Question

If $\text{A}$ and $\text{B}$ are two events such that $\text{P(AUB)}$$=$$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$, $\text{P(A∩B)}$$=$$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$, and $\text{P(B')}$$=$$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$, then $\text{P(A)}$$=?$

• A. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
• B. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
• C. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• D. $\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$

Answer 1

The correct option is C

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

Explanation for the correct option :

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

Given data,

$\text{P(AUB)}$$=$$\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$

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

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

We know that,$\mathbf{\text{P(B')}}$ $\mathbf{=}$ $\mathbf{1}$$\mathbf{-}$$\mathbf{\text{P}}\left(\text{B}\right)$

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

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

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

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

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

$=$$\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)$$+$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$$-$$\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)$

$=$ $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$6$}\right.$

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

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