Probability and sets You are given two events $A$ and $B$ with $P(A \text { or } B) = 0.4$ and $P(A \text{ or } B’) = 0.8$. Determine $P(A)$. Note: $B’ = \text{ not }B = $ the complement of $B$.
Answer: $0.2$
Not sure how to go about solving this problem. I tried $P(A \cup B) = P(A) + P(B) - P(A \text{ and } B) = 0.8$ and $P(A\cup B’) = P(A) + P(B’) - P(A \text{ and } B’) = 0.8$. But I am stuck here. 
 A: Note that $$A=(A\cup B)\cap(A\cup B') $$ and $$\Omega=(A\cup B)\cup(A\cup B') $$ where $\Omega$ is the universal set. Then apply the Probability of union formula to get $$1=\mathbb{P}(\Omega)=\mathbb{P}((A\cup B)\cup(A\cup B'))=\mathbb{P}(A\cup B)+\mathbb{P}(A\cup B')-\mathbb{P}((A\cup B)\cap(A\cup B'))\\
\Rightarrow 1=0.4+0.8-\mathbb{P}(A)\Rightarrow \mathbb{P}(A)=0.2$$
A: We have
$$P(A\cup B)=P(A)+P(B)-P(A\cap B)\tag{1}$$
$$P(A\cup B')=P(A)+P(B')-P(A\cap B')\tag{2}$$
and $$P(B')=1-P(B)\tag{F1}$$ and $$P(A)=P(A\cap(B\cup B'))=P((A\cap B)\cup(A\cap B'))=P(A\cap B)+P(A\cap B')\tag{F2}$$
so adding $(1)$ and $(2)$ and taking in consideration the facts $(F1)$ and $(F2)$ we find:
$$P(A)=P(A\cup B)+P(A\cup B')-1=0.2$$
A: It is best to draw a picture (Venn Diagram). So draw a big rectangle, and within it two intersecting circles, labelled $A$ and $B$. 
We work with complements. 
The complement of $A\cup B$ is the part of the world which is outside both $A$ and $B$. This region is easy to locate. The probability of the complement of $A\cup B$ is $0.6$. Write $0.6$ in the appropriate region, and lightly shade that region.
The complement of $A\cup B'$ may take some thinking to identify. It is the region which is outside $A$ but in $B$. Some algebra may help: $(A\cup B')'=A'\cap B''=A'\cap B$.  The probability of $A \cup B'$ is $0.8$, so its complement has probability $0.2$. Write $0.2$ in the appropriate region, and lightly shade that region.
Note that we have shaded in all but $A$. So the region outside $A$ has probability $0.6+0.2$. It follows that the region inside $A$ has probability $1-(0.6+0.2)$. 
A: Just add the 2 equations. Use $P(B) + P(B')=1$ and $P(A\cap B) + P(A\cap B') = P(A)$
$\blacksquare$
