Question

A and B are two mutually exclusive and exhaustive events with $\mathrm{P}\left(\mathrm{B}\right)=3\mathrm{P}\left(\mathrm{A}\right).$ What is the value of $P\left(\overline{B}\right)$

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B

$\frac{1}{4}$

Explanation for the correct option:

Finding the probability:

As a result, because $A$ and $B$ are mutually exclusive and exhaustive events,

$$\begin{gathered} \mathrm{P}(\mathrm{A}\cap \mathrm{B})=0 \\ \mathrm{P}(\mathrm{A}\cup \mathrm{B})=1 \end{gathered}$$

We are aware of this $\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})$

$\begin{array}{l}\Rightarrow 1=\mathrm{P}\left(\mathrm{A}\right)+3\mathrm{P}\left(\mathrm{A}\right)\left[\mathrm{Given}\mathrm{P}\left(\mathrm{B}\right)=3\mathrm{P}\left(\mathrm{A}\right)\right]\\ \Rightarrow 4\mathrm{P}\left(\mathrm{A}\right)=1\\ \Rightarrow \mathrm{P}\left(\mathrm{A}\right)=\frac{1}{4}\\ \therefore \mathrm{P}\left(\mathrm{B}\right)=3\mathrm{P}\left(\mathrm{A}\right)=3\times \frac{1}{4}\\ =\frac{3}{4}\\ \\ \Rightarrow \mathrm{P}\left(\overline{B}\right)=1-\mathrm{P}\left(B\right)\mathbf{[}\mathbf{\because }\mathbf{P}\mathbf{\left(}\overline{\mathbf{B}}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{]}\\ =1-\frac{3}{4}\\ =\frac{1}{4}\end{array}$

Hence, the correct option is B.