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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 \m$i$d C)p(C \m$i$d A \cap B)$?2011-12-09

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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 $p(A\cap B\cap C)$.