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(D) Given $A$ and $B$ are independent events,$P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{5}$.$(a)$ Since $A$ and $B$ are independent,$P(A|B) = P(A) = \f

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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 $P(B) = \frac{1}{5}$.$(a)$ Since $A$ and $B$ are independent,$P(A|B) = P(A) = \frac{1}{2}$. Thus,option $(a)$ is true.$(b)$ $P(A|A \cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)}$.Since $A \subset (A \cup B)$,$A \cap (A \cup B) = A$.$P(A \cup B) = P(A) + P(B) - P(A)P(B) = \frac{1}{2} + \frac{1}{5} - \frac{1}{10} = \frac{5+2-1}{10} = \frac{6}{10} = \frac{3}{5}$.$P(A|A \cup B) = \frac{1/2}{3/5} = \frac{1}{2} 	imes \frac{5}{3} = \frac{5}{6}$. Thus,option $(b)$ is true.$(c)$ By De Morgan's Law,$A' \cup B' = (A \cap B)'$.$P(A \cap B | (A \cap B)') = \frac{P((A \cap B) \cap (A \cap B)')}{P((A \cap B)')} = \frac{P(\phi)}{P((A \cap B)')} = 0$. Thus,option $(c)$ is true.Since all options are true,the correct 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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