First slide

Operations on Sets

Question

$If A and B are events such that P (\bar{A} \cup \bar{B}) = \frac{3}{4}, P (\bar{A} \cap \bar{B}) = \frac{1}{4} and P (A) = \frac{1}{3}, then P (\bar{A} \cap B) is$

Moderate

A

$\frac{5}{12}$
B

$\frac{3}{8}$
C

$\frac{5}{8}$
D

$\frac{1}{4}$

Solution

$\begin{array}{l} P (\bar{A} \cup \bar{B}) = \frac{3}{4} \Rightarrow P (\bar{A \cap B}) = \frac{3}{4} \Rightarrow P (A \cap B) = \frac{1}{4} \\ P (\bar{A} \cap \bar{B}) = \frac{1}{4} \Rightarrow P (\bar{A \cup B}) = \frac{1}{4} \Rightarrow P (A \cup B) = \frac{3}{4} \\ P (A) = \frac{1}{3} ∴ P (\bar{A}) = \frac{2}{3} \\ Now, P (\bar{A} \cap B) = P (B) - P (A \cap B) \\ Also, P (A \cup B) = P (A) + P (B) - P (A \cap B) \\ \Rightarrow \frac{3}{4} = \frac{1}{3} + P (B) - \frac{1}{4} \\ ∴ P (B) = \frac{3}{4} - \frac{1}{12} = \frac{2}{3} \\ ∴ P (\bar{A} \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{5}{12} \end{array}$

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