Events $A$ and $B$ satisfy $P(A\cup B)=1/2,P(A\cap B)=1/4$ and $P(A\setminus B)=P(B\setminus A)$

Question

Suppose that events $A$ and $B$ satisfy $P(A\cup B)=1/2,P(A\cap B)=1/4$ and $P(A\setminus B)=P(B\setminus A)$. Find $P(A)$.

So I guess $P(A\cup B)=P(A)+P(B)-P(A\cap B)$, would give me an equality that leave $P(A)$ and $P(B)$ but how do I find $P(A)$ or $P(B)$? Thanks

Answer 1

We know that \begin{align} P(A\cup B) &= P(A\cap B) + P(A\setminus B) + P(B\setminus A) \\ &= P(A\cap B) + 2P(A\setminus B) \\ &= \frac{1}{2} \end{align} Since $P(A\cap B) = \frac{1}{4}$, this tells us that $P(A\setminus B) = \frac{1}{8}$. So $P(A) = P(A\cap B) + P(A\setminus B)= \frac{1}{4} + \frac{1}{8} = \frac{3}{8}$.

Answer 2

Note that $$P(A\setminus B)+P(A\cap B)+P(B\setminus A)=P(A\cup B)$$ Since $P(A\setminus B)=P(B\setminus A)$ is given, we can write $$2P(A\setminus B)=P(A\cup B)-P(A\cap B)=\frac12-\frac14=\frac14$$ $$P(A\setminus B)=\frac18$$ Then add this to $P(A\cap B)$ to get $P(A)=\frac18+\frac14=\frac38$.