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Thread: Help on Baye's theorem problem

  1. #1
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    Help on Baye's theorem problem

    A company request materials from three stores X, Y and Z. Store X supplies 45% of the material, store Y supplies 30% of the materials while store Z supplies the remainder. Investigation has shown that 12% of the materials supplied from store X is defective, 15% of the materials supplied by store Y is defective while only 6% of the materials supplied by stores Z is defective.

    If an item is returned after being purchased due to a defect, use Baye’s theorem to calculate the probability that the defective materials was supplied by store
    i. X
    ii.Z
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    Re: Help on Baye's theorem problem

    Quote Originally Posted by benedec View Post
    A company request materials from three stores X, Y and Z. Store X supplies 45% of the material, store Y supplies 30% of the materials while store Z supplies the remainder. Investigation has shown that 12% of the materials supplied from store X is defective, 15% of the materials supplied by store Y is defective while only 6% of the materials supplied by stores Z is defective. If an item is returned after being purchased due to a defect, use Baye’s theorem to calculate the probability that the defective materials was supplied by store
    i. X ii.Z
    P(D)=P(D\cap X)+P(D\cap Y)+P(D\cap Z)
    P(D)=P(D|X)P(X)+P(D|Y)P(Y)+P(D|Z)P(Z).

    Now you want to find P(X|D)=\frac{P(D\cap X)}{P(D)}
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    Re: Help on Baye's theorem problem

    Thank you Plato. In a layman terms, can you elaborate more
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    Re: Help on Baye's theorem problem

    Quote Originally Posted by benedec View Post
    Thank you Plato. In a layman terms, can you elaborate more
    Are you studying probability or not?
    If you are you have a textbook and a set of notes.
    If you are not then there no point in posting a probability question.
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    Re: Help on Baye's theorem problem

    Reasons I asked it seems the formula used is not same as mine, this is what I tried

    P (defective | X) = P (defective | X) P(warehouse X)
    P (defective | X) P(X)+ P (defective | Y) P(Y )+ P (defective | Z) P(Z)
    = 1/3 x 0.12 = 0.33 x 0.12
    1/3 x 0.12 + 1/3 x 0.15 + 1/3 x 0.06 0.33 x 0.12 + 0.33 x 0.15 + 0.33 x 0.66
    = 0.0396 = 0.0396 = 0.129
    0.0396 + 0.0495 + 0.2178 0.3069
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    Re: Help on Baye's theorem problem

    Quote Originally Posted by benedec View Post
    Reasons I asked it seems the formula used is not same as mine, this is what I tried

    P (defective | X) = P (defective | X) P(warehouse X)
    P (defective | X) P(X)+ P (defective | Y) P(Y )+ P (defective | Z) P(Z)
    = 1/3 x 0.12 = 0.33 x 0.12
    1/3 x 0.12 + 1/3 x 0.15 + 1/3 x 0.06 0.33 x 0.12 + 0.33 x 0.15 + 0.33 x 0.66
    = 0.0396 = 0.0396 = 0.129
    0.0396 + 0.0495 + 0.2178 0.3069

    Now you want to find P(X|D)=\frac{P(D| X)P(X)}{P(D|X)P(X)+P(D|Y)P(Y)+P(D|Z)P(Z)}
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