(i) Let $A$ and $B$ be two independent event and $P(A) = 2/5$ and $P(B) = 0.7$. Calculate $P(A \cup B)$
(ii) If $A$ and $B$ are mutually exclusive find $P(B|A)$ and $P(A|B)$. Is it possible that $A$ and $B$ be independent? Why?
(iii) If $P(A \cap B)>0$ then find $P(A \cup B|A\cap B)$
(iv) If $A$ and $B$ are independent and $P(A\cap B)>0$, calculate $P(A\cap B|A \cup B)$ when assumption in part $(i)$ holds
I solved part 1 by doing $P(A \cup B) = P(A) + P(B) - P(A\cap B)$. Since they are independent we know $P(A\cap B) = P(A)P(B)$ so we have $P(A \cup B) = 0.4+0.7-(0.4\cdot 0.7) = 0.82$.
Hints:
(ii) $P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(\emptyset)}{P(B)}.$
(iii) $P(A \cup B \mid A \cap B) = \frac{P((A \cup B) \cap (A \cap B))}{P(A \cap B)} = \frac{P(A \cap B)}{P(A \cap B)}$.
(iv) $P(A \cap B \mid A \cup B) = \frac{P(A \cap B)}{P(A \cup B)} = \frac{P(A)P(B)}{P(A \cup B)}$.