\(|\Omega|={8\choose 3}=\dfrac{8!}{3!5!}=\dfrac{6\cdot7\cdot8}{6}=7\cdot8=56\)
There are two possibilities of choosing three sides of the polygon, so when they are extended, they form a triangle containing the polygon (pictures in the attachment). Multiplying it by the number of sides, gives us 16 possibilities in total.
\(|A|=16\\\\
P(A)=\dfrac{|A|}{|\Omega|}\\
P(A)=\dfrac{16}{56}=\dfrac{2}{7}\)
