(2) A line segment of length $3 \mathrm{~cm}$ is randomly divided into three segments. The probability that these three segments can form a triangle is $\qquad$.
(2) $\frac{1}{4}$ Hint: Let the lengths of two segments be $x$ and $y$, then the length of the third segment is $3-x-y$. When a line segment of length $3 \text{ cm}$ is arbitrarily divided into three segments, it can be represented by the point $(x, y)$. The entire set of outcomes of the experiment is
$$
\Omega=\{(x, y) \mid 0 < x < 3, 0 < y < 3, 0 < 3-x-y\}
$$
The event that the three segments can form a triangle is
$$
A=\{(x, y) \mid x+y>3-x-y, x+3-x-y>y, y+3-x-y>x\}
$$
The area $S_{A}=\frac{9}{8}$, so the probability that the three segments can form a triangle $P(A)=\frac{S_{A}}{S_{\Omega}}=\frac{1}{4}$.