What is the probability that 3 angles chosen at random from the angles $1^{\circ}, 2^{\circ}, 3^{\circ}, \ldots 179^{\circ}$ are the 3 angles of a scalene triangle?
It is known that the mathematical probability is given by the ratio of the number of favorable cases to the total number of possible cases.

First, we determine the number of favorable cases. Let the angles of the triangle be $x, y, z$ and let $x < y < z < 179^{\circ}$. Since the sum of the angles of a triangle is $180^{\circ}$, we have

$$
x + y = 180^{\circ} - z
$$

but according to our condition, $y < z$, so

$$
x + y < 180^{\circ} - y
$$

which gives

$$
x + 2y < 180^{\circ}
$$

or

$$
\frac{x}{2} + y < 90^{\circ}
$$

Keeping this inequality and the condition $x < y$ in mind, we arrive at the following favorable cases:

| If | $x=1^{\circ}$, | then | $y$ | possible | values: | $2^{\circ}$, | $3^{\circ}$, | $\ldots$ | $88^{\circ}$, | $89^{\circ}$, |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| " | $x=2^{\circ}$, | " | " | " | " | $3^{\circ}$, | $4^{\circ}$, | $\ldots$ | $87^{\circ}$, | $88^{\circ}$, |
| " | $x=3^{\circ}$, | " | " | " | " | $4^{\circ}$, | $5^{\circ}$, | $\ldots$ | $87^{\circ}$, | $88^{\circ}$, |
| " | $x=4^{\circ}$, | " | " | " | " | $5^{\circ}$, | $6^{\circ}$, | $\ldots$ | $86^{\circ}$, | $87^{\circ}$, |
| " | $x=5^{\circ}$, | " | " | " | " | $6^{\circ}$, | $7^{\circ}$, | $\ldots$ | $86^{\circ}$, | $87^{\circ}$, |
| . | . | . | . | . | . | . | . | . | . | . |
| " | $x=57^{\circ}$, | " | " | " | " | $58^{\circ}$, | $59^{\circ}$, | $60^{\circ}$, | $61^{\circ}$, | . |
| " | $x=58^{\circ}$, | " | " | " | " | $59^{\circ}$, | $60^{\circ}$, |  |  |  |
| " | $x=59^{\circ}$, | " | " | " | " | $60^{\circ}$. |  |  |  |  |

Thus, the number of possible values for $y$ is:

$$
\begin{gathered}
88 + 86 + 85 + 83 + 82 + \ldots + 4 + 2 + 1 = (88 + 85 + 82 + \ldots + 4 + 1) + (86 + 83 + 80 + \ldots + 5 + 2) = \\
= \frac{30}{2}(88 + 1) + \frac{29}{2}(86 + 2) = 1335 + 1276 = 2611
\end{gathered}
$$

Since the number of favorable cases is equal to the number of possible values for the angle $y$, the number of favorable cases is 2611.

The total number of possible cases is obtained by determining the number of combinations of three elements that can be formed from 179 elements. But

$$
\binom{179}{3} = \frac{179 \cdot 178 \cdot 177}{1 \cdot 2 \cdot 3} = 939929
$$

so the sought probability is:

$$
\frac{2611}{939929}
$$

(Gusztáv Kertész, Pécs.)

The problem was also solved by: Baranyó A., Bartók I., Braun I., Deutsch E., Deutsch I., Enyedi B., Haar A., Kiss J., Kürti I., Liebner A., Neidenbach E., Pénzes Z., Pfeifer Gy., Pichler S., Pivnyik I., Preisich G., Raab R., Riesz K., Riesz M., Róth A., Schwemmer I., Schwarz Gy., Stern D., Szmodics H., Weisz P.