Is it true that $P(A_1 \cap A_2\mid B) = P(A_1\mid B)P(A_2\mid B)$ for any event $B$ if $A_1$ and $A_2$ are independent?

Question

Let's assume there are independent events $A_1$ and $A_2$. Is it true that for any event $B$ the following equation holds $P(A_1 \cap A_2\mid B) = P(A_1\mid B)P(A_2\mid B)$?

Answer 1

how can we prove it?

In general, without independence between $A_1,A_2$ you get

$$\mathbb{P}[A_1,A_2|B]=\frac{\mathbb{P}[A_1,A_2,B]}{\mathbb{P}[B]}=\mathbb{P}[A_1|A_2,B]\cdot \frac{\mathbb{P}[A_2,B]}{\mathbb{P}[B]}=$$

$$=\mathbb{P}[A_1|A_2,B]\cdot\mathbb{P}[A_2|B]$$

Thus if you assume independence, your statement is true as

$$=\mathbb{P}[A_1|A_2,B]=\mathbb{P}[A_1|B]$$