \over 5} * {3 \over 4} \right) + {1 \over 4} * \left({5 \over 5} * {4 \over 4} \right)}

    = {{3 \over 40} \over {1 \over 40} + {3 \over 40} + {3 \over 20} + {1 \over 4}}<br />
= {{3 \over 40} \over {2 \over 4}}<br />
= {3 \over 20}

    So that should be a 15% chance of having gotten the 2 white balls from bowl 3. Right?

    Is that at all correct?
    Last edited by sadmath; September 26th 2009 at 02:12 PM. Reason: prettied it up with math tags
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