# Bayes rule for two conditioned events

I only saw Bayes rule with one conditioned event and now I need to work with two events and I am not sure if this is a right way to do it:

$$p(A \mid B, C) = \frac{p(B \mid A, C)\ p(A \mid C)}{p(B \mid C)} = \frac{p(B, A \mid C)}{p(B\mid C)}$$

I honestly do not know if this makes sense, but I needed to have $$p(B\mid C)$$ in denominator for the problem that I am solving, so I thought this might be the way to do it.

Why is the above true or false?

This is valid. The key is that $$B, C = B \cap C$$ is a single event (that is, a subset of the sample space), so your result follows from the definition of conditional probability. (Note that we'll use the same trick with the event $$A, B = A \cap B$$.) $$p(A \mid B, C) = \frac{p(A, B, C)}{p(B, C)} = \frac{p(A, B \mid C) p(C)}{p(B \mid C) p(C)}$$ This approach can also be used to show that your middle term, $$\frac{p(B \mid A, C) p(A \mid C)}{p(B \mid C)},$$ is also equivalent to each of these expressions.