Question

If $4P\left( A \right)=6P\left( B \right)=10P\left( A\cap B \right)=1$ then $P\left( \dfrac{B}{A} \right)$
A. $\dfrac{2}{5}$
B. $\dfrac{3}{5}$
C. $\dfrac{7}{10}$
D. $\dfrac{19}{60}$

Answer 1

For the type of probability question we will use the product formulas. As product formulas are the way of writing the particular probability case into a conditional probability. As $P\left( \dfrac{B}{A} \right)$ is the conditional probability which can be written as $\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$ .

Complete step by step answer:
Moving ahead with the question in step wise manner;
As we had to find out the value of $P\left( \dfrac{B}{A} \right)$ , which means the probability of event ‘B’ that depends on another event ‘A’. It is also known as "the probability of ‘B’ given ‘A’", and according to conditional probability we can write it as $\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$ .
So by solving it we will get the answer. In order to solve it we need to get the relation between $P\left( A\cap B \right)$ and $P\left( A \right)$ . Which we can get through the given condition $4P\left( A \right)=6P\left( B \right)=10P\left( A\cap B \right)=1$ . So as according to the given condition we have relation between $P\left( A\cap B \right)$ and $P\left( A \right)$ as; $4P\left( A \right)=10P\left( A\cap B \right)$ .
So as we want $P\left( A \right)$ in denominator so divide the given relation by $P\left( A \right)$ , so we will get;
$\begin{aligned} & 4P\left( A \right)=10P\left( A\cap B \right) \\\ & 4=10\dfrac{P\left( A\cap B \right)}{P\left( A \right)} \\\ \end{aligned}$
Since we want only $\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$ with no other constant or value with it, so let us divide the above relation by 10 in both side of equation to get the isolated expression as $\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$ , so we will get;
$\begin{aligned} & 4=10\dfrac{P\left( A\cap B \right)}{P\left( A \right)} \\\ & \dfrac{4}{10}=\dfrac{10}{10}\times \dfrac{P\left( A\cap B \right)}{P\left( A \right)} \\\ \end{aligned}$
So on further simplifying we will get;
$\dfrac{2}{5}=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$
So we got $\dfrac{2}{5}=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$ and as we had mentioned that $\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$ is equal to $P\left( \dfrac{B}{A} \right)$ , which is a product formula.
So we can say that $P\left( \dfrac{B}{A} \right)$ is equal to $\dfrac{2}{5}$ .

So, the correct answer is “Option A”.

Note: Conditional probability is one of the easiest ways to find out the probability with such a given condition. According to conditional probability $P\left( \dfrac{B}{A} \right)=\dfrac{P\left( A\cap B \right)}{P\left( A \right)}$ , and $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$ .