Let $A$ and $B$ be events with $P(A)=\frac{1}{3}, P(B)=\frac{1}{4}$ and $P(A \cup B)=\frac{1}{2}$. Then which of the following statement is incorrect?

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Aicharya · Accepted Answer

Given, $P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(A \cup B)=\frac{1}{2}$ $	herefore \quad P\left(A^Cight)=1-P(A)=1-\frac{1}{3}=\frac{2}{3}$ $P(A) P(B)=\frac{1}{3} \cdot \frac{1}{4}=\frac{1}{12}$ and $\quad P(A \cap B)=P(A)+P(B)-P(A \cup B)$ $=\frac{1}{3}+\frac{1}{4}-\frac{1}{2}=\frac{1}{12}$ $\Rightarrow A$ and $B$ are independent. $\begin{aligned} P(A / B) & =\frac{P(A \cap B)}{P(B)}=\frac{1 / 12}{1 / 4}=1 / 3 \ P\left(A^C \cap Bight) & =P(B)-P(A \cap B)\end{aligned}$ $=\frac{1}{4}-\frac{1}{12}=\frac{1}{6}$ $P\left(A \cap B^Cight)=P(A) \cdot P\left(B^Cight)$ $=\frac{1}{3} \cdot\left(1-\frac{1}{4}ight)=\frac{1}{3} 	imes \frac{3}{4}=1 / 4$ $	herefore P\left(A^C \cap Bight)=1 / 3$ is incorrect among above options.

Let $A$ and $B$ be events with $P(A)=\frac{1}{3}, P(B)=\frac{1}{4}$ and $P(A \cup B)=\frac{1}{2}$. Then…

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