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Thread: Just checking

  1. #1
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    Just checking

    I have P(B|A) and P(A), there isn't a quick way to get P(B) off this is there?

    A and B are not independent.

    Thanks,

    Brendan
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  2. #2
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    Hello, Brendan!

    I have $\displaystyle P(B|A)$ and $\displaystyle P(A).$

    There isn't a quick way to get $\displaystyle P(B)$ off this, is there? . no

    $\displaystyle \,A$ and $\displaystyle \,B$ are not independent.

    From Bayes' Theorem, we have: .$\displaystyle P(B|A) \;=\;\dfrac{P(A \cap B)}{P(A)} $

    . . Hence: .$\displaystyle P(A \cap B) \;=\;P(B|A)\cdot P(A)$


    But without more information, such as $\displaystyle P(A \cup B)$,

    . . we cannot isolate $\displaystyle P(B).$
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  3. #3
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    Thanks, I assumed that was the case but just wanted to be sure.
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