Question

If $\text{A}$ and$\text{B}$ are two events such that $\text{P(A)≠ 0,}$$\text{P(B)≠ 1}$then $\text{P}\left(\frac{\text{A'}}{\text{B'}}\right)\text{=?}$

• A. $1-\text{P}\left(\frac{\text{A}}{\text{B}}\right)$
• B. $\text{1-P}\left(\frac{\text{A'}}{\text{B}}\right)$
• C. $\frac{\text{[1-P(AUB)]}}{\text{P(B')}}$
• D. $\frac{\text{P(A')}}{\text{P(B')}}$

Answer 1

The correct option is C

$\frac{\text{[1-P(AUB)]}}{\text{P(B')}}$

Explanation for the correct option :

Step $\mathbf{1}$. Given data Finding the value of $\mathit{P}\left(\frac{A\text{'}}{B\text{'}}\right)\mathbf{:}$

$\text{P(A)≠ 0}$ and

$\text{P(B)≠ 1}$

Step $\mathbf{2}$. Finding the value of $\mathbf{\text{P}}\mathbf{}\mathbf{\left(}\frac{\mathbf{A}\mathbf{\text{'}}}{\mathbf{B}\mathbf{\text{'}}}\mathbf{\right)}$

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

$=$ $\frac{\text{P}\left(\text{A∪B}\right)\text{'}}{\text{P}\left(\text{B'}\right)}$ $\left[\because \text{P}\left(\text{A'∩B'}\right)=\text{P}\left(\text{A∪B}\right)\text{'}\right]$

$=$ $\frac{\left[\text{1-P}\left(\text{A∪B}\right)\right]}{\mathrm{P}\left(\text{B'}\right)}$ $\left[\because \text{P}\left(\text{A'∩B'}\right)=\left(1-P\left(A\cup B\right)\right)\right]$

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

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