2. If a line segment of length 3 cm is randomly divided into three segments, the probability that these segments can form a triangle is $\qquad$ .
2. $\frac{1}{4}$.

Let the lengths of two segments be $x$ and $y$. Then the length of the third segment is $3-x-y$. Represent the segments using the point $(x, y)$.
Thus, the region formed is
$$
\Omega=\{(x, y) \mid 03-x-y, \\
& x+3-x-y>y, y+3-x-y>x\},
\end{aligned}
$$

The area $S_{A}=\frac{9}{8}$.
Therefore, the probability of forming a triangle is
$$
P(A)=\frac{S_{A}}{S_{\Omega}}=\frac{1}{4} .
$$