$A$ and $B$ are mutually exclusive
$$
\begin{aligned}
& A \cap B=\phi \Rightarrow P(A \cap B)=0 \\
& \text { and } P(A)=\frac{1}{4} \text { and } P(B)=\frac{3}{7}
\end{aligned}
$$
Now, $P\left(\frac{A}{A \cup B}\right)=\frac{P(A \cap(A \cup B))}{P(A \cup B)}$
$$
\left[\because P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}\right]
$$
$$
\begin{aligned}
& \because \quad A \cap(A \cup B)=A \\
& \therefore P(A \cap(A \cup B))=P(A)=\frac{1}{4}
\end{aligned}
$$
$$
\text { and } \begin{aligned}
P(A \cup B) & =P(A)+P(B)-P(A \cap B) \\
& =\frac{1}{4}+\frac{3}{7}-0=\frac{7+12}{28}=\frac{19}{28}
\end{aligned}
$$
$\begin{aligned} \therefore \text { Required probability } & =\frac{\frac{1}{4}}{\frac{19}{28}} \\ & =\frac{28}{19 \times 4}=\frac{7}{19}\end{aligned}$