Question

If $A$ and $B$ be two events associated with a random experiment such that $P\left(A\right)=0.3,P\left(B\right)=0.4$ and $P(A\cup B)=0.6$, then value of $P\left(\frac{\overline{A}}{B}\right)+P\left(\frac{A}{\overline{B}}\right)$ is

• A. $1$
• B. $\frac{3}{4}$
• C. $\frac{2}{5}$
• D. $\frac{13}{12}$
• E. $\frac{11}{12}$

Answer 1

The correct option is A $1$

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

$\Rightarrow P\left(\overline{A}|B\right)+P\left(A|\overline{B}\right)$

$=\frac{P(\overline{A}\cap B)}{P\left(B\right)}+\frac{P(A\cap \overline{B})}{P\left(\overline{B}\right)}$

$=\frac{P\left(\overline{A}\right)P\left(B\right)}{P\left(B\right)}+\frac{P\left(A\right)P\left(\overline{B}\right)}{P\left(\overline{B}\right)}$

$=P\left(\overline{A}\right)+P\left(A\right)$

$=1-P\left(\overline{A}\right)+P\left(A\right)$ since $P\left(\overline{A}\right)=1-P\left(\overline{A}\right)$

$=1$