Question

If $\text{A}$ and $\text{B}$ are two events of a random experiment, such that $\text{P}\left(\text{A∪B}\right)\text{=}\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5,\text{P}\left(\text{A'∪B'}\right)\text{=}{\displaystyle \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$10,$}\right.}$}\right.$ and $\text{P(B) =}\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\text{,}$ then the value of $\text{P(A)}$ equals

• A. $\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$
• B. $\raisebox{1ex}{$8$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$
• C. $\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$
• D. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

Answer 1

The correct option is C

$\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$

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

Given data,

$\text{P}\left(\text{A∪B}\right)\text{=}\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

$\text{P(A'UB') =}\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$ and

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

We know that,

$\mathbf{\text{P}}\left(\text{A∩B}\right)\mathit{+}\mathbf{\text{P}}\left(\text{A∩B}\right)\mathbf{\text{'}}\mathbf{=1}....\left(1\right)$

As $\text{P(A'UB') = P(A∩B)'}$

Equation $\left(1\right)$ becomes

$\begin{array}{rcl}\text{P}\left(A\cap B\right)& =& 1-\left(\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)\\ & =& \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\end{array}$

Step2 . Find the value of $\mathbf{\text{P(A)}}$

We know that,

⇒ $\text{P(A) = P}\left(A\cup B\right)+P(A\cap B)-P\left(B\right)$; $\mathbf{\text{P(AUB) = P}}\left(A\right)\mathbf{+}\mathbf{P}\left(B\right)\mathbf{-}\mathbf{P}\mathbf{(}\mathbf{A}\mathbf{\cap }\mathbf{B}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}$

$\text{=}\left(\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)+\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)-\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)$

$=\left(\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\right)$

$\mathbf{\text{∴P}}\mathbf{\left(}\mathbf{\text{A}}\mathbf{\right)}\mathbf{=}\raisebox{1ex}{$\mathbf{7}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{10}$}\right.$

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