How to prove independence is not a transitive relation? Suppose $A$ and $B$ are independent, and $B$ and $C$ are independent.
Is $B$ independent of $A \cap C$?  Is $B$ independent of $A \cup C$? If so, prove. If not so, give a counterexample.
Could someone give some hints on how to prove them? I am quite confused about them.
 A: The probability of rain here (in Boston) is pretty close to independent of the probability of rain in north London and the rain in south London. But those two events are probably correlated.
Edit in response to comments. True enough - I answered the question in the title, not in the post. Fortunately, there are two correct answers. Here's another that may be useful.
Flip two coins. $A$ is first coin heads, $C$ is second coin heads and $B$ is that the coins agree. Then each event is independent of each of the other two and each has probability $1/2$.  $A \cap C$ say both coins are heads, so $B$ is true - hence not independent of $A \cap C$ . $A \cup C$ says there's at least one head, so two tails can't happen and the probability of $B$ is just $1/3$, so $B$ is not independent of $A \cup C$ .
In fact, in this example you can take any one of the three events and think of it as the middle one that the other two are independent of.
A: Let $\Omega=\{1,2,3,4,5,6\}$ with $\mathbb P(E)=\frac{|E|}6$ for $E\subset\Omega$. Let $A=\{1,2,3\}$, $B=\{1,4\}$, and $C=\{1,5,6\}$. Then $$\mathbb P(A\cap B) = \mathbb P(A)\mathbb P(B) = \mathbb P(B\cap C) =\mathbb P(B)\mathbb P(C) = \frac16,$$ so $A$ and $B$ are independent and $B$ and $C$ are independent. But $$\mathbb P(B\cap(A\cap C)) = \frac16\ne \frac1{18}=\mathbb P(B)\mathbb P(A\cap C) $$ and $$\mathbb P(B\cap(A\cup C))=\frac16\ne \frac5{18}=\mathbb P(B)\mathbb P(A\cup C), $$ so $B$ is not independent of $A\cap C$ or $A\cup C$.
A: Here is a visual counterexample:



A: I flip two coins. Let $A$ be the event that the first coin lands heads up, and let $B$ be the event the second coin lands heads up. Let $C$ be the same event as $A$!
$A$ and $B$ are independent, and $B$ and $C$ are independent, but $A$ and $C$ are as far from independent as it gets.
