Proving $\displaystyle P(A|B \ \mathrm{and} \ C) = \frac{P(A|C)P(B|A \ \mathrm{and} \ C)}{P(B|C)}$

Question

Problem

Prove that $\displaystyle P(A \mid B \cap C) = \frac{P(A\mid C) \cdot P(B\mid A \cap C)}{P(B\mid C)}$.

Thoughts

I'm having some trouble interpreting $\displaystyle P(A\mid B \cap C)$, and whether it means $\displaystyle P((A\mid B) \cap C)$ or $\displaystyle P(A\mid (B \cap C))$.

Does the former make sense at all? The probability of (A given B) and C?

And in either case, I don't really know where to start with this one, as I don't know what to assume as given.

It certainly looks like a case of Bayes' theorem, but the second probability in the numberator throws me off.

Answer 1

First of all, $P(A|B\cap C)$ means $P(A|(B\cap C))$.

Then, you can use the definition of conditional probability: $P(A|B)=\frac{P(A \cap B)}{P(B)}$

LHS: $P(A|B\cap C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}$

RHS: $\frac{P(A|C)P(B|A\cap C)}{P(B|C)}=\frac{\frac{P(A\cap C)}{P(C)}\frac{P(B\cap A\cap C)}{P(A\cap C)}}{\frac{P(B\cap C)}{P(C)}}=\frac{P(B\cap A\cap C)}{P(B\cap C)}=\frac{P(A\cap B\cap C)}{P(B\cap C)}$

So, LHS=RHS.

Hope that helps!