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I have a question regarding Bayes theorem, and independent intersected events as the conditional. How can this probability be decomposed using Bayes?

$P(H \mid E \cap R)$ where $E$ and $H$ are independent events.

Does that equal $P(H\mid E)\cdot P(H\mid R)$?

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  • $\begingroup$ No. Let $R$ also be independent of $H$. Then you're asking if $P(H)=P(H)\cdot P(H)$ $\endgroup$
    – T.J. Gaffney
    Oct 29, 2014 at 3:22

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$P(H|E,R) = P(H,E,R)/P(E,R)$

$ P(H,E,R) = P(H|E,R)/P(E,R)$

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