Question

It is given that the events $A$and $B$ are such that$P\left(A\right)=\frac{1}{4}$, $P\left(\frac{A}{B}\right)=\frac{1}{2}$ and$P\left(\frac{B}{A}\right)=\frac{2}{3}$. Then,$P\left(B\right)$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is C

$\frac{1}{3}$

Explanation for correct option:

Calculating the probability of an event:

Given data, $P\left(A\right)=\frac{1}{4}$,$P\left(\frac{A}{B}\right)=\frac{1}{2}$ and $P\left(\frac{B}{A}\right)=\frac{2}{3}$

Now we know that,

$P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}\dots \left(1\right)$

$P\left(\frac{B}{A}\right)=\frac{P(B\cap A)}{P\left(A\right)}\dots \left(2\right)$

Now, substituting the given values in the formula, we get

$$\begin{gathered} \frac{2}{3}=\frac{P(B\cap A)}{P\left(A\right)} \\ \frac{2}{3}=\frac{P\left(B\cap A\right)}{{\displaystyle \frac{1}{4}}} \\ P\left(B\cap A\right)=\frac{1}{6} \end{gathered}$$

Substituting this value in equation $\left(1\right)$ we get,

$$\begin{gathered} P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)} \\ \Rightarrow \frac{1}{2}=\frac{{\displaystyle \frac{1}{6}}}{P\left(B\right)} \\ \Rightarrow P\left(B\right)=\frac{1}{6}\times \frac{2}{1} \\ \Rightarrow P\left(B\right)=\frac{1}{3} \end{gathered}$$

Hence, option (C) is the correct answer.