The numerator of a fraction is 3 less than its denominator.

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 obtained is $\frac{x-3}{x+1}$.

According to the given condition,

$\frac{x-3}{x+1}=\frac{x-3}{x}-\frac{1}{15}$

$\Rightarrow \frac{x-3}{x}-\frac{x-3}{x+1}=\frac{1}{15}$

$\Rightarrow \frac{(x-3)(x+1)-x(x-3)}{x(x+1)}=\frac{1}{15}$

$\Rightarrow \frac{x^{2}-2 x-3-x^{2}+3 x}{x^{2}+x}=\frac{1}{15}$

$\Rightarrow \frac{x-3}{x^{2}+x}=\frac{1}{15}$

$\Rightarrow x^{2}+x=15 x-45$

$\Rightarrow x^{2}-14 x+45=0$

$\Rightarrow x^{2}-9 x-5 x+45=0$

$\Rightarrow x(x-9)-5(x-9)=0$

$\Rightarrow(x-5)(x-9)=0$

$\Rightarrow x-5=0$ or $x-9=0$

$\Rightarrow x=5$ or $x=9$

When x = 5,

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

When x = 9,

$\frac{x-3}{x}=\frac{9-3}{9}=\frac{6}{9}=\frac{2}{3}$      (This fraction is neglected because this does not satisfies the given condition.)

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