Three friends, Xenia, Yvett, and Zita, each arrive at an internet café that is open from noon to 8 PM, at independently and randomly chosen times, each staying for one hour. What is the probability that all three of them meet?
I. solution. Let's assume that Xénia, Yvett, and Zita arrive at the internet café $x, y$, and $z$ hours after 12 o'clock. Since each of them stays exactly one hour and the café is open until 8 PM, they all need to arrive by 7 PM, i.e., $0 \leq x \leq 7, 0 \leq y \leq 7$, and $0 \leq z \leq 7$. We can represent the arrival times of the three girls as ordered triples of points $P(x; y; z)$ in space, and the measure of the event space as the volume of the set of points. In this case, the entire event space is a cube with an edge length of 7 units, whose volume is 343 cubic units.

![](https://cdn.mathpix.com/cropped/2024_05_02_2e51b95f23022e6ad5d2g-1.jpg?height=545&width=616&top_left_y=394&top_left_x=749)

The conditions for the girls to meet are:

$$
|x-y| \leq 1 \Longleftrightarrow x-1 \leq y \leq x+1, \quad|y-z| \leq 1 \Longleftrightarrow y-1 \leq z \leq y+1
$$

and

$$
|z-x| \leq 1 \Longleftrightarrow z-1 \leq x \leq z+1
$$

The condition for the three girls to meet is that the point $P(x; y; z)$ representing their arrival times is located within the cube with an edge length of 7 units, between the planes $y=x-1$ and $y=x+1$, $z=y-1$ and $z=y+1$, and $x=z-1$ and $x=z+1$. This region represents the favorable cases and can be divided into a cube with a side length of 1 unit and 3 oblique prisms with a square base, each with a height of 6 units. The diagram shows one such square-based oblique prism (prism ABCDLMNK).

Thus, the volume of the region is $1 + 3 \cdot 6 = 19$ cubic units. The desired probability is: $p = \frac{19}{343}$.

II. solution. Divide one hour into $n$ small time units. Since the girls each stay for one hour at the internet café, they cannot arrive in the last hour, leaving 7 hours, or $7n$ time units. From these, we select one for each girl to enter the café. The total number of cases can be easily counted: independently, each can choose any of the $7n$ time points, so this is $(7n)^3 = 343n^3$ cases.

The favorable cases are when all three arrive within the same hour. Since the café opens at noon, if the last to arrive does so within the first hour, they will definitely meet, which is $n^3$ cases, as each of the three can choose any of the first $n$ time points.

For the other cases, consider when the last to arrive arrives: this can be in $6n$ different ways. If this last person arrives alone, the other two can choose from the previous $n-1$ time points, and it matters who was the last, which can happen in 3 ways, so this is $3(n-1)^2$ cases. If two arrive last at the same time, the third can arrive in $(n-1)$ time points, and again, it must be multiplied by 3 because the two who arrived last can be any of the three, which is $3(n-1)$ cases. When all three arrive in the same time unit, this is one case. The sum of these must be multiplied by $6n$, because the arrival time of the last person can be any of $6n$.

This is a total of

$$
6n\left(3(n-1)^2 + 3(n-1) + 1\right) = 18n^3 - 18n^2 + 6n
$$

cases.

The probability that all three meet is the number of favorable cases divided by the total number of cases:

$$
\frac{n^3 + 18n^3 - 18n^2 + 6n}{343n^3} = \frac{19n^3 - 18n^2 + 6n}{343n^3} = \frac{19}{343} - \frac{18}{343n} + \frac{6}{343n^2}
$$

We need to examine what this expression approaches as $n$ approaches infinity, since in reality there are infinitely many time points in the 7 hours when they can arrive. Since the last two terms approach zero, the entire expression approaches $\frac{19}{343}$, so this is the probability that all three will meet.