How to prove that conditional independence does not imply independence? I am trying to prove that conditional independence does not imply independence, ie that $P(A|C)P(B|C)=P(A \cap B|C) \nRightarrow P(A \cap B)=P(A)P(B)$
I guess I need a counter-example but I am struggling to find a way  of homing in onto one. 
So far I have tried drawing Venn diagrams, and I can see that there is no reason why the sizes of the relevant intersections should multiply as implied by the above, but I am not sure how to proceed from there.

If $C = A^c,$ then as long as the $P(A) \neq 1,$ then the conditional independence equation will just be $0=0$ while we've learned nothing about whether $A$ and $B$ are independent.
That is, take any two $A, B$ that aren't independent, then $P(A) \neq 1$ (why?) and setting $C=A^c$ implies $A|C, B|C$ are independent.
