We have, $P(A)=\frac{1}{4}, P\left(\frac{A}{B}\right)=\frac{1}{4}$ and $P\left(\frac{B}{A}\right)=\frac{1}{2}$
Now, $\quad P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{1}{4}$
and $\quad P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}=\frac{1}{2}$

$$
\Rightarrow \quad P(A \cap B)=\frac{1}{2} \times P(A)=\frac{1}{2} \times \frac{1}{4}=\frac{1}{8}
$$
Substitute the value of $P(A \cap B)$ in Eq. (i), we get
$$
P(B)=\frac{1}{8} \times 4=\frac{1}{2}
$$
(l) $P\left(\frac{\bar{A}}{\bar{B}}\right)=\frac{P(\bar{A} \cap \bar{B})}{P(\bar{B})}=\frac{P(A \cup B)^{\prime}}{P(\bar{B})}=\frac{1-P(A \cup B)}{1-P(B)}$
$$
\begin{aligned}
& =\frac{1-[P(A)+P(B)-P(A \cap B)]}{1-P(B)} \\
& =\frac{1-\left(\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\right)}{1-\frac{1}{2}}=\frac{1-\frac{5}{8}}{\frac{1}{2}}=\frac{3}{4}
\end{aligned}
$$
(II) $P(A \cap B)=\frac{1}{8}[\because A$ and $B$ are not mutually exclusive $]$
(III)
$$
\text { (III) } \begin{aligned}
P\left(\frac{A}{B}\right) & +P\left(\frac{A}{\bar{B}}\right) \\
& =\frac{P(A \cap B)}{P(B)}+\frac{P(A \cap \bar{B})}{P(\bar{B})}
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{P(A \cap B)}{P(B)}+\frac{P(A)-P(A \cap B)}{1-P(B)} \\
& =\frac{1}{4}+\frac{\frac{1}{4}-\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}
\end{aligned}
$$
Hence, only (I) is correct.