Is the equation true that $P(A\cap B|C) = P(A|C)P(B|C)$ while A, B are independent.

Since I got into grad school to study computer sicence, I have been a T.A. and mid-term procter.

Plus, I should check students answering sheets.

To make answers, I would like to check the equation true which is

$P(A\cap B|C) = P(A|C)P(B|C)$ while A, B are independent.

I think that is true.

But I am still little bit not sure.

If that is true, could you prove why?

Thank you for helping in advance.

• What does the "comma " mean? – SchrodingersCat Nov 3 '15 at 13:48
• A intersection B. – Woonghee Lee Nov 3 '15 at 13:57
• Try it out on $C=A\triangle B$. Then the LHS takes value $0$ but the RHS can easily take a positive value. – drhab Nov 3 '15 at 14:10

Let $A$ and $B$ be the events that two independently tossed coins come up heads. Let $C$ be the event "exactly one coin comes up heads". Then the LHS is $0$, while the RHS is $1/4$.
• No, that is not true. Independence of $A$ and $B$ doesn't imply that $A$ and $B$ are conditionally independent on $C$ - $C$ could in principle tell you everything about $A$ if you know $B$ as well. – Daniel Littlewood Nov 12 '15 at 18:17