Exactly one point

Note that a circle of radius 2 is 4 times as big as a circle of radius 1. So, each point has a $\frac{1}{4}$ of being inside the inner circle (or within one from the center), and $\frac{3}{4}$ probability of not being within one from the center.

So, for each point, the probability that it is the only point within one unit of the center is given by:

$P_{\text{each point}} = \frac{1}{4}(\frac{3}{4})^3$

since there is a $\frac{1}{4}$ probability that it is within one from the center of the circle, and there is a $\frac{3}{4}$ probability that each of the other three isn't within one unit from the center.

And, there are four points to consider, so the total probability is given by

$P = 4P_{\text{each point}} = 4\frac{1}{4}(\frac{3}{4})^3 = \frac{27}{64}$

$27+64 = \boxed{91}$