Anna and Béla agree to meet between 5 and $\frac{1}{2} 6$ in the afternoon. What is the probability that the one who arrives first does not wait more than 10 minutes for the other, if both adhere to what they agreed upon?
Let $E$ be the event that the earlier arriving person does not wait longer than 10 minutes, that is, $\frac{1}{6}$ of an hour. Let one person arrive $x$ hours after 5 PM, and the other $y$ hours after 5 PM. Since they adhere to what they agreed upon, $0 \leq x \leq \frac{1}{2}$ and $0 \leq y \leq \frac{1}{2}$. Any possible event can be represented by a point $(x ; y)$ on the rectangular coordinate system of the square. The area of the square is $\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$. The event $E$ occurs if and only if the point $(x ; y)$ on the square satisfies $|x-y| \leq \frac{1}{6}$, that is, $y \leq x+\frac{1}{6}$ and $y \geq x-\frac{1}{6}$. Therefore, the probability of the event $E$ is proportional to the area of the intersection of the strip between the lines $y=x+\frac{1}{6}$ and $y=x-\frac{1}{6}$ and the square. This area is $\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{3}\right)^{2}=\frac{5}{36}$.

The probability of the event $E$ is: $P(E)=\frac{5}{36}: \frac{1}{4}=5 / 9$.

Based on the work of Zoltán Szilasi (Debrecen, KLTE Gyak. Gimn., 9th grade) and Tamás Birkner (Budapest, Deutsche Schule, 5th grade)

![](https://cdn.mathpix.com/cropped/2024_05_02_12bf2aa6c06498063932g-1.jpg?height=422&width=504&top_left_y=781&top_left_x=797)