Question

If $\text{A}$ and $\text{B}$ are two events such that $\text{P(AUB)}$$=$$\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$, $\text{P(A∩B)}$$=$$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$, $\text{P(A’)}$$=$$\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$, then $\text{P(A’∩B)}$ is equal to

• A. $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$12$}\right.$
• B. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$
• C. $\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$
• D. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Answer 1

The correct option is A

$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$12$}\right.$

Explanation for the correct option :

Step $\mathbf{1}$. Find $\mathbf{\text{P(A)}}$

Given data

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

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

$\text{P(A’)}$$=$$\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

We know that,

$\text{P(A’)}$$=$$1$$-$$\text{P(A)}$

$\Rightarrow$ $\text{P(A)}$ $=$$1$$-$$\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

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

Step $\mathbf{2}$. Find $\mathbf{\text{P(B)}}$

We know that,

$\text{P(AUB)}$ $=$ $\text{P(A)}$$+$$\text{P(B)}$$-$$\text{P(A∩B)}$

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

$\therefore$ $\text{P(B)}$ $=$ $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Step $\mathbf{3}$. Find $\mathbf{\text{P(A'∩B)}}$

We know that,

$\text{P(A'∩B)}$$=$ $\text{P(B)}$$-$$\text{P(A∩B)}$

$=$ $\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$$-$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)$

$\therefore$ $\mathbf{\text{P(A'∩B)}}$$\mathbf{=}$ $\raisebox{1ex}{$\mathbf{5}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{12}$}\right.$

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