1. (mathematics) The probabilities of finding Kuzia the flea at vertices $A, B, C$ and $D$ of the square $A B C D$ are $p_{1}, p_{2}, p_{3}$ and $p_{4}$ respectively $\left(p_{1}+p_{2}+p_{3}+p_{4}=1\right)$. The flea jumps to the nearest vertex of the square, and which of the two it jumps to is completely random and equally likely. Find the probability of finding Kuzia at vertex $B$ after her 2019th jump.
Solution. Let $P_{A}(n), P_{B}(n), P_{C}(n), P_{D}(n)$ be the probabilities of finding the flea at vertices $A, B, C$, and $D$ after $n$ jumps. Then

$$
P_{A}(1)=\frac{1}{2}\left(p_{2}+p_{4}\right)=P_{C}(1), P_{B}(1)=\frac{1}{2}\left(p_{1}+p_{3}\right)=P_{D}(1)
$$

Figure 1 shows the probabilities of events after the first jump.

![](https://cdn.mathpix.com/cropped/2024_05_06_25c5d6100f66704b4a85g-1.jpg?height=382&width=386&top_left_y=1194&top_left_x=869)

Figure 1

After the second jump,

$$
\begin{aligned}
& P_{A}(2)=\frac{1}{2} P_{B}(1)+\frac{1}{2} P_{D}(1)=\frac{1}{2}\left(p_{1}+p_{3}\right)=P_{B}(1)=P_{D}(1)=P_{C}(2) \\
& P_{B}(2)=\frac{1}{2} P_{A}(1)+\frac{1}{2} P_{C}(1)=\frac{1}{2}\left(p_{2}+p_{4}\right)=P_{A}(1)=P_{C}(1)=P_{D}(2)
\end{aligned}
$$

Figure 2 shows the probabilities of events after the second jump.

![](https://cdn.mathpix.com/cropped/2024_05_06_25c5d6100f66704b4a85g-1.jpg?height=391&width=392&top_left_y=2072&top_left_x=889)

Figure 2

The picture at $n=2$ can be obtained from the picture at $n=1$ by rotating it $90^{\circ}$ counterclockwise, and a $180^{\circ}$ rotation leaves the probability distribution unchanged. Therefore, for all odd $n$ the probability distribution coincides with the one shown in Figure 1, and for even $n$ - in Figure 2. Consequently, for $n=2019$ the probability $P_{B}(2019)=P_{B}(1)=\frac{1}{2}\left(p_{1}+p_{3}\right)=0.2$.