Probability of meeting This is a basic probability question. 

Persons A and B decide to arrive and meet sometime between 7 and 8 pm. Whoever arrives first will wait for ten minutes for the other person. If the other person doesn't turn up inside ten minutes then the person waiting will leave. What is the probability that they will meet? I am assuming uniform distribution for arrival time between 7 pm and 8 pm for both of them.
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.
I hope your are doing well, I tried a new method to resolve the meeting problem but I didn't find the same result.

We have to calculate $P(|X-Y| &lt; 1/6)$ where X and Y are independents uniformly distributed between 0 and 1.

So : $P(|X-Y| &lt; 1/6)  =  \int_{0}^{1} \int_{0}^{1} |x - y| * dx*dy$ 

Then : $P(|X-Y| &lt; 1/6)  =  \int_{0}^{1} \int_{y - \frac{1}{6}}^{y + \frac{1}{6}}  dx*dy$ 

Finally : $P(|X-Y| &lt; 1/6)  = \int_{0}^{1} \frac{1}{3} *dy = \frac{1}{3} $

Please help me to find my error.

Thank you