Target Exam

CUET

Subject

Maths. Section B1

Chapter

Probability

Question:

For any two events $A$ and $B$, if $P(\bar{A}) = \frac{1}{2}, P(\bar B) = \frac{2}{3}$ and $P(A \cap B) = \frac{1}{4}$, then $P\left(\frac{\bar{A}}{B}\right)$ equals:

Options:

$\frac{3}{8}$

$\frac{8}{9}$

$\frac{5}{8}$

$\frac{1}{4}$

Correct Answer:

$\frac{1}{4}$

Explanation:

$P(\bar{A}) = \frac{1}{2} \Rightarrow P(A) = \frac{1}{2}$

$P(\bar{B}) = \frac{2}{3} \Rightarrow P(B) = \frac{1}{3}$

$P(A \cap B) = \frac{1}{4}$

$P(\bar{A} \cap B) = P(B) - P(A \cap B)$

$= \frac{1}{3} - \frac{1}{4} = \frac{1}{12}$

$P\left(\frac{\bar{A}}{B}\right) = P(\bar{A} \mid B)$

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

$= \frac{\frac{1}{12}}{\frac{1}{3}} = \frac{1}{4}$

The value is $\frac{1}{4}$.