Question

It is given that the events $A$ and $B$ are such that $P\left( A \right) = \dfrac{1}{4}$, $P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{2}$ and $P\left( {\dfrac{B}{A}} \right) = \dfrac{2}{3}$. Then, $P\left( B \right)$ is equal to
A) $\dfrac{1}{2}$
B) $\dfrac{1}{6}$
C) $\dfrac{1}{3}$
D) $\dfrac{2}{3}$

Answer 1

Hint: We are given the probabilities of a few conditions. We are to use the formula of conditional probability, which is,
$P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}}$
Using this formula, we could find the value of $P\left( {A \cap B} \right)$. By substituting the value in the other given condition we can get the required value.

Complete step by step answer:
Given, two events $A$ and $B$.
And, $P\left( A \right) = \dfrac{1}{4}$, $P\left( {\dfrac{A}{B}} \right) = \dfrac{1}{2}$ and $P\left( {\dfrac{B}{A}} \right) = \dfrac{2}{3}$.
Now, we know, $P\left( {\dfrac{A}{B}} \right) = \dfrac{{P\left( {A \cap B} \right)}}{{P\left( B \right)}} - - - \left( 1 \right)$
And, $P\left( {\dfrac{B}{A}} \right) = \dfrac{{P\left( {B \cap A} \right)}}{{P\left( A \right)}} - - - \left( 2 \right)$
Substituting the given values in $\left( 2 \right)$, we get,
$ \Rightarrow \dfrac{2}{3} = \dfrac{{P\left( {B \cap A} \right)}}{{\dfrac{1}{4}}}$
Now, multiplying both sides by $\dfrac{1}{4}$, we get,
$ \Rightarrow \dfrac{2}{3} \times \dfrac{1}{4} = P\left( {B \cap A} \right)$
$ \Rightarrow P\left( {B \cap A} \right) = \dfrac{1}{6}$
Now, we know, the intersection of two events is the same irrespective of the way we write.
That is, $P\left( {A \cap B} \right) = P\left( {B \cap A} \right)$
Therefore, we can clearly say that, $P\left( {A \cap B} \right) = \dfrac{1}{6}$.
Now, substituting the values of $P\left( {\dfrac{A}{B}} \right)$ and $P\left( {A \cap B} \right)$ in $\left( 1 \right)$, we get,
$ \Rightarrow \dfrac{1}{2} = \dfrac{{\dfrac{1}{6}}}{{P\left( B \right)}}$
Now, multiplying both sides with $6$, we get,
$ \Rightarrow \dfrac{1}{2} \times 6 = \dfrac{1}{{P\left( B \right)}}$
$ \Rightarrow 3 = \dfrac{1}{{P\left( B \right)}}$
Taking the inverse on both sides, we get,
$ \Rightarrow P\left( B \right) = \dfrac{1}{3}$
Therefore, $P\left( B \right)$ is $\dfrac{1}{3}$, the correct option is option (C).

Note:
To solve this problem, the concept of conditional probability is used. In probability theory, $A$ conditional probability is a measure of the probability of an event occurring, given that another event (by assumption, presumption, assertion or evidence) has already occurred. If the event of interest is $A$ and the event $B$ is known or assumed to have occurred, the probability is called "the conditional probability of $A$ given $B$", or "the probability of under the condition $B$”. In this problem, we used the simplest notation, that is, $P\left( {\dfrac{A}{B}} \right)$.