Question

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

Answer 1

Hint: Here, to solve the given problem we use the conditional probability concept.

Given,
$$4P(A)=6P(B)=10P(A\cap B)=1\to (1)$$
Now, from equation 1, let us find ‘$P(A)$’, ‘$P(B)$’and ‘$P(A\cap B)$’ values.
$4P(A)=1\Rightarrow P(A)=\frac{1}{4}$
$6P(B)=1\Rightarrow P(B)=\frac{1}{6}$
$10P(A\cap B)=1\Rightarrow P(A\cap B)=\frac{1}{10}$
Here, we need to find the value of $P(B/A)$ i.e.., the probability of the event B after the
 occurrence of event A.
So, to find the $P(B/A)$ let us consider the concept of conditional probability i.e..,
$P(B/A)=\frac{P(A\cap B)}{P(A)}\to (2)$
Let us substitute the obtained values of $P(A\cap B)$ and $P(A)$ in equation 2, we get

$$\begin{gathered} \Rightarrow P(B/A)=\frac{P(A\cap B)}{P(A)} \\ \Rightarrow P(B/A)=\frac{\frac{1}{10}}{\frac{1}{4}} \\ \Rightarrow P(B/A)=\frac{4}{10} \\ \Rightarrow P(B/A)=\frac{2}{5} \end{gathered}$$
Hence, the obtained value of $P(B/A)$ is$\frac{2}{5}$.
Hence the correct option for the given question is ‘A’.
Note: As, to find the conditional probability of $P(B/A)=\frac{P(A\cap B)}{P(A)}$i.e.., the
 probability of the event B after the occurrence of event A .The probability is defined only after the occurrence of event A i.e.., $P(A)$ should be greater than zero.