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Math Help - Conditional probability

  1. #1
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    Conditional probability

    Suppose you have an urn with 2 yellow balls, 3 red balls, 6 white balls, and 12 blue balls. You randomly draw 4 balls from the urn without replacement. Given that at least one of the balls is yellow, what is the probability that all 4 balls will be different colors?

    I know that you have to use Bayes' rule, with P(B|A) = 1 because if you have 4 colors, one of them must be yellow. But I'm not sure about the different probabilities.

    I think P(A) might be 2 choose 1 + 3 choose 1 + 6 choose 1 + 12 choose 1 all divided by 23 choose 4, but I"m not sure. And I don't know how to get P(>= 1 yellow).Any one have any idea?

    Thanks!
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  2. #2
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    Hello, horan!

    Suppose you have an urn with 2 yellow balls, 3 red balls, 6 white balls, and 12 blue balls.
    You randomly draw 4 balls from the urn without replacement.
    Given that at least one of the balls is yellow,
    what is the probability that all 4 balls will be different colors?
    Let 4DC = "4 different colors" . . .and AL1Y = "at least one yellow"

    According to Bayes' Theorm, we want . P(4DC\,|\,AL1Y) \;=\;\frac{P(4DC \wedge AL1Y)}{P(AL1Y)} .[1]

    There are: . {21\choose4} \:=\:8855 possible drawings.


    Numerator

    As you pointed out: (4DC \wedge AL1Y) \:=\:4DC

    The number of ways to get 4 colors is: . {2\choose1}{3\choose1}{6\choose1}{12\choose1} \:=\:432 ways.
    Hence: . P(4DC \wedge AL1Y) \:=\:\frac{432}{8855}


    Denominator

    The opposite of "at least 1 yellow" is "no yellow."
    There are: . {21\choose4} \:=\:5985 ways to get no yellows.
    So there are: . 8855-5985 \:=\:2870 ways to get at least one yellow.

    Hence: . P(AL1Y) \:=\:\frac{2870}{8855}


    Substitute into [1]: . P(4DC\,|\,AL1Y) \;=\;\frac{\:\frac{432}{8855}\:}{\:\frac{2870}{885  5}\:} \;=\;\frac{432}{2870} \;=\;\frac{216}{1435}

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