208. On a line, points $0, \pm 1, \pm 2, \pm 3, \ldots$ are marked.

![](https://cdn.mathpix.com/cropped/2024_05_21_a55bead9955e32b1e916g-169.jpg?height=160&width=711&top_left_y=1341&top_left_x=293)

Fig. 80.

(Fig. 80). A particle, upon reaching point $n$, moves to point $n+1$ with probability $\frac{1}{2}$ and to point $n-1$ with probability $\frac{1}{2}$ in one unit of time. Initially, the particle is at point 0. Find:

a) the probability $x$ that the particle will eventually visit point 1;

b) the probability $y$ that the particle will eventually visit point -1;

c) the probability $z$ that it will eventually return to point 0 (i.e., visit point 0 at a moment different from the initial one).
208. Since all points are completely equal, $x$ is the probability that a particle that has left point $n-1$ will ever visit $n$. Similarly, the probability that a particle that has left point $n+1$ will visit $n$ is $y$.

- Suppose the particle starts from point 0 and consider the events:

$A$: "the particle will ever return to point 0", $B_{1}$: "the particle will move to point 1 on the first step",

$B_{2}$: "the particle will move to point -1 on the first step".

By the formula of total probability (property 7 §)

$$
\mathbf{P}(A)=\mathbf{P}\left(B_{1}\right) \mathbf{P}\left(A / B_{1}\right)+\mathbf{P}\left(B_{2}\right) \mathbf{P}\left(A_{1} B_{2}\right)
$$

If condition $B_{1}$ is satisfied, it means that the particle is in point 1 after one step. For event $A$ to occur under this condition, the particle must, having left 1, eventually reach 0. The probability of this is $y$. Thus, $\mathbf{P}\left(A / B_{1}\right)=y$. Similarly, $\mathbf{P}\left(A / B_{2}\right)=x$. Further, $\mathbf{P}(A)=z, \mathbf{P}\left(B_{1}\right)=\mathbf{P}\left(B_{2}\right)=\frac{1}{2}$. Substituting these values into formula (1), we get

$$
z=\frac{1}{2} y+\frac{1}{2} x
$$

Furthermore, it is obvious that $x=y$, from which $z=x$.

To calculate $x$, consider the event $C$: "a particle that has left 0 will eventually reach 1". By property 7

$$
x=\mathbf{P}(C)=\mathbf{P}\left(B_{1}\right) \mathbf{P}\left(C / B_{1}\right)+\mathbf{P}\left(B_{2}\right) \mathbf{P}\left(C / B_{2}\right)
$$

We have $\mathbf{P}\left(B_{1}\right)=\mathbf{P}\left(B_{2}\right)=\frac{1}{2}, \mathbf{P}\left(C / B_{1}\right)=1$. Further, $\mathbf{P}\left(C / B_{2}\right)$ is the probability that a particle from -1 will eventually reach 1. Let's calculate this probability. So, the particle is at -1. We need to find the probability of event $D$: "the particle will eventually visit 1". Introduce the event $F$: "the particle will eventually visit 0". Obviously, $F D=D$ (indeed, the occurrence of $D$ is equivalent to the simultaneous occurrence of both events $F$ and $D$). Therefore, by property 6

$$
\mathbf{P}(D)=\mathbf{P}(F D)=\mathbf{P}(F) \mathbf{P}(D / F)
$$

But $\mathbf{P}(F)=x, \mathbf{P}\left(D_{/} F\right)=x$. Therefore,

$$
\mathbf{P}(D)=x^{2}
$$

Finally, from formulas (3), (4) we derive

$$
\begin{gathered}
x=\frac{1}{2}+\frac{1}{2} x^{2} \\
x^{2}-2 x+1=0 \\
(x-1)^{2}=0
\end{gathered}
$$

from which $x=1$. Therefore, $y=z=1$.