10. Lottery. (From 7th grade, 3 points) It so happened that the Absent-Minded Scholar had only 20 rubles left, and he needed to buy a bus ticket to go home. The ticket costs 45 rubles. Near the bus stop, instant lottery tickets are sold for exactly 10 rubles each. With a probability of \( p = 0.1 \), a ticket contains a win of 30 rubles, and there are no other prizes. The Scholar decided to take a risk, as he had nothing to lose. Find the probability that the Absent-Minded Scholar will be able to win enough money to buy a bus ticket.
Solution. As a result of one or several consecutive games, the Scientist may end up with no money at all (state 0), 10 rubles, 20, 30, 40 rubles, or he may win enough money to buy a bus ticket (state A). The possible transitions between states are schematically represented by a graph (Fig. 7). The probabilities of transitions from one state to another are labeled on the corresponding edges of the graph. Let \( x_{1} \) be the probability of eventually transitioning from state 10 to state A. Similarly, let \( x_{2}, x_{3} \), and \( x_{4} \) be the probabilities of reaching state A from states 20, 30, and 40, respectively. We are interested in \( x_{2} \).

![](https://cdn.mathpix.com/cropped/2024_05_06_11f8b78b86004d31b3e4g-06.jpg?height=572&width=734&top_left_y=642&top_left_x=1186)

Fig. 7. Graph of transitions from state to state

It is impossible to transition directly from state 10 to state A, but one can reach state 30 with probability \( p \), and from there to A. Therefore,

\[
x_{1} = q \cdot 0 + p x_{3}
\]

where \( q = 1 - p \) is the probability that the ticket will not contain a win.

Similarly analyzing states 20, 30, and 40, we obtain three more equations, which together with the first one form the system

\[
\left\{\begin{array}{l}
x_{1} = p x_{3} \\
x_{2} = p x_{4} + q x_{1} \\
x_{3} = p + q x_{2} \\
x_{4} = p + q x_{3}
\end{array}\right.
\]

Substituting the expressions for \( x_{1} \) and \( x_{4} \) into the second equation:

\[
x_{2} = p \left( p + q x_{3} \right) + q p x_{3} = p^{2} + 2 p q x_{3}
\]

Now substituting the expression for \( x_{3} \) and we get

\[
x_{2} = p^{2} + 2 p q \left( p + q x_{2} \right)
\]

from which

\[
x_{2} = \frac{p^{2}(1 + 2 q)}{1 - 2 p q^{2}} = \frac{p^{2}(1 + 2 - 2 p)}{1 - 2 p (1 - p)^{2}} = \frac{3 p^{2} - 2 p^{3}}{1 - 2 p + 4 p^{2} - 2 p^{3}}
\]

For \( p = 0.1 \), we find: \( x_{2} = \frac{14}{419} \approx 0.033 \).

Answer: approximately \( 0.033 \).