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I am comfortable with the basic form of Bayes Rule from Probability Theory: $$P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ but not sure how to prove this extension: $$P(A|E,F)=\frac{P(A,F|E)}{P(F|E)}$$

Thanks!

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 The proof of the extension is the same as for the original: Plug in the definition of conditional probability and simplify. This is not a research-level question, so I've voted to close. – Andreas Blass Nov 3 2011 at 15:43
As Andreas wrote, the second one is a straightforward application of the first one. Your problem may be that you are writing things like $E,F$ and $A,F$ when these should be $E \cap F$ and $A \cap F$. – Thierry Zell Nov 3 2011 at 16:45
stats.stackexchange.com or math.stackexchange.com might be more helpful, as they have a broader scope than MO. – David Roberts Nov 3 2011 at 23:53

closed as too localized by Andreas Blass, Yemon Choi, David Roberts, François G. Dorais♦ Nov 3 2011 at 23:55

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