Conditional Probability and Independence Basics Suppose $A$ and $B$ are independent events, with $C$ a third arbitrary event.  Does this imply that $P(A \cap B |C) = P(A|C)P(B|C)?$
I've tried proving this using the standard definitions:
$$P(A\cap B|C) = \frac{P(A\cap B \cap C)}{P(C)} = \frac{P(C | A \cap B)P(A)P(B)}{P(C)}$$
but I don't know how to proceed from here so I'm starting to think this might not be true.  Any counterexamples?
 A: Independence does not imply conditional independence:  The simplest counter-example here is to take $C = \bar{A} \cup \bar{B}$ (i.e., the event $C$ occurs so long as $A$ and $B$ don't both occur).  We then have the trivial result:
$$\begin{equation} \begin{aligned}
\mathbb{P}(A \cap B | C) 
&= \mathbb{P}(A \cap B | \bar{A} \cup \bar{B}) \\[6pt]
&= \mathbb{P}(A \cap B | \overline{A \cap B}) \\[6pt]
&= \frac{\mathbb{P}(\varnothing)}{\mathbb{P} (\overline{A \cap B})} = 0. \\[6pt]
\end{aligned} \end{equation}$$
Now, let $A$ and $B$ be independent events with $0<\mathbb{P}(A)<1$ and $0<\mathbb{P}(B)<1$.  Then we have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(A | C) \cdot \mathbb{P}(B | C) 
&= \mathbb{P}(A | \overline{A} \cup \bar{B}) \cdot \mathbb{P}(B | \bar{A} \cup \bar{B})  \\[6pt]
&= \frac{\mathbb{P}(A \cap \bar{B})}{\mathbb{P}(\bar{A} \cup \bar{B})} \cdot \frac{\mathbb{P}(\bar{A} \cap B)}{\mathbb{P}(\bar{A} \cup \bar{B})}  \\[6pt]
&= \frac{\mathbb{P}(A) \cdot \mathbb{P}(\bar{B})}{\mathbb{P}(\bar{A} \cup \bar{B})} \cdot \frac{\mathbb{P}(\bar{A}) \cdot \mathbb{P}(B)}{\mathbb{P}(\bar{A} \cup \bar{B})} > 0. \\[6pt]
\end{aligned} \end{equation}$$
