If A and B are two events such that  $P\left(A\right|B)=0.6,P(B\left|A\right)=0.3,P\left(A\right)=0.1thenP\left(\overline{A}\cap \overline{B}\right)=$

$P\left(A\right|B)=0.6,P(B\left|A\right)=0.3,P\left(A\right)=0.1$

$,P\left(B\right|A)=0.3\Rightarrow \frac{P\left(A\cap B\right)}{P\left(A\right)}=0.3$

$P\left(A\cap B\right)=0.3\left(0.1\right)=0.03$

$,P\left(A\right|B)=0.6\Rightarrow \frac{P\left(A\cap B\right)}{P\left(B\right)}=0.6$

$0.03=\left(0.6\right)P\left(B\right)\Rightarrow P\left(B\right)==0.05$

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)=0.1+0.05-0.03=0.12$

$P\left(\overline{A}\cap \overline{B}\right)=P\left(\overline{A\cup B}\right)=1-P\left(A\cup B\right)=1-0.12=0.88$