An equilateral triangle with side length $8$ has three black equilateral triangles of side length $2$ at its vertices and a centered black equilateral triangle of side length $3$, with sides parallel to the large triangle. A circular coin of diameter $1$ is placed at random so that it lies entirely inside the large triangle. What is the probability that the coin covers part of a black region?

1. The possible centers of the coin form the inward offset of the large triangle by $\frac12$. Its area is $\frac{\sqrt3}{4}(8-\sqrt3)^2=\frac{67\sqrt3-48}{4}$.
2. The central black triangle has side $3$. Its $\frac12$-neighborhood (the triangle, a strip of width $\frac12$ along each side, and three circular sectors at the corners that together form a disk of radius $\frac12$) has area $\frac{9\sqrt3}4+9\cdot\frac12+\pi\left(\frac12\right)^2=\frac{9\sqrt3}4+\frac92+\frac\pi4$.
3. At each corner, the possible centers within $\frac12$ of its black triangle form a triangle of area $\frac{\sqrt3}{4}(2-\frac1{\sqrt3})^2=\frac{13\sqrt3}{12}-1$.
4. These four regions do not overlap. Along the line from a vertex of the large triangle to its center, the corner region reaches distance $\sqrt3+\frac12$ from the vertex. The nearest point of the central region is at distance $\frac8{\sqrt3}-\sqrt3-\frac12=\frac5{\sqrt3}-\frac12$ from the vertex ($\frac8{\sqrt3}$ is the distance from the vertex to the center and $\sqrt3$ is the circumradius of the central triangle). Since $\frac5{\sqrt3}-\frac12-\left(\sqrt3+\frac12\right)=\frac2{\sqrt3}-1>0$, there is a gap between them.
5. The total area of the favorable centers is $\frac{9\sqrt3}4+\frac92+\frac\pi4+3\left(\frac{13\sqrt3}{12}-1\right)=\frac{22\sqrt3+6+\pi}4$.
6. Therefore, the probability is $\dfrac{(22\sqrt3+6+\pi)/4}{(67\sqrt3-48)/4}=\boxed{\dfrac{22\sqrt3+6+\pi}{67\sqrt3-48}}$.