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Math Help - Law of total probability - need some clarification

  1. #1
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    Law of total probability - need some clarification

    Two machines, A and B, operate in some factory.
    Machine A produces 10% of the factory's products.
    Machine B produces 90% of the factory's products.

    1% of machine A's products are flawed.
    5% of machine B's products are flawed.

    One product randomly selected.
    what is the probability that it's flawed?


    So, this is what I did:
    let C=\left \{ product\hspace{4} x\hspace{4} is\hspace{4} flawed \right \}
    let A=\left \{ x\hspace{4} produced\hspace{4} in\hspace{4} A \right \}
    let B=\left \{ x\hspace{4} produced\hspace{4} in\hspace{4} B \right \}
    Now, according to the law of total probability, we'll get that:
    P(C)=P(A)\cdot P(C|A)+P(B)\cdot P(C|B)=10\%\cdot (10\%\cdot 1\%)+90\%\cdot (90\%\cdot 5\%)=4.05\%

    but the answer is:
    P(C)=P(A)\cdot P(C|A)+P(B)\cdot P(C|B)=(10\%\cdot 1\%)+(90\%\cdot 5\%)=4.6\%

    Why were P(A) and P(B) omitted?
    Last edited by Stormey; December 7th 2013 at 02:49 AM.
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    Re: Law of total probability - need some clarification

    Quote Originally Posted by Stormey View Post
    Two machines, A and B, operate in some factory. Machine A produces 10% of the factory's products. Machine B produces 90% of the factory's products.
    1% of machine A's products are flawed. 5% of machine B's products are flawed.
    One product randomly selected.
    what is the probability that it's flawed?
    So, this is what I did:
    let C=\left \{ product\hspace{4} x\hspace{4} is\hspace{4} flawed \right \}
    let A=\left \{ x\hspace{4} produced\hspace{4} in\hspace{4} A \right \}
    let B=\left \{ x\hspace{4} produced\hspace{4} in\hspace{4} B \right \}
    the answer is:
    P(C)=P(A)\cdot P(C|A)+P(B)\cdot P(C|B)=(10\%\cdot 1\%)+(90\%\cdot 5\%)=4.6\%
    P(C)=P(C\cap A)+P(C\cap B)
    P(C\cap A)=P(C|A)P(A)=(0.1)(0.01),
    That is, probability it is flawed given A made it times probability that A made it.
    Last edited by Plato; December 7th 2013 at 04:01 AM.
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    Re: Law of total probability - need some clarification

    Hi Plato, thanks for the help.

    Quote Originally Posted by Plato View Post
    That is, probability it is flawed given A made it times probability that A made it.
    that is exactly my question...
    I'll try to rephrase.

    the probability tree looks like this:



    "probability it is flawed given A made it" (meaning P(C|A) ), if we follow the tree from its root along its branches, is 0.1\times 0.01

    "probability that A made it" (meaning P(A) ), again, according the tree it's 0.1

    so "probability it is flawed given A made it times probability that A made it" should be  (0.1\times 0.01)\times 0.1
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    Re: Law of total probability - need some clarification

    Quote Originally Posted by Stormey View Post
    Hi Plato, thanks for the help.


    that is exactly my question...
    I'll try to rephrase.

    the probability tree looks like this:



    "probability it is flawed given A made it" (meaning P(C|A) ), if we follow the tree from its root along its branches, is 0.1\times 0.01

    "probability that A made it" (meaning P(A) ), again, according the tree it's 0.1

    so "probability it is flawed given A made it times probability that A made it" should be  (0.1\times 0.01)\times 0.1
    I really donít understand what you are asking.
    While in general I am no fan of probability trees, your's makes the answer quite clear.
    In order for a part to be flawed is must come from either machine A or machine B.

     \begin{align*} \mathcal{P}(C)&=\mathcal{P}(C\cap A)+ \mathcal{P}(C\cap B)\\ &= \mathcal{P}(A)\mathcal{P}(C|A)+ \mathcal{P}(B)\mathcal{P}(C|B)\\&=(0.1)(0.01)+(0.9  )(0.05)\end{align*}
    Thanks from Stormey
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    Re: Law of total probability - need some clarification

    Quote Originally Posted by Plato View Post
    I really donít understand what you are asking.
    While in general I am no fan of probability trees, your's makes the answer quite clear.
    In order for a part to be flawed is must come from either machine A or machine B.

     \begin{align*} \mathcal{P}(C)&=\mathcal{P}(C\cap A)+ \mathcal{P}(C\cap B)\\ &= \mathcal{P}(A)\mathcal{P}(C|A)+ \mathcal{P}(B)\mathcal{P}(C|B)\\&=(0.1)(0.01)+(0.9  )(0.05)\end{align*}
    OK, I get it now,
    I misunderstood the law of total probability, and it confused me.

    Thanks.
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    Re: Law of total probability - need some clarification

    Here's how I would do it:

    Assume the machines produce 1000 products

    Machine A produces 10% of the factory's products.
    so machine A produces 100 of the products.

    Machine B produces 90% of the factory's products.
    And machine B produces 900.

    1% of machine A's products are flawed.
    1% of 100 is 1. Machine A produces 1 flawed product.

    5% of machine B's products are flawed.
    5% of 900 is 45. Machine B produces 45 flawed products.

    One product randomly selected.
    what is the probability that it's flawed?
    Of the 1000 products 45+ 1= 46 are flawed. The probability any one is flawed is \frac{46}{1000}= 0.046.
    Thanks from Stormey
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