Question

If $\text{A}$ and $\text{B}$ are two events such that $\text{P(A)}$ $\text{=}$ $\frac{1}{3}$, $\text{P(B)}$ $\text{=}$ $\frac{1}{4}$ and $\text{P(A∩B)}$ $\text{=}$ $\frac{1}{5}$, then $P\left(\frac{B\text{'}}{A\text{'}}\right)$ $\text{=}$?

• A. $\frac{37}{40}$
• B. $\frac{37}{45}$
• C. $\frac{23}{40}$
• D. None of these

Answer 1

The correct option is A

$\frac{37}{40}$

Explanation for the correct option :

Step $\mathbf{1}$. Given data

$\text{P(A)}$ $\text{=}$ $\frac{1}{3}$

$\text{P(B)}$ $\text{=}$ $\frac{1}{4}$

$\text{P(A∩B)}$ $\text{=}$ $\frac{1}{5}$

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

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

$\text{=}$ $\left(\frac{1}{3}\right)$ $+$ $\left(\frac{1}{4}\right)$ $-$ $\left(\frac{1}{5}\right)$

$\text{=}$ $\frac{\left(20+15-12\right)}{60}$

Therefore, $\mathrm{\text{P}}\mathrm{\left(}\mathrm{\text{A∪B}}\mathrm{\right)}$ $\text{=}$ $\frac{23}{60}$

Step $\mathbf{3}$. Finding the value of $\mathbf{\text{P(A')}}$

We know that, $\text{P(A')}$ $\text{=}$ $1$$-$ $\text{P(A)}$

$\text{=}$ $1$$-$$\left(\frac{1}{3}\right)$

Therefore, $\text{P(A')}$ $\text{=}$ $\frac{2}{3}$

Step $\mathbf{4}$. Finding the value of $P\left(\frac{B\text{'}}{A\text{'}}\right)$

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

$\text{=}$ $\frac{\left(1-\text{P}\left(\text{A∪B}\right)\right)}{\text{P}\left(\text{A'}\right)}$

$\text{=}$ $\frac{\left(1-\left({\displaystyle \frac{23}{60}}\right)\right)}{\left({\displaystyle \frac{2}{3}}\right)}$

$\text{=}$ $\frac{\left({\displaystyle \frac{37}{60}}\right)}{\left({\displaystyle \frac{2}{3}}\right)}$

$\text{=}$ $\frac{37}{40}$

Therefore, $\mathit{P}\left(\frac{B\text{'}}{A\text{'}}\right)$ $\mathbf{\text{=}}$ $\frac{\mathbf{37}}{\mathbf{40}}$

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