If $P(A) =\frac{3}{5}, P(B) =\frac{1}{2}$ and $P(A∩B) =\frac{1}{4}$, then $P(\bar A|\bar B)$ is |
$\frac{3}{40}$ $\frac{3}{10}$ $\frac{17}{20}$ $\frac{17}{40}$ |
$\frac{3}{10}$ |
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, $P(\overline{A} \mid \overline{B}) = \frac{3/20}{1/2} = \frac{3}{10}$ |