All the events are independent, and I want to find the probability of:
$(A \cap B) \cup (A \cap C)$
My first thought was doing the following:
$(A \cap B) \cup (A \cap C) \implies$
$P((A \cap B) \cup (A \cap C))$
$= P(A)*P(B) + P(A) * P(C) - P(A)*P(B)*P(A)*P(C)$
However, today I thought about a different approach:
$(A \cap B) \cup (A \cap C)$
$= A \cap (B \cup C) \implies$
$P(A)*P(B \cup C)$
$= P(A) * (P(B) + P(C) - P(B)P(C))$
$ = P(A) * P(B) + P(A) * P(C) - P(A) * P(B) * P(C)$
The last one has P(A) multiplied only one time on the last term, while the first one has P(A) multiplied two times.
My guess is that, even if A, B and C are independent, A and A are not independent, because they are the same event, Which may make the first thought wrong, since I just assumed $(A \cap B)$ and $(A \cap C))$ were independ, which is not the case.
What I actually had to do was:
$(A \cap B) \cup (A \cap C) \implies$
$P((A \cap B) \cup (A \cap C))$
$= P(A)*P(B) + P(A)*P(C) - P((A \cap B) \cap (A \cap C)) $
$= P(A)*P(B) + P(A)*P(C) - P(A \cap C)*P((A \cap B) |(A \cap C)) $
$= P(A)*P(B) + P(A)*P(C) - P(A)*P(C)*P(B)$
Because $P((A \cap B) |(A \cap C)) = P(B)$
Sorry for making the example so long, but I wanted to make my thoughts as clear as possible.
Is my last thought correct? Also, I'm not sure about the last algebrical step. Is It correct? Thanks in advance.