If $A$ and $B$ are any two events such that $P\left(\frac{A}{B}\right)=0.6$, $P\left(\frac{B}{A}\right)=0.3$ and $P\left(A\right)=0.1$ then the value of $P\left(\overline{A}\cap \overline{B}\right)$ is 

Given that $A,B$ are any two events such that $P\left(\frac{A}{B}\right)=0.6$, $P\left(\frac{B}{A}\right)=0.3$ and $P\left(A\right)=0.1$

We know that $P\left(\frac{A}{B}\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}$ and $P\left(\frac{B}{A}\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}$

Hence, 

  $\begin{array}{rcl}\frac{P\left({\displaystyle \frac{A}{B}}\right)}{P\left({\displaystyle \frac{B}{A}}\right)}& =& \frac{P\left(A\cap B\right)}{P\left(B\right)}\times \frac{P\left(A\right)}{P\left(A\cap B\right)}\\ \frac{0.6}{0.3}& =& \frac{0.1}{P\left(B\right)}\\ P\left(B\right)& =& 0.05\end{array}$

and 

     $\begin{array}{rcl}P\left(A\cap B\right)& =& P\left(A\right)·P\left(\frac{B}{A}\right)\\ & =& \left(0.1\right)·\left(0.3\right)\\ & =& 0.03\end{array}$

Consider $P\left(\overline{A}\cap \overline{B}\right)$

$\begin{array}{rcl}P\left(\overline{A}\cap \overline{B}\right)& =& P\left(\overline{A\cup B}\right)\\ & =& 1-P\left(A\cup B\right)\\ & =& 1-P\left(A\right)-P\left(B\right)+P\left(A\cap B\right)\\ & =& 1-0.1-0.05+0.03\\ & =& \boxed{0.88}\end{array}$