Kolya and Zhenya agreed to meet at the metro in the first hour of the day. Kolya arrives at the meeting place between noon and one o'clock, waits for 10 minutes, and leaves. Zhenya does the same.

a) What is the probability that they will meet?

b) How will the probability of meeting change if Zhenya decides to arrive before half past twelve, while Kolya still arrives between noon and one o'clock?

c) How will the probability of meeting change if Zhenya decides to arrive at any time from 12:00 to 12:50, while Kolya still arrives between 12:00 and 13:00?
a) Let's graphically represent the arrival times. We will plot Zhenya's arrival time on the X-axis and Kolya's arrival time on the Y-axis.

![](https://cdn.mathpix.com/cropped/2024_05_08_293b4772a418c01a5915g-11.jpg?height=783&width=1234&top_left_y=2067&top_left_x=412)

The figure outlined in black represents the set of arrival times for both that result in a meeting. The probability is the ratio of the times of arrival that satisfy the meeting condition to the set of all possible arrivals. Therefore, we need to divide the area of the highlighted part by the area of the $6 \times 6$ square.

b) Let's graphically represent the arrival times as in part a). In this case, the possible arrival time for Zhenya will be in the interval from 0 to 30 minutes. The area of the highlighted figure relative to the area of the entire figure $6 \times 3$ is $5.5: 18 = 11 / 36$. Thus, the probability remains unchanged.

![](https://cdn.mathpix.com/cropped/2024_05_08_293b4772a418c01a5915g-12.jpg?height=777&width=1218&top_left_y=323&top_left_x=426)

c) From the figure corresponding to the new condition, it is clear that the probability is $9.5: 30 = 19 / 60 > 11 / 36$.

![](https://cdn.mathpix.com/cropped/2024_05_08_293b4772a418c01a5915g-12.jpg?height=778&width=1234&top_left_y=1183&top_left_x=412)

## Answer

a) ${ }^{11} / 36 ;$ b) ${ }^{11 / 36}$; c) ${ }^{19} / 60$.