Independence of intersections Let $(\Omega,\mathfrak{A},P)$ be a probability space and $A,B,C\in\mathfrak{A}$ some events where $A$ and $B$ are independent. I'm a bit confused now as I intuitively think that this also implies independence of $A\cap C$ and $B\cap C$. Am I right? Unfortunately, I immediately get stuck when I try to use the definition.
 A: Consider $A = \{ 1,2 \}$, $B = \{1,4\}$ and $C = \{1\}$ on $\Omega = \{1,2,3,4\}$, with $P( X ) = \frac{|X|}{|\Omega|}$
$$P(A \cap B ) = 1/4$$
$$P(A) P(B) = 1/4$$ 
So A and B are independent, yet 
$$P( (A \cap C) \cap (B\cap C) ) = 1/4$$
But, note that
$$P(A \cap C) P (B\cap C)  = 1/16.$$
Therefore, these events are not independent
A: Two events $A$ and $B$ are independent $\iff$ $P(A \cap B) = P(A)P(B)$. This alone does not tell us anything about the third event $C$.
As a counterexample, consider flipping two coins. Call the event that the first one comes up heads $A$ (so $P(A)=.5$), the event that the second one comes up heads $B$ (so $P(B)=.5$), and the event that they both come up tails $C$ (so $P(C)=.25$). $A$ and $B$ are certainly independent (our second coin does not care how our first coin came up). But clearly $P(A \cap C)=0 \neq P(A)P(C)$ (since the first coin cannot come up both heads and tails), so $A$ and $C$ are not independent. Similarly for $B$ and $C$.
A: As @Michael commented, a counter-example can be posed as follows: Consider a regular die and throw it twice consecutively. Considering the events:
$ A :=\{ \text{get 2 at first} \}$, $ B :=\{ \text{get 5 at second} \}$, and $ C :=\{ \text{the sum of the two results is 7} \}$. You can see that:
$$P(A)=P(B)=P(C)=\frac{1}{6},$$
$$P(A\cap B)=P(A\cap C)=P(B\cap C) = \frac{1}{36},$$
$$P(A\cap B\cap C) = \frac{1}{36}.$$
Thus, these events, although pair-independent, they are not independent mutually:
$$P(A\cap B\cap C)\neq P(A)P(B)P(C).$$
Please, let me know if this helps or if you find that something is missing.
