What is the probability of choosing r independent vectors in $\mathbb{R}^n$ in the unit sphere? I was trying to compute the probability of choosing $r \leq n$ indepedent vectors $a_i \in \mathbb{R}^n$ such that they are independent.
I was told that the probability that they are not independent is zero if they lie in the unit sphere. However, was unsure of how to compute such a probability or if it was true.
The intuition I have is that there are infinite directions I could choose, so the probability that I choose the same vector pointing in the same direction is zero. However, that doesn't seem rigorous enough. I feel there might be a caveat I need to take care of to make sure its correct. 
Is that correct and why? If we didn't choose them in the unit sphere would it still be true?
 A: Since $r \leq n$, then yes, the probability of independence is $1$.
You'll have linear dependence if any of the vectors is in the span of the other $r - 1$ vectors. But the span of $r - 1$ vectors is a hyperplane; a set of measure $0$ in $\Bbb R^n$. This hyperplane has dimension $r - 1$ and intersects the unit sphere in a region whose dimension is at most $r - 2$, and thus has measure $0$, with respect to the surface of the sphere (which has dimension $n - 1$). 
Thus, the $r$th vector is almost surely not in the span of the other $r - 1$. 
I think the idea behind using the unit sphere is that you're restricting your attention to unit vectors. If we had been able to choose vectors in all of $\Bbb R^n$ instead of just the unit sphere, it wouldn't really change anything. Roughly, the unit sphere parameterizes the 'directionality' that vectors can have, and this is all that affects linear independence. If we have a set of not-necessarily-unit vectors, we can replace each vector $v$ with the unit vector $\frac{v}{||v||}$ in the same direction. This won't affect linear (in)dependence at all, so we might as well ignore length differences, and look at the unit sphere.
I think focusing on the sphere makes the argument a little tougher to digest, personally. It's clear that hyperplanes of dimension $r-1 < n$ "take up no volume" in $\Bbb R^n$, and so we're almost certain to pick vectors that aren't in the span of what we have. The same idea is still true on the sphere, but we have to use a different measure. The intersection of our span with the surface of the unit sphere is still "too small" with respect to the entire surface to give a non-negligible probability, but it's not as obvious, in my opinion.
So, it seems to me that you need to utilize some kind of measure on $\Bbb R^n$ or the unit sphere to more rigorously justify the assertion.

EDIT for a specific example:
I'm certainly not an expert, but higher-end probability relies on measure theory. Just think about being on the unit $2$-sphere $S^2 = \{(x, y, z):x^2 + y^2 + z^2 = 1\}$. Any (nontrivial proper) subspace of $\Bbb R^3$ will intersect our sphere in either a pair of antipodal points (call the set of such a pair $P$) or a great circle (call it $C$), corresponding to the span of $1$ or $2$ vectors intersecting $S^2$ (the intersection of a line or a plane with the unit sphere). 
The probability of picking a point in $P$ is essentially $$\frac{\text{the area of } P}{\text{the area of }S^2},$$ and it's analogous for picking a point on a great circle. However, $0$- and $1$- dimensional objects have $0$ as their $2$-dimensional area, leading to a $0$ probability of choosing such a point.
Measure theory just makes $n$-dimensional area rigorous, but it does more than that, in the context of probability theory. But, the take-away is that probabilities are measures, and when it comes to problems like this, we really need the measure-theoretic view: just counting discrete possibilities isn't good enough.
Hopefully, however, the intuition here is helpful, because it's all I can provide.
