Question

Consider two events A and B such that $P\left(A\right)=\frac{1}{4}$, $P\left(\frac{B}{A}\right)=\frac{1}{2}$, $P\left(\frac{A}{B}\right)=\frac{1}{4}$. For each of the following statements, which is true.

$\mathrm{I}.\mathrm{P}\left(\frac{\mathrm{A}\text{'}}{\mathrm{B}\text{'}}\right)=\frac{3}{4}$

$\mathrm{II}.$ The events A and B are mutually exclusive

$\mathrm{III}.\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)+\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}\text{'}}\right)=1$

• A. $\mathrm{I}\mathrm{only}$
• B. $\mathrm{I}\&\mathrm{II}$
• C. $\mathrm{I}\&\hspace{0.17em}\mathrm{III}$
• D. $\mathrm{II}\&\mathrm{III}$

Answer 1

The correct option is A

$\mathrm{I}\mathrm{only}$

Explanation for correct option.

Given,

$A\&B$are two event such that $P\left(A\right)=\frac{1}{4}$,$P\left(\frac{B}{A}\right)=\frac{1}{2}$, $P\left(\frac{A}{B}\right)=\frac{1}{4}$.

Step1. explanation for statement I

$\mathrm{I}.\mathrm{P}\left(\frac{\mathrm{A}\text{'}}{\mathrm{B}\text{'}}\right)=\frac{3}{4}$

Given, $P\left(\frac{A}{B}\right)=\frac{1}{4}$

Therefore,

$\begin{array}{rcl}P\left(\frac{A\text{'}}{B\text{'}}\right)& =& 1-P\left(\frac{A}{B}\right)\\ & =& 1-\frac{1}{4}\\ & =& \frac{3}{4}\end{array}$

Statement I is true.

Step2. Explanation for statement II

$\mathrm{II}.$ The events A and B are mutually exclusive.

$$\begin{gathered} \Rightarrow P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)} \\ \Rightarrow \frac{1}{2}=\frac{P(A\cap B)}{{\displaystyle \frac{1}{4}}} \\ \Rightarrow P(A\cap B)=\frac{1}{8} \end{gathered}$$

Therefore ‘A’ & ‘B’ are not mutually exclusive events.

Step3. Explanation for Statement III.

$\begin{array}{rcl}& & P\left(\frac{A}{B}\right)+P\left(\frac{A}{B\text{'}}\right)\\ & =& \frac{1}{4}+\left[\frac{P\left(A\right)-P(A\cap B)}{P(B\text{'})}\right]\\ & =& \frac{1}{4}+\frac{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{8}}}{{\displaystyle \frac{1}{2}}}\\ & =& \frac{1}{4}+\frac{1}{4}\\ & =& \frac{1}{4}\end{array}$

But, $\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)+\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}\text{'}}\right)=1$

Statement III is also wrong.

Hence, the correct option is $\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{.}$