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Math Help - Conditional Probability (ii)

  1. #1
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    Conditional Probability (ii)

    box A contains 2 red balls,
    box B contains 2 white balls,
    box C contains 1 red ball and 1 white ball,

    A box is selected at random (with equal probabilities) and one ball is taken at random from that box.

    (i) Compute the probability of selecting a white ball using Bayes Rule.

    A(1/3) _____ Red(1/6)
    .........|____ Red(1/6)

    B(1/3)______White(1/6)
    ........|_____White(1/6)

    C(1/3)______Red(1/6)
    ........|_____White(1/6)

    so.. P(White) = [(1/3)*(1/6)] + [(1/3)*(1/6)] + [(1/3)*(1/6)]
    = 1/6

    Correct?

    (ii)Compute the conditional probability that B has been selected given that the ball is white.

    P(B|White) = P(B intersect White) / P(White)

    what is my P(B intersect White)?

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  2. #2
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by nameck View Post
    box A contains 2 red balls,
    box B contains 2 white balls,
    box C contains 1 red ball and 1 white ball,

    A box is selected at random (with equal probabilities) and one ball is taken at random from that box.

    (i) Compute the probability of selecting a white ball using Bayes Rule.

    A(1/3) _____ Red(1/6)
    .........|____ Red(1/6)

    B(1/3)______White(1/6)
    ........|_____White(1/6)

    C(1/3)______Red(1/6)
    ........|_____White(1/6)

    so.. P(White) = [(1/3)*(1/6)] + [(1/3)*(1/6)] + [(1/3)*(1/6)]
    = 1/6

    Correct?

    W=event of choosing a white ball
    A=selecting from box A
    B=selecting from box B
    C=selecting from box C

    P(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C).
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  3. #3
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    Quote Originally Posted by danneedshelp View Post
    w=event of choosing a white ball
    a=selecting from box a
    b=selecting from box b
    c=selecting from box c

    p(w)=p(w|a)p(a)+p(w|b)p(b)+p(w|c)p(c).
    p(w|a) = 0 ?
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  4. #4
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by nameck View Post
    p(w|a) = 0 ?
    Si
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    Quote Originally Posted by Danneedshelp View Post
    W=event of choosing a white ball
    A=selecting from box A
    B=selecting from box B
    C=selecting from box C

    P(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C).
    P(W|A)P(A) = 0
    P(W|B)P(B) = (1/6 + 1/6)(1/3) = 2/9
    P(W|C)P(C) = (1/6)(1/3) = 1/18

    P(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)
    ......= 0 + 2/9 + 1/18
    ......= 5/18

    correct?
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  6. #6
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by nameck View Post
    P(W|A)P(A) = 0
    P(W|B)P(B) = (1/6 + 1/6)(1/3) = 2/9
    P(W|C)P(C) = (1/6)(1/3) = 1/18

    P(W)=P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)
    ......= 0 + 2/9 + 1/18
    ......= 5/18

    correct?
    No, P(W|A) represents the probability of selecting a white ball, given you are selecting from box A. Since there are no white balls in box A, the probability P(W|A)=0. So, following that train of thought we have

    P(W|A)P(A)=(0/2)(1/3)=0
    P(W|B)P(B)=(2/2)(1/3)=(1/3)
    P(W|C)P(C)=(1/2)(1/3)=(1/6).
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  7. #7
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    understood..
    what bout question (ii)?
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  8. #8
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by nameck View Post
    understood..
    what bout question (ii)?
    The below equation is from wikipedia:Bayes' theorem - Wikipedia, the free encyclopedia



    In terms of your problem, you need to compute

    P(B|W)=\frac{P(W|B)P(B)}<br />
{P(W|A)P(A)+P(W|B)P(B)+P(W|C)P(C)} (we are partitioning the boxes).

    Notice you have already figured out both the numerator and denominator.
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