Explanation

(c)
Given that, $P\left( \frac{B}{A} \right) = \frac{1}{2}\ \Rightarrow \ \frac{P(B \cap A)}{P(A)} = \frac{1}{2}$
$\Rightarrow \ P(B \cap A) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}$
$P\left( \frac{A}{B} \right) = \frac{1}{4}\ \Rightarrow \ \frac{P(A \cap B)}{P(B)} = \frac{1}{4}$,
$\Rightarrow P(B) = 4P(A \cap B) \Rightarrow P(B) = \frac{1}{2}$
$\therefore P(A \cap B) = \frac{1}{8} = \ \frac{1}{2} \bullet \frac{1}{4} = P(A) \bullet P(B)$
$\because$ Events $A\ and\ B$ are independent.
Now, $P\left( \frac{A^{'}}{B} \right) = \frac{P\left( A^{'} \cap {\ B}^{'} \right)}{P(B)}$
$= \frac{P\left( A^{'} \right)P(B')}{P(B)} = \frac{3}{4}$
and $P\left( \frac{B^{'}}{A^{'}} \right) = \frac{P\left( B^{'} \cap {\ A}^{'} \right)}{P\left( A^{'} \right)}$
$= \frac{P\left( B^{'} \right)P(A')}{P\left( A^{'} \right)} = \frac{1}{2}$