Why does $P(A \cap B \mid C) = P(B \mid C) \cdot P(A \mid B \cap C)$?

Question

I was watching a lecture and there was an equation containing $P(A \cap B \mid C)$ and the instructor said that this is equal to $P(B \mid C) \cdot P(A \mid B \cap C)$ and I don't understand this step.

Answer 1

Just translate into equations. Recall that, by definition, if $P(Y) \ne 0$,

$$P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}$$

Then, assuming $P(C), P(B\cap C)$ are nonzero,

$$\begin{align*} P(A \cap B \mid C) &= \frac{P(A \cap B \cap C)}{P(C)}\\ P(B \mid C) \cdot P(A \mid B \cap C) &= \frac{\color{blue}{P(B \cap C)}}{P(C)} \cdot \frac{P(A \cap B \cap C)}{\color{blue}{P(B \cap C)}} \\ &= \frac{P(A \cap B \cap C)}{P(C)} \\ &= P(A \cap B \mid C) \end{align*}$$

since the blue factors cancel.