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(D) Given $A$ and $B$ are independent events,$P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$.$1$. For option $A$: Since $A$ and $B$ are independent,$P(A

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All of the above

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(D) Given $A$ and $B$ are independent events,$P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$.$1$. For option $A$: Since $A$ and $B$ are independent,$P(A/B) = P(A) = \frac{1}{2}$. Thus,option $A$ is correct.$2$. For option $B$: $P(A/(A \cup B)) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A) + P(B) - P(A \cap B)}$.Since $P(A \cap B) = P(A)P(B) = \frac{1}{2} 	imes \frac{1}{5} = \frac{1}{10}$,$P(A \cup B) = \frac{1}{2} + \frac{1}{5} - \frac{1}{10} = \frac{5+2-1}{10} = \frac{6}{10} = \frac{3}{5}$.So,$P(A/(A \cup B)) = \frac{1/2}{3/5} = \frac{1}{2} 	imes \frac{5}{3} = \frac{5}{6}$. Thus,option $B$ is correct.$3$. For option $C$: $P((A \cap B)/(A' \cup B')) = \frac{P((A \cap B) \cap (A' \cup B'))}{P(A' \cup B')}$.By De Morgan's Law,$A' \cup B' = (A \cap B)'$.So,$(A \cap B) \cap (A \cap B)' = \emptyset$.Therefore,the numerator is $P(\emptyset) = 0$. Thus,option $C$ is correct.Since all options are correct,the answer is $D$.

If $A$ and $B$ are two independent events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$,then which of the following is true?

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