I am trying to prove why the intersection of two events $A, B$ that are independent of C is also independent of C so that the following equality holds: $$P(A\cap B\cap C)= P(A\cap B)P(C)$$ Intuitively, it looks true but why is that?

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    $\begingroup$ If $A$ and $B$ are each independent of $C$, it need not follow that $A \cap B$ is independent of $C$... Right? $\endgroup$ – GEdgar Aug 1 '15 at 18:38
  • $\begingroup$ @GEdgar That's what I am trying to prove since I am not sure it's correct. $\endgroup$ – mgus Aug 1 '15 at 18:49
  • $\begingroup$ You can read David's example... Or you can look up the topic "pairwise independent" in an elementary probability text. $\endgroup$ – GEdgar Aug 1 '15 at 18:52

This is not true. Say $\Omega=\{(0,0),(0,1),(1,0),(1,1)\}$. Say each point of $\Omega$ has probability $1/4$. Let $A=\{(0,0),(1,0)\}$, $B=\{(0,0),(0,1)\}$, and $C=\{(0,0),(1,1)\}$.

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    $\begingroup$ This means that in such a problem I should have some information about $P(A\cap B|C)$ in order to calculate $P(A\cap B\cap C)$ given that I know $P(C)$, right? Or is there any other way? $\endgroup$ – mgus Aug 1 '15 at 18:58

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