Anya, Borya, and Vasya decided to go to the "Fir Tree". They agreed to meet at the bus stop, but they don't know who will arrive at what time. Each of them can arrive at a random time between 15:00 and 16:00. Vasya is the most patient of all: if he arrives and there is neither Anya nor Borya at the bus stop, he will wait for 15 minutes, and if he doesn't wait for anyone, he will go to the "Fir Tree" alone. Borya is less patient: he will wait only 10 minutes. Anya is the most impatient: she won't wait at all. However, if Borya and Vasya meet, they will wait for Anya until 16:00. What is the probability that they will all go to the "Fir Tree" together?
Since Anya will not wait at all, the boys will go to the "Fir Tree" together only if Anya comes last. Clearly, these events are independent, so $P(\{\text{all three will go to the "Fir Tree" together}\}) = P(\{\text{Anya will come last}\}) P(\{\text{Borya and Vasya will meet}\})$.

The first probability is clearly $1 / 3$.

The second probability is found geometrically (see the figure):

![](https://cdn.mathpix.com/cropped/2024_05_08_293b4772a418c01a5915g-01.jpg?height=595&width=649&top_left_y=963&top_left_x=710)

The figure outlined in black is the set of arrival times for both, during which they will meet. Thus, the probability is the ratio of the area of the highlighted part to the area of the $12 \times 12$ square. As a result, we get $\frac{12^{2}-\frac{9^{2}}{2}-\frac{10^{2}}{2}}{12^{2}} \approx 0.3715$.

Therefore, the desired probability is $1 / 3 \cdot 0.3715 \approx 0.124$.

## Answer

$\approx 0.124$.

## Problem