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Question: If \[4P(A) = 6P(B) = 10P(A \cap B) = 1\] then \(P(\frac{B}{A}) = \) -------. A.\(\frac{2}{5}\) B...

If 4P(A)=6P(B)=10P(AB)=14P(A) = 6P(B) = 10P(A \cap B) = 1 then P(BA)=P(\frac{B}{A}) = -------.
A.25\frac{2}{5}
B.35\frac{3}{5}
C.710\frac{7}{{10}}
D.1960\frac{{19}}{{60}}

Explanation

Solution

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

Given,
4P(A)=6P(B)=10P(AB)=1(1)4P(A) = 6P(B) = 10P(A \cap B) = 1 \to (1)
Now, from equation 1, let us find ‘P(A)P(A)’, ‘P(B)P(B)’and ‘P(AB)P(A \cap B)’ values.
4P(A)=1P(A)=144P(A) = 1 \Rightarrow P(A) = \frac{1}{4}
6P(B)=1P(B)=166P(B) = 1 \Rightarrow P(B) = \frac{1}{6}
10P(AB)=1P(AB)=11010P(A \cap B) = 1 \Rightarrow P(A \cap B) = \frac{1}{{10}}
Here, we need to find the value of P(B/A)P(B/A) i.e.., the probability of the event B after the
occurrence of event A.
So, to find the P(B/A)P(B/A) let us consider the concept of conditional probability i.e..,
P(B/A)=P(AB)P(A)(2)P(B/A) = \frac{{P(A \cap B)}}{{P(A)}} \to (2)
Let us substitute the obtained values of P(AB)P(A \cap B) and P(A)P(A) in equation 2, we get

P(B/A)=P(AB)P(A) P(B/A)=11014 P(B/A)=410 P(B/A)=25  \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)P(B/A) is25\frac{2}{5}.
Hence the correct option for the given question is ‘A’.
Note: As, to find the conditional probability of P(B/A)=P(AB)P(A)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)P(A) should be greater than zero.