Conditional probability with independent events If A, B and C are independent events with $P(A) = 0.3$, $P(B) = 0.2$ and $P(C) = 0.4$,
compute $P(A \cap B|B \cup C)$.
$$P(A \cap B|B \cup C) = P((A \cap B) \cap (B \cup C))/P(B \cup C) = P(A \cap B)/P(B \cup C)$$
I can't really understand how the statement in numerator $P((A \cap B) \cap (B \cup C))$ can become $P(A \cap B)$. I guess there is some kind of rule when all events are independent? Sorry for being a noob.
The link below might better clarify what i mean

 A: Don't apologize for this.
Think of it graphically, via terms of a Venn diagramm:
http://i.imgur.com/Yzr1TKM.gif 
This doesn't have to do with independence, by the way.
Look at the intersection $P(A∩B)$, right? Now, look at the union $P(B∪C)$! It's obviously simply the parts of $B$ and $C$ put together. Now we look at the intersection of those two areas. Since $P(A∩B)$ is completely contained in $P(B∪C)$, the intersection IS $P(A∩B)$.
   Or, look at the diagram: We have the part labeled $P(A∩B)$, of course, then we have the part labeled $P(A∩B∩C)$, but everything else is not contained in the original $P(A∩B)$ and therefore not part of the intersection.
There you have it: One is contained in the other.
A: An alternative way to see it is to use the distributive law
$$
X \cap (Y \cup Z) = (X \cap Y) \cup (X \cap Z),
$$
with $X = (A \cap B)$, $Y = B$, and $Z = C$. So, we have
\begin{align}
(A \cap B) \cap (B \cup C)
 &=  ((A \cap B) \cap B) \cup ((A \cap B) \cap C) \\
 &= (A \cap B) \cup (A \cap B \cap C) \\
 &= (A \cap B).
\end{align}
So, ${\rm Pr}\{(A \cap B) \cap (B \cup C)\} = {\rm Pr}\{A \cap B\}$.
