Answer

Assume that we break the stick at points \(X\) and \(Y\). Assume \(X < Y\). Then for the stick to form a triangle, the three lengths \(X, Y-X\) and \(1-Y\) should satisfy the following three inequalities

\[ \begin{aligned} X+(Y-X) &> 1-Y\newline (Y-X) + (1-Y) &> X\newline X + (1-Y) &> Y-X \end{aligned} \]

which is nothing but the triangluar region between the points \((0, 0.5), (0.5, 0.5)\) and \((0.5, 1)\) and has the area of \(1/8\). We should also consider the case \(Y < X\) and by symmetry, the area is same. Now, \(X\) and \(Y\) comprise of the entire square region \(X \leq 1\) and \(Y \leq 1\). Hence the required probability is \(2 \times 1/8 = 1/4\).