96*. A piece is broken off from each of three identical rods; the break points of all three rods are chosen at random. What is the probability that a triangle can be formed from the three resulting pieces?
96. The possible outcomes of the experiment are determined by a triplet of numbers $x, y, z$, each chosen randomly between 0 and $l$, where $l$ is the length of the considered rods. Considering these numbers as coordinates of a point in space, we can represent the set of all outcomes as a collection of points within a cube OABCDEFG with side length $l$ (Fig. 84). In this case, the probability that a point $(x, y, z)$ falls within some part of this cube is determined as the ratio of the volume of the considered part to the volume of the entire cube.

Now let us determine where the points corresponding to favorable outcomes of the experiment will be located. For three segments $x, y$, and $z$ to form a triangle, they must satisfy the well-known inequalities:

$$
x+y>z, \quad x+z>y, \quad y+z>x
$$

It is easy to verify that the set of points satisfying the equation $x+y=z$ will fill a plane passing through points $O, E$, and $G$, and all points for which $x+y>z$ will be located on the same side of this plane as point $F$. Similarly, all points for which $x+z=y$ will fill a plane passing through points $O, B$, and $G$; all points for which $y+z=x$ will fill a plane passing through points $O, B$, and $E$. Thus, all points corresponding to favorable outcomes [i.e.,

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-326.jpg?height=373&width=348&top_left_y=563&top_left_x=679)

Fig. 84. satisfying all three inequalities (*)] will be located within the region OBGEF, bounded by the three specified planes and three faces of the cube. Note now that the volume of the cube differs from the volume of the body OBGEF by three times the volume of the tetrahedron $E O B A$ (since the tetrahedra $E O B A$, $G O E D$, and $G O B C$ are equal in volume). The volume of $E O B A$ is

$$
\frac{1}{3} E A \cdot \text{area } O A B=\frac{1}{3} l \cdot \frac{1}{2} l^{2}=\frac{l^{3}}{6}
$$

Therefore, the volume of the body $O B G E F$ is

$$
l^{3}-3 \frac{l^{3}}{6}=\frac{l^{3}}{2}
$$

and the desired probability is

$$
\frac{l^{3}}{2}: l^{3}=\frac{1}{2}
$$