One Cut and a Fixed Third Side

Let $X$ be the position of the cut measured from one end, so $X\sim\mathrm{Unif}(0,8)$. The three sides are then $5$, $X$ and $8-X$. A triangle exists exactly when each side is shorter than the sum of the other two.

• $5 < X + (8-X) = 8$: this always holds.
• $X < 5 + (8-X)$: this gives $X < \tfrac{13}{2}$.
• $8-X < 5 + X$: this gives $X > \tfrac{3}{2}$.

So a triangle forms if and only if $\tfrac32 < X < \tfrac{13}{2}$. This interval has length $5$ out of a total length of $8$. Equivalently, each piece must be shorter than $\tfrac{13}{2}$, that is, shorter than half the perimeter $13$.

Generalisation. Take a fixed side $a$ and a stick of length $b \ge a$ cut uniformly. The condition becomes $|2X-b| < a$, so the probability is $a/b$. If instead $b<a$, the first inequality fails and the probability is $0$.