Let A and B be events with P(A)=frac{1}{3}, P | vedclass

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(C) Given,$P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(A \cup B)=\frac{1}{2}$.First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A

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$P(A^C \cap B)=\frac{1}{3}$

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(C) Given,$P(A)=\frac{1}{3}, P(B)=\frac{1}{4}, P(A \cup B)=\frac{1}{2}$.First,we find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:$P(A \cap B) = P(A) + P(B) - P(A \cup B) = \frac{1}{3} + \frac{1}{4} - \frac{1}{2} = \frac{4+3-6}{12} = \frac{1}{12}$.Check for independence: $P(A) \cdot P(B) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$. Since $P(A \cap B) = P(A) \cdot P(B)$,$A$ and $B$ are independent. (Option $A$ is correct).Calculate $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/12}{1/4} = \frac{1}{3}$. (Option $B$ is correct).Calculate $P(A^C \cap B) = P(B) - P(A \cap B) = \frac{1}{4} - \frac{1}{12} = \frac{3-1}{12} = \frac{2}{12} = \frac{1}{6}$. (Option $C$ is incorrect,as it states $1/3$).Calculate $P(A \cap B^C) = P(A) - P(A \cap B) = \frac{1}{3} - \frac{1}{12} = \frac{4-1}{12} = \frac{3}{12} = \frac{1}{4}$. (Option $D$ is correct).Thus,the incorrect statement is $P(A^C \cap B) = \frac{1}{3}$.

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 statements is incorrect?

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