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Math Help - Probability Question Proving?

  1. #1
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    Probability Question Proving?

    Suppose that A, B and C are independent events and P(C) ≠ 0. Prove:
    a) P (A ∩ B | C) = P(A | C) P(B | C)
    b) P (A ∪ B | C) = P(A | C) + P(B | C) - P (A ∩ B | C)


    My book only this formula to proved the above question.
    p(A | B) =(A ∩ B)/ P(B)

    I dont know how to deal with 3 constants?
    Any help?

    Thanks
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  2. #2
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    Re: Probability Question Proving?

    Hello, blackZ!

    I'll do the first one . . .


    \text{Suppose that }A, B, C\text{ are independent events and }P(C) \ne 0.

    \text{Prove: }\:a)\;P(A\cap B\,|\,C) \:=\: P(A\,|\,C)\cdot P(B\,|\,C)

    Since A,B,C are independent,
    . . we have, for example: P(A \cap B) \:=\:P(A)\cdot P(B)


    On the left side we have:

    . . P(A \cap B\,|\,C) \;=\;\frac{P(A \cap B \cap C)}{P(C)}

    . . . . . . . . =\;\frac{P(A)\cdot P(B)\cdot P(C)}{P(C)} \;=\;P(A)\cdot P(B)


    On the right side we have:

    . . P(A\,|\,C)\cdot P(B\,|\,C) \;=\;\frac{P(A\cap C)}{P(C)}\cdot \frac{P(B\cap C)}{P(C)}

    . . . . . . . . =\;\frac{P(A)\cdot P(C)}{P(C)} \cdot\frac{P(B)\cdot P(C)}{P(C)} \;=\;P(A)\cdot P(B)

    \text{ta-}DAA!

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  3. #3
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    Re: Probability Question Proving?

    Thanks for the (a).
    For (b), I simplified the right side to P(A)+P(B) - P(A)P(B)

    I am not sure what UNION and CONDITIONAL probability means when A,B and C are independent? Is it the same as INTERSECTION and CONDITIONAL?

    Thanks
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  4. #4
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    Re: Probability Question Proving?

    Quote Originally Posted by blackZ View Post
    Thanks for the (a).
    For (b), I simplified the right side to P(A)+P(B) - P(A)P(B)
    I am not sure what UNION and CONDITIONAL probability means when A,B and C are independent? Is it the same as INTERSECTION and CONDITIONAL?
    \begin{align*}\mathcal{P}[(A\cup B)\cap C]&= \mathcal{P}[(A\cap C)+(B\cap C)]\\&= \mathcal{P}(A\cap C)+ \mathcal{P}(B\cap C)- \mathcal{P}(A\cap B\cap C)\\&= \mathcal{P}(A)\mathcal{P}(C)+ \mathcal{P}(B)\mathcal{P}(C) -\mathcal{P}(A) \mathcal{P}(B)\mathcal{P}(C)\\&= [\mathcal{P}(A)+ \mathcal{P}(B) -\mathcal{P}(A) \mathcal{P}(B)]\mathcal{P}(C)\\&= [\mathcal{P}(A\cup B)]\mathcal{P}(C)   \end{align*}
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  5. #5
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    Re: Probability Question Proving?

    Quote Originally Posted by Plato View Post
    \begin{align*}\mathcal{P}[(A\cup B)\cap C]&= \mathcal{P}[(A\cap C)+(B\cap C)]\\&= \mathcal{P}(A\cap C)+ \mathcal{P}(B\cap C)- \mathcal{P}(A\cap B\cap C)\\&= \mathcal{P}(A)\mathcal{P}(C)+ \mathcal{P}(B)\mathcal{P}(C) -\mathcal{P}(A) \mathcal{P}(B)\mathcal{P}(C)\\&= [\mathcal{P}(A)+ \mathcal{P}(B) -\mathcal{P}(A) \mathcal{P}(B)]\mathcal{P}(C)\\&= [\mathcal{P}(A\cup B)]\mathcal{P}(C)   \end{align*}
    Thats more confusing. I dont know how the conditional probability | will go in there.
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  6. #6
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    Re: Probability Question Proving?

    Hello again, blackZ!

    For (b), I simplified the right side to: P(A)+P(B) - P(A)\!\cdot\!P(B) . Good!

    The left side is: . P(A \cup B\,|\,C) \;=\;\frac{P\big([A \cup B] \cap C\big)}{P(C)} .[1]

    We know that:
    . . P(A \cup B) \:=\:P(A) + P(B) - P(A \cap B)
    . x . . . . . . . =\; P(A) + P(B) - P(A)\!\cdot\!P(B)


    Then [1] becomes:

    . \frac{\big[P(A) + P(B) - P(A)\!\cdot\!P(B)\big]\cdot P(C)}{P(C)} \;=\; P(A) + P(B) - P(A)\!\cdot\!P(B)

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  7. #7
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    Re: Probability Question Proving?

    Quote Originally Posted by blackZ View Post
    Thats more confusing. I dont know how the conditional probability | will go in there.
    OK, I thought you could work through it.
    \begin{align*}\mathcal{P}[(A\cup B)\cap C]&= \mathcal{P}[(A\cap C)+(B\cap C)]\\&= \mathcal{P}(A\cap C)+ \mathcal{P}(B\cap C)- \mathcal{P}(A\cap B\cap C)\\&= \mathcal{P}(A|C)\mathcal{P}(C)+ \mathcal{P}(B|C)\mathcal{P}(C) -\mathcal{P}(A\cap B|C)\mathcal{P}(C) \end{align*}

    Divide by both sides by \mathcal{P}(C)
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