Let the denominator of the required fraction be $x$.
Numerator of the required fraction $=x-3$
$\therefore$ Original fraction $=\frac{x-3}{x}$
If 1 is added to the denominator, then the new fraction obtaincd is $\frac{x-3}{x+1}$
According to the given condition,
$\frac{x-3}{x+1}=\frac{x-3}{x}-\frac{1}{15}$
$\begin{array}{l}
\Rightarrow \frac{x-3}{x}-\frac{x-3}{x+1}=\frac{1}{15} \\
\Rightarrow \frac{(x-3)(x+1)-x(x-3)}{x(x+1)}=\frac{1}{15} \\
\Rightarrow \frac{x^{2}-2 x-3-x^{2}+3 x}{x^{2}+x}=\frac{1}{15} \\
\Rightarrow \frac{x-3}{x^{2}+x}=\frac{1}{15} \\
\Rightarrow x^{2}+x=15 x-45 \\
\Rightarrow x^{2}-14 x+45=0 \\
\Rightarrow x^{2}-9 x-5 x+45=0 \\
\Rightarrow x(x-9)-5(x-9)=0 \\
\Rightarrow(x-5)(x-9)=0 \\
\Rightarrow x-5=0 \text { or } x-9=0 \\
\Rightarrow x=5 \text { or } x=9
\end{array}$

When $x=5$

$\frac{x-3}{x}=\frac{5-3}{5}=\frac{2}{5}$

When $x=9$, $\frac{x-3}{x}=\frac{9-3}{9}=\frac{6}{9}=\frac{2}{3} \quad$

(This fraction is neglected because this does not satisfies the given condition.)

Hence, the required fraction is $\frac{2}{5}$.