$\text{ If }A,B\text{ are two events such that }P(A\cup B)=0.6,P\left(A\right)=P\left(B\right),P\left(\frac{B}{A}\right)=0.8,\text{ then the }$$\text{ value of }P\left(\right(A\cap \overline{B})\cup (\overline{A}\cap B\left)\right)\text{ is }$

$\begin{array}{l}\text{ Given that }P(A\cup B)=0.6,P\left(A\right)=P\left(B\right)\\ \text{ And }P\left(\frac{B}{A}\right)=0.8\\ \Rightarrow \frac{P(A\cap B)}{P\left(A\right)}=0.8\\ \Rightarrow P(A\cap B)=0.8P\left(A\right)\\ \text{ We have, }P(A\cup B)=P\left(A\right)+P\left(B\right)-P(A\cap B)\\ \Rightarrow 0.6=P\left(A\right)+P\left(B\right)-0.8P\left(A\right)\\ \Rightarrow P\left(A\right)=\frac{1}{2}=0.5\\ \therefore P(A\cap B)=\left(0.8\right)P\left(A\right)\\ =\left(0.8\right)\left(0.5\right)\end{array}$

$\begin{array}{l}=0.4\\ P\left(\right(A\cap \overline{B})\cup (\overline{A}\cap B\left)\right)=P(A\cup B)-P(A\cap B)\\ =0.6-0.4\\ =0.2\end{array}$