$A$ and $B$ are events such that $P(A \cup B)=3 / 4, P(A \cap B)=1 / 4, P(\bar{A})=2 / 3$ then $P(\bar{A} \cap B)$ is

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Aicharya · Accepted Answer

$P(A \cup B)=P(A)+P(B)-P(A \cap B) \Rightarrow \frac{3}{4}=1-P(\bar{A})+P(B)-\frac{1}{4}$ $\Rightarrow 1=1-\frac{2}{3}+P(B) \Rightarrow P(B)=\frac{2}{3} ; \quad$ Now, $P(\bar{A} \cap B)=P(B)-P(A \cap B)=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$

$A$ and $B$ are events such that $P(A \cup B)=3 / 4, P(A \cap B)=1 / 4, P(\bar{A})=2 / 3$ then $P(\bar{A}…

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