If P(A) =3/5, P(B) =1/2 and P(A∩B) =1/4, then P(bar A|bar B) is

The correct answer is Option (2) → $\frac{3}{10}$

Given:

$P(A) = \frac{3}{5}, \quad P(B) = \frac{1}{2}, \quad P(A \cap B) = \frac{1}{4}$

We are asked to find: $P(\overline{A} \mid \overline{B})$

Use the conditional probability formula:

$P(\overline{A} \mid \overline{B}) = \frac{P(\overline{A} \cap \overline{B})}{P(\overline{B})}$

Note that:

$P(\overline{B}) = 1 - P(B) = 1 - \frac{1}{2} = \frac{1}{2}$

Now compute $P(\overline{A} \cap \overline{B})$:

$P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B)$

$P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{5} + \frac{1}{2} - \frac{1}{4}$

Take LCM of 20:

$P(A \cup B) = \frac{12}{20} + \frac{10}{20} - \frac{5}{20} = \frac{17}{20}$

So, $P(\overline{A} \cap \overline{B}) = 1 - \frac{17}{20} = \frac{3}{20}$

Finally,

$P(\overline{A} \mid \overline{B}) = \frac{3/20}{1/2} = \frac{3}{10}$