Picking random points in the volume of sphere with uniform probability

Let's say your sphere is centered at the origin $(0,0,0)$.

For the distance $D$ from the origin of your random pointpoint, note that you want $P(D \le r) = \left(\frac{r}{R_s}\right)^3$. Thus if $U$ is uniformly distributed between 0 and 1, taking $D = R_s U^{1/3}$ will do the trick.

For the direction, a useful fact is that if $X_1, X_2, X_3$ are independent normal random variables with mean 0 and variance 1, then $$\frac{1}{\sqrt{X_1^2 + X_2^2 + X_3^2}} (X_1, X_2, X_3)$$ is uniformly distributed on (the surface of) the unit sphere. You can generate normal random variables from uniform ones in various ways; the Box-Muller algorithm is a nice simple approach.

So if you choose $U$ uniformly distributed between 0 and 1, and $X_1, X_2, X_3$ iid standard normal and independent of $U$, then $$\frac{R_s U^{1/3}}{\sqrt{X_1^2 + X_2^2 + X_3^2}} (X_1, X_2, X_3)$$ would produce a uniformly distributed point inside the ball of radius $R_s$.

An alternative method in $3$ dimensions:

Step 1: Take $x, y, $ and $z$ each uniform on $[-r_s, r_s]$.

Step 2: If $x^2+y^2+z^2\leq r_s^2$, stop. If not, throw them away and return to step $1$.

Your success probability each time is given by the volume of the sphere over the volume of the cube, which is about $0.52$. So you'll require slightly more than $2$ samples on average.

If you're in higher dimensions, this is not a very efficient process at all, because in a large number of dimensions a random point from the cube is probably not in the sphere (so you'll have to take many points before you get a success). In that case a modified version of Nate's algorithm would be the way to go.

Nate and Kevin already answered the two I knew. Recalling this and this, I think that another way to generate a uniform distribution over the sphere surface would be to generate a uniform distribution over the vertical cylinder enclosing the sphere, and then project horizontally.

That is, generate $z \sim U[-R,R]$, $\theta \sim U[0,2\pi]$, and then $x=\sqrt{R^2-z^2} \cos(\theta)$, $y=\sqrt{R^2-z^2} \sin(\theta)$. This (if I'm not mistaken) gives a uniform distribution over the sphere surface. Then, apply Nate's recipe to get a uniform distribution over the sphere volume.

This method is a little simpler (and more efficient) than the accepted answer, though it's not generalizable to other dimensions.