We will use constructive counting to solve this. There are $2$ cases: Either all $3$ points are adjacent, or exactly $2$ points are adjacent. If all $3$ points are adjacent, then we have $8$ choices. If we have exactly $2$ adjacent points, then we will have $8$ places to put the adjacent points and $4$ places to put the remaining point, so we have $8 \cdot 4$ choices. The total amount of choices is $\left(\begin{array}{l}8 \textbackslash\textbackslash{} 3\end{array}\right)=8 \cdot 7$. Thus, our answer is $\frac{8+8 \cdot 4}{8 \cdot 7}=\frac{1+4}{7}=$ (D) $\frac{5}{7}$. 
From a regular octagon, a triangle is formed by connecting three randomly chosen vertices of the octagon. What is the probability that at least one of the sides of the triangle is also a side of the octagon? (2018 AMC 8 Problems, Question \#23) 

[1] $\frac{2}{7}$ 
[2] $\frac{5}{42}$ 
[3] $\frac{11}{14}$ 
[4] $\frac{5}{7}$ 
[5] $\frac{6}{7}$ 