Question

If $\text{A}$ and $\text{B}$ are two independent events such that $\text{P(A) =}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{, P(B)=}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\text{,}$then

• A. $\text{P(A/B) = 1/2}$
• B. $\text{P(A/AUB) = 5/6}$
• C. $\text{P(A∩B/A'UB')=0}$
• D. All of these

Answer 1

The correct option is D

All of these

Step $\mathbf{1}$. Find the value of $\mathbf{\text{P}}\left(\raisebox{1ex}{$\text{A}$}\!\left/ \!\raisebox{-1ex}{$B$}\right.\right)$:

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

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

We know that, $\mathbf{\text{P}}\left(\raisebox{1ex}{$\text{A}$}\!\left/ \!\raisebox{-1ex}{$\text{B}$}\right.\right)$ $\mathbf{=}$ $\frac{\mathbf{\text{P}}\left(\text{A∩B}\right)}{\mathbf{\text{P}}\left(\text{B}\right)}$

$=$ $\frac{\text{P}\left(\text{A}\right)\text{·P}\left(\text{B}\right)}{\text{P}\left(\text{B}\right)}$ $\left[\text{∵ P}\left(\text{A∩B}\right)\text{=P}\left(\text{A}\right)\text{·P}\left(\text{B}\right)\right]$

$=$ $\text{P}\left(\text{A}\right)$

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

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

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

$=$ $\text{P}\left(\text{A}\right)\text{/P}\left(\text{AUB}\right)$ $....\left(1\right)$ $\left[\text{∵ P}\left(\text{A∩(AUB)}\right)\text{=P}\left(\text{A}\right)\right]$

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

$=$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$$+$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)$$-$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$$\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)$ $\left[\text{∵ P}\left(\text{A∩B}\right)\text{=P}\left(\text{A}\right)\text{·P}\left(\text{B}\right)\right]$

$=$ $\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$10$}\right.$

$\therefore$ $\text{P}\left(\text{AUB}\right)$ $=$ $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$

Therefore, equation $\left(1\right)$ becomes

$\text{P(A/AUB)}$ $=$ $\frac{\left({\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}{\left({\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}\right)}$

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

Step $\mathbf{3}$. Find the value of$\mathbf{\text{P(A∩B/A'UB')}}$

$\mathbf{\text{P(A∩B/A'UB')}}$ $\mathbf{=}$ $\frac{\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{A∩B}}\mathbf{\right)}\mathbf{\text{∩P}}\mathbf{\left(}\mathbf{\text{A'UB'}}\mathbf{\right)}}{\mathbf{P}\mathbf{\left(}\mathbf{\text{A'UB'}}\mathbf{\right)}}$

$\therefore$ $\text{P(A∩B/A'UB')}$ $=$ $0$

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