Question

If A and B are two events such that $\text{P(A)}$$=$$\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$, $\text{P(B)}$$=$$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$, and $\text{P(AUB)}$$=$$\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$, then $\text{P(A/B)}$$=?$

• A. $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
• B. $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• C. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
• D. None of these

Answer 1

The correct option is A

$\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

Explanation for the correct option :

Step $\mathbf{1}$.Find the value of $\mathbf{\text{P}}\left(\text{A∩B}\right)$

Given data,

$\text{P(A)}$ $=$ $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$

$\text{P(B)}$ $=$ $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$ and

$\text{P(AUB)}$$=$ $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

We know that,

So, $\text{P}\left(\text{A∩B}\right)$ $=$ $\text{P(A)}$ $+$ $\text{P(B)}$ $-$ $\text{P}\left(\text{A∪B}\right)$ $\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{A∪B}}\mathbf{\right)}\mathbf{\text{= P(A) + P(B) - P(A∩B)}}$

$=$ $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$ $+$ $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$ $-$ $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

$=$ $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$

$\therefore$$\text{P}\left(\text{A∩B}\right)$ $=$ $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

Step $2$. Find the value of $\mathbf{\text{P(A/B)}}$

We know that, $\mathbf{\text{P(A/B)}}$ $\mathbf{=}$ $\mathbf{\text{P}}\left(\text{A∩B}\right)\mathbf{\text{/P}}\left(\text{B}\right)$

$=$ $\raisebox{1ex}{$\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)$}\!\left/ \!\raisebox{-1ex}{$\left({\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)$}\right.$

$\mathbf{\therefore }$ $\mathbf{\text{P(A/B)}}$$\mathbf{=}$ $\raisebox{1ex}{$\mathbf{2}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{5}$}\right.$

Hence, option $\mathbf{"}\mathbf{\text{A"}}$ is correct.