# Does independence of B with individual events imply independence with the union of these events?

I feel this is simple enough to have been asked before, but I simply couldn't find it if so. I apologise if this is the case.

If B is an event and for each k, $\{B,A_k\}$ is a pair of independent events, then is $\{B,\bigcup_{k=1}^nA_k\}$ a pair of independent events?

Clearly if one can prove it for $A_1$ and $A_2$ then induction is easy, so this is what I've been trying to do. By expanding the union of events I have reduced it to determining whether $\{B\cap A_1\cap A_2\}$ is independent, but here, annoyingly, I've come to a halt.

• It is rather healthy that you "have come to a halt" since the result does not hold. Counterexamples are everywhere on the web, say there. – Did Mar 29 '15 at 17:18
• Perfect, thank you. I suspected this might be the case but couldn't think of a counterexample. – George Moore Mar 29 '15 at 17:22

Let $x,y$ be two random numbers, each is either $0$ or $1$.
Let $A_1$ be the event $x=1$, $A_2$ is the event $y=1$ and $B$ the event $x+y$ is even. $B$ is independent of both $A_1$ and $A_2$ but not their union.
If the $A_k$ are mutually disjoint, then "yes", because in that case $$P(B\cap(\cup A_k)) = P(\cup (B \cap A_k)) = \sum P(B \cap A_k) = \sum P(B)P(A_k) = P(B) P(\cup A_k)$$ and hence $B$ and $\cup A_k$ are mutually independent.