4. Let $x$ and $y$ be the lengths of two segments on the left side. The length of the third segment is $L-(x+y)$. The sample space consists of all pairs $(x, y)$ for which

$$
\left\{\begin{array} { l } 
{ 0 \leq x \leq L } \\
{ 0 \leq y \leq L } \\
{ 0 \leq L - ( x + y ) \leq L }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
0 \leq x \leq L \\
0 \leq y \leq L \\
0 \leq x+y \leq L
\end{array}\right.\right.
$$

On the coordinate plane, they form a triangle with vertices $(0,0),(L, 0),(0, L):$

![](https://cdn.mathpix.com/cropped/2024_05_06_ec5437cc5bb4b0805b1eg-05.jpg?height=329&width=519&top_left_y=1439&top_left_x=241)

The area of this triangle is $L^{2} / 2$. To solve the problem, it is more convenient to calculate the probability of unfavorable events, which are determined by the system of inequalities

$$
\left\{\begin{array} { l } 
{ x \leq \frac { 5 } { 1 2 } L } \\
{ y \leq \frac { 5 } { 1 2 } L } \\
{ L - ( x + y ) \leq \frac { 5 } { 1 2 } L }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x \leq \frac{5}{12} L \\
y \leq \frac{5}{12} L \\
x+y \geq \frac{7}{12} L
\end{array}\right.\right.
$$

On the coordinate plane, they also form a right triangle with legs $L / 4$:

![](https://cdn.mathpix.com/cropped/2024_05_06_ec5437cc5bb4b0805b1eg-06.jpg?height=544&width=729&top_left_y=768&top_left_x=352)

The area of this triangle is $\frac{1}{32} L^{2}$. The probability of unfavorable events is the ratio of the areas

$$
q=\frac{\frac{1}{32} L^{2}}{\frac{1}{2} L^{2}}=\frac{1}{16}
$$

And the desired probability is

$$
p=1-q=1-\frac{1}{16}=\frac{15}{16}
$$
Answer: $\frac{15}{16}$.