Let X and Y be two events such that P(X mid Y | vedclass

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(B) Given $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{1}{2}$.Since $P(X \cap Y) = \frac{1}{6}$,we have $\frac{1/6}{P(Y)} = \frac{1}{2} \Rightarrow

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$(AB)$

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(B) Given $P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)} = \frac{1}{2}$.Since $P(X \cap Y) = \frac{1}{6}$,we have $\frac{1/6}{P(Y)} = \frac{1}{2} \Rightarrow P(Y) = \frac{1}{3}$.Given $P(Y \mid X) = \frac{P(X \cap Y)}{P(X)} = \frac{1}{3}$.Since $P(X \cap Y) = \frac{1}{6}$,we have $\frac{1/6}{P(X)} = \frac{1}{3} \Rightarrow P(X) = \frac{1}{2}$.Now,$P(X \cup Y) = P(X) + P(Y) - P(X \cap Y) = \frac{1}{2} + \frac{1}{3} - \frac{1}{6} = \frac{3+2-1}{6} = \frac{4}{6} = \frac{2}{3}$. Thus,$(A)$ is correct.Check for independence: $P(X) \cdot P(Y) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}$.Since $P(X \cap Y) = \frac{1}{6} = P(X) \cdot P(Y)$,$X$ and $Y$ are independent. Thus,$(B)$ is correct.$P(X^C \cap Y) = P(Y) - P(X \cap Y) = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$. Thus,$(D)$ is incorrect.

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Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}$,$P(Y \mid X)=\frac{1}{3}$,and $P(X \cap Y)=\frac{1}{6}$. Which of the following is (are) correct?$(A)$ $P(X \cup Y)=\frac{2}{3}$$(B)$ $X$ and $Y$ are independent$(C)$ $X$ and $Y$ are not independent$(D)$ $P(X^C \cap Y)=\frac{1}{3}$