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(C) Since $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B) = \frac{1}{3} 	imes \frac{1}{6} = \frac{1}{18}$.For option $A$: $P(A / B) = P(A)

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$P(A / B^{\prime})=\frac{1}{3}$

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(C) Since $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B) = \frac{1}{3} 	imes \frac{1}{6} = \frac{1}{18}$.For option $A$: $P(A / B) = P(A) = \frac{1}{3}$,so $P(A / B) = \frac{2}{3}$ is $FALSE$.For option $B$: $P(A / (A \cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A) + P(B) - P(A \cap B)} = \frac{1/3}{1/3 + 1/6 - 1/18} = \frac{1/3}{6/18 + 3/18 - 1/18} = \frac{1/3}{8/18} = \frac{1}{3} 	imes \frac{18}{8} = \frac{6}{8} = \frac{3}{4}$,so $P(A / (A \cup B)) = \frac{1}{4}$ is $FALSE$.For option $C$: Since $A$ and $B$ are independent,$A$ and $B^{\prime}$ are also independent. Thus,$P(A / B^{\prime}) = P(A) = \frac{1}{3}$. This is $TRUE$.For option $D$: Since $A$ and $B$ are independent,$A^{\prime}$ and $B^{\prime}$ are also independent. Thus,$P(A^{\prime} / B^{\prime}) = P(A^{\prime}) = 1 - P(A) = 1 - \frac{1}{3} = \frac{2}{3}$,so $P(A^{\prime} / B^{\prime}) = \frac{1}{3}$ is $FALSE$.

Let $A$ and $B$ be two independent events such that $P(A)=\frac{1}{3}$ and $P(B)=\frac{1}{6}$. Then,which of the following is $TRUE$?

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