94*. A rod of length $l$ is broken into three parts at two randomly chosen points. What is the probability that the length of none of the resulting pieces will exceed a given value $a$?
94. This problem is a generalization of the previous one. Indeed, problem 93 can obviously be formulated as follows: what is the probability that none of the three parts of a rod broken at two randomly chosen points will exceed half the total length of the rod? In this problem, the length $\frac{l}{2}$ is replaced by an arbitrary length $a$.

As in the first solution to problem 93, all possible outcomes of the experiment in question will be characterized by two numbers: $A K=x$ and $A L=y$, where $K$ and $L$ are the points of break, and $A$ is the left end of the rod. In this case, the set of all outcomes can be represented as the set of points of the square OMNP with side $l$; the probability that the point $(x, y)$ will fall inside any part of this square will be equal to the ratio of the area of this part to the area of the entire square.

Let us determine which part of the square is filled by points corresponding to unfavorable outcomes of the experiment. First of all, it is obvious that for $a \leqslant \frac{l}{3}$ all outcomes will be unfavorable: at least one of the three parts of the rod will certainly be no less than $\frac{l}{3}$. Now let us consider separately the case $\frac{l}{3} \leqslant a \leqslant \leqslant \frac{l}{2}$ (Fig. 82, a) and $a \geqslant \frac{l}{2}$ (Fig. 82, b).

For the left piece of the rod to be greater than $a$, it is necessary that $x>a, y>a$; points satisfying these conditions fill the square $Y R Z N$ with side $l$ - $a$ in Fig. 82, a and b. Similarly, for the right piece of the rod to be greater than $a$, it is necessary that $x>l-a$ or $y-x>a$; the first condition is satisfied by the points of the triangle VMW, and the second by the points of the triangle $S P T$.

Now we only need to calculate what area remains unshaded in Fig. 82, a and b. In the first case (Fig. 82, a), these will be two isosceles right triangles, the legs of which, obviously, are equal to $P S-U S-Y T=l-a-2(l-2 a)=3 a-l$; thus, the unshaded area is equal to $(3 a-l)^{2}$. In the second case (Fig. 82, b), from the area of the entire square, we need to subtract the areas of two squares with side $l-a$ and two isosceles right triangles with legs $l-a$; thus, the unshaded area is equal to $l^{2}-3(l-a)^{2}$. Dividing this area by $l^{2}$ (the area of the entire square), we get that the probability that none of the three pieces of the rod will exceed $a$ is

$$
\begin{array}{r}
0, \text { if } 0 \leqslant a \leqslant \frac{l}{3}, \\
\left(3 \frac{a}{l}-1\right)^{2}, \text { if } \frac{l}{3} \leqslant a \leqslant \frac{l}{2}, \\
1-3\left(1-\frac{a}{l}\right)^{2}, \text { if } \frac{l}{2} \leqslant a \leqslant l
\end{array}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-324.jpg?height=499&width=497&top_left_y=378&top_left_x=44)
a)

![](https://cdn.mathpix.com/cropped/2024_05_21_68334347b64402497ba5g-324.jpg?height=486&width=485&top_left_y=396&top_left_x=538)

b)

Fig. 82.