Proving the conditional probability

Question

Can someone help me with this?

For any three events $A$, $B$ and $C$, prove the following identity: $$Pr(A \cap B\vert C) = Pr(A\vert B \cap C)Pr(B\vert C)$$

I really have no idea how to do this and where to start..

Answer 1

Note that $$P(A\vert B)=\frac{P(A\cap B)}{P(B)}$$So, $$P(A\cap B\vert C)=\frac{P(A\cap B\cap C)}{P(C)}$$$$P(A\vert B\cap C)=\frac{P(A\cap B\cap C)}{P(B\cap C)}$$$$P(B\vert C)=\frac{P(B\cap C)}{P(C)}$$Hence, $$P(A\vert B\cap C)\cdot P(B\vert C) = \frac{P(A\cap B\cap C)}{P(B\cap C)}\cdot \frac{P(B\cap C)}{P(C)} = \frac{P(A\cap B\cap C)}{P(C)} = P(A\cap B\vert C)$$

Answer 2

Hint:

Use the relation $$P(A\vert B)=\frac{P(A\cap B)}{P(B)}$$

on both the LHS and the RHS.

Answer 3

Apply $P(A \mid B)=\dfrac{P(A \cap B)}{P(B)}$

LHS$=P(A\cap B \mid C)=\dfrac{P(A \cap B \cap C)}{P(C)}$

$\text{RHS}=P(A \mid B \cap C)\cdot P(B \mid C)=\dfrac{P(A \cap B \cap C)}{P(B \cap C)} \cdot \dfrac{P(B \cap C)}{P(C)}= \dfrac{P(A \cap B \cap C)}{P(C)} \blacksquare$