Expanding $P(Z \cap X \mid Y)$

Question

I can't see exactly why $P(Z \cap X \mid Y) = P(Z \mid X \cap Y) \mathop{P}(X \mid Y)$. I have been told (and it is also my intuition) that this has to do with the definition of conditional probability, but I can't see how this works in detail. Can someone spell this out for me?

Answer 1

This is simply because \begin{align*} \Pr(Z \cap X \mid Y) &= \frac{\Pr(Z \cap X \cap Y)}{\Pr(Y)} \\ &= \frac{\Pr(Z \cap X \cap Y)}{\Pr(X \cap Y)} \frac{\Pr(X \cap Y)}{\Pr(Y)} \\ &= \Pr(Z \mid X \cap Y) \Pr(X \mid Y). \end{align*}