Probability of meeting

My favourite way of solving uniform distribution related problems is to try and think of them in terms of relative volume/area/length. For instance, take a continuous uniform distribution on $[0, 1]$. The probability that $X \in (a, b)$, where $0 \le a \le b \le 1$, and $X$ is a random variable drawn from that distribution, is just the length of the interval $(a, b)$.

Let's do another example. Let $X$ be a uniform distribution over $(0, 5)$. What is the probability that $X \in (a, b)$, where $0 \le a \le b \le 5$?. It is just $\frac{\text{length of interval $(a,b)$}}{\text{length of interval (0, 5)}} = \frac{\text{length of interval $(a,b)$}}{5}$.

This generalizes as well. Let there be a uniform distribution over a square $S$. What is the probability that a randomnly drawn point will be in a particular region $H$?. The answer is $\frac{\text{area of $H$}}{\text{area of $S$}}$.

Let's apply this technique to the current problem.

The above picture represents the problem. It is easy to see that the distribution in the question is equivalent to placing a uniform distribution over the square in the diagram, where the times of arrival are the co-ordinates of a randomnly drawn point from the distribution. The shaded are is the favourable area. It represents points $(x, y)$ which are within $10$ of each other. Thus what we need is the area of that region divided by the area of the whole region.