12. Buses. (From 8th grade, 2 points) At the stop near the Absent-Minded Scholar's house, buses of two routes stop: No. 152 and No. 251. Both of them head to the metro station. The interval between buses No. 152 is exactly 5 minutes, and the interval between buses of route No. 251 is exactly 7 minutes. The intervals are strictly observed, but the two routes are not coordinated with each other and their schedules do not depend on each other. At a completely random moment in time, the Absent-Minded Scholar comes to the stop and gets on the first bus that arrives to get to the metro. What is the probability that the Scholar will get on bus No. 251?
Solution. Let's assume that the Absent-Minded Scholar arrived at the bus stop at time 0 minutes. Let $x-5$ and $x$ be the moments when buses of route 152 arrived at the stop, and the Scholar did not catch the first of these buses but could have caught the second, i.e., $x-5<0 \leq x$. From this inequality, we get that $0 \leq x<5$.

![](https://cdn.mathpix.com/cropped/2024_05_06_11f8b78b86004d31b3e4g-09.jpg?height=825&width=668&top_left_y=524&top_left_x=291)

Fig. 9.

Since the Scholar arrives at a random moment (without any preference), we can assume that after the moment of his arrival, which we took as the starting point, bus No. 152 can arrive at the stop at any moment within the interval from 0 to 5 minutes, with no moment being more preferable than any other.

Reasoning similarly for the buses of route 251, we see that after the Scholar's arrival, bus No. 251 can appear at any moment within the interval from 0 to 7 minutes: $0 \leq y<7$.

Thus, the set of elementary events can be represented as a rectangle $O A B C$, bounded by the coordinate axes and the lines $x=5$ and $y=7$ (Fig. 9). Each elementary event is a point with coordinates $(x ; y)$, which is randomly chosen from this rectangle.

The event "The Scholar will board bus No. 251" will occur only if bus No. 251 arrives before bus No. 152, i.e., if $y<x$. Inside the rectangle, this inequality defines the triangle $O D C$. Therefore, the probability of this event is

$$
\mathrm{P}(O D C)=\frac{S_{O D C}}{S_{O A B C}}=\frac{25}{2 \cdot 35}=\frac{5}{14} .
$$

Answer: $5 / 14$.