2. (20 points) Two points are randomly and independently chosen on a line segment, and the segment is then divided into three parts at these points. Question: What is the probability that the three new segments can form a triangle?


Suppose the initial line segment is the interval $[0,1]$, and the two randomly chosen points are $x, y (0 < x < y < 1)$. The three new segments are $x, y - x, 1 - y$. For these three segments to form a triangle, they must satisfy the triangle inequality:
\begin{array}{l}
x + (y - x) > 1 - y , \\
x + ( 1 - y ) > y - x , \\
( y - x ) + ( 1 - y ) > x
\end{array} \Rightarrow \left\{\begin{array}{l}
y>\frac{1}{2}, \\
y-x>\frac{1}{2}, \\
y-xx$ ). Therefore, the probability that the three new segments can form a triangle is $\frac{1}{4}$.