Solution:

The set of points $X$ such that the circle of radius $1$ centered at $X$ lies entirely inside $\triangle ABC$ is itself a triangle, $A'B'C'$, such that $AB$ is parallel to $A'B'$, $BC$ is parallel to $B'C'$, and $CA$ is parallel to $C'A'$, and furthermore $AB$ and $A'B'$, $BC$ and $B'C'$, and $CA$ and $C'A'$ are all $1$ unit apart. We can use this to calculate that $A'B'C'$ is an equilateral triangle with height $10$, and hence has area $\frac{100}{\sqrt{3}}$.

On the other hand, the set of points $X$ such that the circle of radius $1$ centered at $X$ contains $O$ is a circle of radius $1$, centered at $O$, and hence has area $\pi$.

The probability that the circle centered at $X$ contains $O$ given that it also lies in $ABC$ is then the ratio of the two areas, that is,
$$\frac{\pi}{\frac{100}{\sqrt{3}}} = \frac{\sqrt{3}\,\pi}{100}.$$