Question

If $A$ and $B$ are two events such that
$P(A/B)=0.6,P(B/A)=0.3$ and $P(A)=0.1$
then $P(\bar{A}\cap \bar{B})$ equals

• A. 0.88
• B. 0.12
• C. 0.6
• D. 0.4

Answer 1

Answer: (A)

We have,
$P(A/B)=0.6,P(B/A)=0.3,P(A)=0.1$
We know that,
$\Rightarrow P(B/A)=\frac{P(B\cap A)}{P(A)}$
$0.3=\frac{P(B\cap A)}{0.1}$
$\Rightarrow P(B\cap A)=0.03$
Or $P(A\cap B)=0.03$
Also,$P(A/B)=\frac{P(A\cap B)}{P(B)}$
$\Rightarrow 0.6=\frac{0.03}{P(B)}$
$\Rightarrow P(B)=0.05$
$\therefore P(\bar{A}\cap \bar{B})=1-P(A\cup B)$
$=1-[P(A)+P(B)-P(A\cap B)]$
$=1-0.1-0.05+0.03=0.88$