From the numbers 1, 2, .., 179, we randomly select three different ones. What is the probability that these three numbers are the measurements in degrees of the three angles of a triangle?
First, we need to clarify the model we are working with. We can choose three different numbers in $\binom{179}{3}$ ways (we do not consider the order), and any two selections are equally likely, so we are using the "classical model". (The result would not change if we considered the order, since then the number of all and good cases would both increase sixfold.) The total number of cases is therefore $\binom{179}{3}$.

Let's count the good cases. Let the three selected angles be $\alpha^{\circ}\beta\left(\right.$ while $\beta \geqq 90^{\circ}-a$ in which case $\gamma=180^{\circ}-(\alpha+\beta) \leqq 180^{\circ}-\left(90^{\circ}+a\right)=$ $90^{\circ}-a \leqq \beta$). For a given $\alpha$, this gives $89^{\circ}-3 a$ possibilities. The value of $a$ can vary between 1 and 29, since the smallest angle is less than $60^{\circ}$, and in this case $89^{\circ}-3 a$ is also positive. Thus, in total,

$$
\sum_{n=1}^{29}(89-3 a)=89 \cdot 29-3 \sum_{a=1}^{29} a=89 \cdot 29-3 \cdot \frac{29 \cdot 30}{2}=1276
$$

possibilities are obtained.

2nd case: $\alpha=2 a-1$ is odd. Then the possible values of $\beta$ are: $2 a, 2 a+1, \ldots, 90^{\circ}-a$. $\left(\beta \leqq 90^{\circ}-a\right.$ according to $\gamma=$ $180^{\circ}-(\alpha+\beta) \geqq 180^{\circ}-\left(89^{\circ}+a\right)=91^{\circ}-a>\beta$ and $\beta \geqq 91^{\circ}-a$ in which case $\gamma<\beta$ would follow.) The value of $a$ can now vary between 1 and 30 (exactly when $\alpha=2 a-1 \leqq 59$ holds), so this case

$$
\sum_{a=1}^{30}(91-3 a)=91 \cdot 30-3 \sum_{a=1}^{30} a=1335
$$

gives possibilities.

In the two cases, we get a total of $1276+1335=2611$ good selections, so the desired probability is:

$$
\frac{2611}{\binom{179}{3}}=\frac{2611}{939929} \approx 0.0027779
$$