10. Divide a line segment of length $a$ into three segments; the probability that these three segments can form a triangle is $\qquad$
10. $\frac{1}{4}$.

Let the lengths of the three segments be $x, y, a-(x+y)$. Then
$$
\left\{\begin{array}{l}
0 < x < a, \\
0 < y < a, \\
0 < a - (x + y), \\
x + a - (x + y) > y, \\
y + a - (x + y) > x
\end{array} \Rightarrow \left\{\begin{array}{l}
x + y > \frac{a}{2}, \\
y < \frac{a}{2}, \\
x < \frac{a}{2}.
\end{array}\right.\right.
$$

The system of inequalities (2) represents the interior (excluding the boundary) of the triangle $\triangle A^{\prime} B^{\prime} C^{\prime}$ (where $A^{\prime}, B^{\prime}, C^{\prime}$ are the midpoints of $OA, OB, AB$ respectively) as shown in Figure 8. Therefore, the probability that these three segments can form a triangle is
$$
P=\frac{S_{\triangle A^{\prime} B^{\prime} C^{\prime}}}{S_{\triangle A O B}}=\frac{1}{4} .
$$