(10) Divide a line segment of length $a$ into three segments. The probability that these three segments can form a triangle is $\qquad$
(10) $\frac{1}{4}$ Hint: Let the lengths of the three segments be $x, y, a-(x+y)$, then
$$
\left\{\begin{array}{l}
0 < x < a, \\
0 < y < a, \\
0 < a-(x+y) < a, \\
x+y > a-(x+y), \\
x+a-(x+y) > y, \\
y+a-(x+y) > x,
\end{array}\right.
$$

which simplifies to
$$
\left\{\begin{array}{l}
x+y>\frac{a}{2}, \\
y<\frac{a}{2}, \\
x<\frac{a}{2} .
\end{array}\right.
$$

This system of inequalities represents the interior (excluding the boundary) of the region $\triangle A^{\prime} B^{\prime} C^{\prime}$, where $A^{\prime}, B^{\prime}, C^{\prime}$ are the midpoints of $O A, O B, A B$ respectively.
Thus, the probability that these three segments can form a triangle is $P=\frac{S_{\triangle A^{\prime} B^{\prime} C^{\prime}}}{S_{\triangle A O B}}=\frac{1}{4}$.