If \[4P(A) = 6P(B) = 10P(A \cap B) = 1\] then $P(\frac{B}{A}) = $ -------.

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

\[4P(A) = 6P(B) = 10P(A \cap B) = 1 \to (1)\]
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

   \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} \\
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.

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