# Calculate $P(A' \cap B)$ and $P(A' \cap B')$ given $P(A) = 1/2$, $P(B) = 1/2$ and $P(A \cup B)= 2/3$

We are given that $$P(A) = P(B) = 1/2$$ and $$P(A \cup B)= 2/3$$.

I found that events $$A$$ and $$B$$ are not mutually exclusive by showing that $$P(A \cup B) \neq P(A) + P(B)$$ (i.e. $$2/3 \neq 1$$)

I also found that the two events are not independent by showing that $$P(A \cap B) \neq P(A)P(B)$$ (i.e. $$1/3 \neq 1/4$$)

Because the events are not mutually exclusive or independent I am having trouble solving for $$P(A'\cap B)$$ and $$P(A' \cap B')$$. I wanted to use the fact that $$P(A'\cap B) = P(B) - P(A\cap B)$$ but the events must be mutually exclusive to use this.

Any suggestions?

First: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ This tells you that $P(A\cap B)=P(A)+P(B)-P(A\cup B)$; you know all the quantities on the right side.
Second: $$P(A^C\cap B)=P(B)-P(A\cap B)$$
Third: $$P(A^C\cap B^C)=1-P(A\cup B).$$