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What would be $p(A\cap B \cap C)$ using the multiplicative Bayes theorem?

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This $p(A \cap B \cap C) = p(A)p(B \mid C)p(C \mid A \cap B)$? –  Peteris Dec 9 '11 at 21:52

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Is the multiplicative Bayes' Theorem any different from the plain old Bayes' Theorem? The Bayes' Theorem I know is not about $p(A\cap B)$; instead, it relates $p(A\mid B)$ and $p(B\mid A)$. So I wouldn't expect Bayes' Theorem to say anything about $p(A\cap B\cap C)$. You might find this Wikipedia page helpful. It does include some formulas involving $p(A\cap B\cap C)$.

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