15. Two Umbrellas ${ }^{2}$ (from 8th grade, 4 points). Every morning, the Absent-Minded Scholar goes to work, and in the evening - home. The Scholar has two umbrellas in total, and he takes one with him only in one of two cases: either it is raining outside, or there is no umbrella where he is going. After some time, the Scholar calculated (and we remember that the Absent-Minded Scholar always calculates everything), that he takes an umbrella with him in $20 \%$ of the cases. Find the probability of rain.
Solution. It is assumed that the probability of rain is the same every day. Let's denote this probability by $x$.

We will consider the situation where the Scientist is traveling from point A to point B (it doesn't matter which is work and which is home). Let's go back one day and consider three possibilities: there were no umbrellas at point B, there was one umbrella at point B, and there were two umbrellas at point B.

Immediately, we notice that there could not have been two umbrellas at point B, as this would mean the Scientist returned to point A without an umbrella when there were no umbrellas in A at all, which contradicts the condition. Therefore, we are left with two possibilities: there were no umbrellas at B (event $B_{0}$ with probability $p_{0}$) or there was one umbrella at B (event $B_{1}$ with probability $p_{1}$).

We will construct an experiment tree tracing the Scientist's movements yesterday from A to B and then back from B to A.

![](https://cdn.mathpix.com/cropped/2024_05_06_a933850cc044137d5d9dg-08.jpg?height=600&width=1002&top_left_y=1482&top_left_x=607)

The letter $R$ denotes the event "rain," and the letter $D$ denotes the event "no rain." For example, the sequence $S B_{0} R D$ in the tree corresponds to the sequence of events "yesterday there were no umbrellas at B, it rained on the way there, but not on the way back." We will write the corresponding probabilities next to the edges and the number of umbrellas left at point B after the Scientist left under each sequence. 3.[^1]

## $\mathrm{XI}$ Internet Olympiad in Probability Theory and Statistics. Main (correspondence) round. $15.12 .77-21.01 .18$

The event "there are no umbrellas at B today" corresponds to the sequences $S B_{0} R R, S B_{0} D R$, and $S B_{1} D R$. For example, the sequence $S B_{0} D R$ means that there were no umbrellas at B yesterday, so the Scientist brought an umbrella with him despite the lack of rain ($D$), but took it back to A because it rained ($R$). Assuming that the probabilities of the events $B_{0}$ "no umbrellas at B" and $B_{1}$ "one umbrella at B" do not change from day to day, we get the system

$$
\left\{\begin{array}{l}
\mathrm{P}\left(B_{0}\right)=\mathrm{P}\left(S B_{0} R R\right)+\mathrm{P}\left(S B_{0} D R\right)+\mathrm{P}\left(S B_{1} D R\right) \\
\mathrm{P}\left(B_{0}\right)+\mathrm{P}\left(B_{1}\right)=1
\end{array}\right.
$$

which is

![](https://cdn.mathpix.com/cropped/2024_05_06_a933850cc044137d5d9dg-09.jpg?height=131&width=1314&top_left_y=723&top_left_x=451)

It remains to express the probability of the event "the Scientist takes an umbrella with him" through the probabilities of independent events $R$ "rain" and $B_{0}$ "no umbrellas at B":

$$
\mathrm{P}\left(R \cup B_{0}\right)=\mathrm{P}(R)+\mathrm{P}\left(B_{0}\right)-\mathrm{P}(R) \cdot \mathrm{P}\left(B_{0}\right)=0.2,
$$

from which

$$
x+p_{0}-x p_{0}=0.2 ; \quad x+\frac{x}{x+1}-\frac{x^{2}}{x+1}=0.2 ; \quad 2 x=0.2 x+0.2 ; x=\frac{1}{9} .
$$

Answer: $1 / 9$.