I choose you!

Using what @Eli Ross suggested, we can note that the hypotenuse of the triangle must be formed by two points which are opposite each other (endpoints of a diameter).

There are $\frac{2010}{2} = 1005$ diameters, and $2010- 2= 2008$ potential right-angle vertices given a diameter (hypotenuse).

Thus, the probability is $\frac{1005 \cdot 2008}{\binom{2010}{3}} = \frac{6\cdot 1005 \cdot 2008}{2010 \cdot 2009 \cdot 2008} = \frac{3}{2009},$ so $A+B = 3 + 2009 = 2012.$

This is a great problem @Daniel Awakens and welcome back to Brilliant! We haven't seen you in awhile and we're glad you're back. Thanks for sharing this one.

I'll let someone else post a solution, but if you need a hint, remember that inscribed angles have angle measure equal to half of the arc. So, for a right triangle, we need a $180^\circ$ arc.

If that's unfamiliar, you can learn more with this practice quiz .