Transforming conditional probability

Question

Does somebody know why the following equation holds?

$$P(A,B|C) = P(A|B,C) P(B|C)$$

Which rule is used? I think it is not Bayes' rule.

Answer 1

You just need to use the rule $P(Y|X) = \frac{P(Y,X)}{P(X)}$: $$ \begin{align} P(A,B|C) &= \frac{P(A,B,C)}{P(C)} \\ &= \frac{P(A|B,C)P(B,C)}{P(C)} \\ &= P(A|B,C)\frac{P(B,C)}{P(C)} \\ &= P(A|B,C)P(B|C) \end{align} $$

Answer 2

It's because: $$ P(A|B , C) = \frac{ P(A \cap B \cap C) }{ P(B \cap C) }=\frac{P((A \cap B)|C) \cdot P(C)}{P(B|C) \cdot P(C)}=\frac{\rm{P}((A\cap B)|C)}{\rm{P}(B|C)} $$