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Math Help - Probability Proof

  1. #1
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    Probability Proof

    Let A and B be events in a sample space with positive probabilities. Prove that P(B|A)>P(B)\iff P(A|B)>P(A)

    The seems relatively easy.

    1st direction: P\Rightarrow Q

    \displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}\cdot\frac{P(A)}{P(A)}>P(B)\cdot\frac{P(B  )}{P(B)}

    \displaystyle\Rightarrow\frac{P(A\cap B)}{P(A)}\cdot\frac{P(A)}{P(B)}>P(B)\cdot\frac{P(A  )}{P(B)}

    \displaystyle\frac{P(A\cap B)}{P(B)}>P(A)\Rightarrow P(A|B)>P(A)

    The second direction is identical so I am not going to post it.

    Is this how it is proven though?
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  2. #2
    Super Member Random Variable's Avatar
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    I think it's even simpler than that.

     \displaystyle P(B|A) = \frac{P(A \cap B)}{P(A)} >  P(B)

    then  \displaystyle P(A|B) = \frac{P(A \cap B)}{P(B)} > \frac{P(A)P(B)}{P(B)} = P(A)
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