In an island tribe, there is a custom - during a ritual dance, the shaman tosses up three thin straight rods of the same length, tied in a shape resembling the letter П. Adjacent rods are connected by a short thread and can freely rotate relative to each other. The rods fall on the sand, forming a random figure. If a self-intersection occurs (the first and third rods cross each other), the tribe will face poor harvests and various troubles in the coming year. If there is no self-intersection, the year will be successful - bountiful and happy. Find the probability that the rods will predict good fortune for the year 2017.
A broken line formed by rods is called $A B C D$. Let $\alpha=\angle A B C$ be the angle between the first and second links, and $\beta=\angle B C D$ be the angle between the second and third (see left figure). We can assume that $0 \leq \alpha \leq \pi$ and then $0 \leq \beta \leq 2 \pi$.

The elementary outcomes are pairs $(\alpha, \beta)$. On the coordinate plane $\alpha O \beta$, they fill a rectangle $G$ (see right figure), with the probability of a point $(\alpha, \beta)$ falling inside a certain figure being proportional to the area of that figure.
![](https://cdn.mathpix.com/cropped/2024_05_08_22585010b2534c7ab9ebg-47.jpg?height=498&width=1288&top_left_y=1792&top_left_x=387)

The first and third links $A B$ and $C D$ intersect at some point $K$ only if the ray $C D$ lies inside the angle $B C A$, and the ray $B A$ lies inside the angle $C B D$, i.e., $0 \leq \angle B C D < \angle B C A$, $0 \leq \angle A B C < \angle C B D$.

Considering that triangles $A B C$ and $C B D$ are isosceles, we get the conditions $0 \leq \beta < \pi / 2 - \alpha / 2$ and $0 \leq \alpha < \pi / 2 - \beta / 2$. On the coordinate plane $\alpha O \beta$, these inequalities define a quadrilateral $F$ inside the rectangle $G$. The probability of self-intersection $P(F) = S_{F} / S_{G} = 1 / 12$. Therefore, the desired probability is $1 - 1 / 12 = 11 / 12$.

## Answer

An implausible legend says that one day Stirling was contemplating Stirling numbers of the second kind and absentmindedly threw 10 fair dice on the table. After another throw, he suddenly noticed that the combination of points included all numbers from 1 to 6. Stirling immediately wondered what the probability of such an event would be. What is the probability that when 10 dice are thrown, each number of points from 1 to 6 appears at least once?

## Solution

Let's find the probability of the event $A$ "some number is missing in the rolled combination." Denote $A_{k}$ as the event "number $k$ is missing." Then

$A = A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \cup A_{5} \cup A_{6}$.

Using the inclusion-exclusion principle, we get:

$P(A) = P(A_{1}) + P(A_{2}) + \ldots + P(A_{6}) - P(A_{1} \cap A_{2}) - P(A_{1} \cap A_{3}) - \ldots - P(A_{5} \cap A_{6}) + P(A_{1} \cap A_{2} \cap A_{3}) + P(A_{1} \cap A_{2} \cap A_{4}) + \ldots + P(A_{4} \cap A_{5} \cap A_{6}) - \ldots - P(A_{1} \cap A_{2} \cap A_{3} \cap A_{4} \cap A_{5} \cap A_{6})$.

All probabilities $P(A_{k})$ are equal to $(5 / 6)^{10}$, and there are $C_{6}^{1} = 6$ of them. The probabilities of pairwise intersections are $(4 / 6)^{10}$, and there are $C_{6}^{2} = 15$ of them, and so on. Therefore,

$P(A) = 6 \cdot (5 / 6)^{10} - 15 \cdot (4 / 6)^{10} + 20 \cdot (3 / 6)^{10} - 15 \cdot (2 / 6)^{10} + 6 \cdot (1 / 6)^{10}$.

Thus, the probability that each number appears at least once is $1 - P(A)$.

## Answer

$1 - 6 \cdot (5 / 6)^{10} + 15 \cdot (4 / 6)^{10} - 20 \cdot (3 / 6)^{10} + 15 \cdot (2 / 6)^{10} - 6 \cdot (1 / 6)^{10} \approx 0.272$.