In right triangle $FGH$ with $\angle H = 90^\circ$, we have $FG = 17$ and $HG = 15$.  Find $\sin G$.
The triangle is shown below:

[asy]
pair F,G,H;
H = (0,0);
G = (15,0);
F = (0,8);
draw(F--G--H--F);
draw(rightanglemark(G,H,F,20));
label("$H$",H,SW);
label("$G$",G,SE);
label("$F$",F,N);
label("$17$",(F+G)/2,NE);
label("$15$",G/2,S);
[/asy]

The Pythagorean Theorem gives us $FH = \sqrt{FG^2 - GH^2} = \sqrt{289 - 225} = 8$, so $\sin G = \frac{FH}{FG} = \boxed{\frac{8}{17}}$.