Two people agree to meet each other on a particular day, between 5 and 6 PM, They arrive independently on a uniform time between 5 and 6 and wait for 15 mintues. What is the probability that they meet each other ?
As nicely described by Eric Angle, we can assume that each arrival time is uniformly distributed in the interval $[0,1]$.
Let $X$ be the arrival time of A, and $Y$ the arrival time of B. We want $\Pr(|X-Y|\le 1/4)$.
1) Draw the square that has corners $(0,0)$, $(0,1)$, $(1,1)$, and $(0,1)$.
2) Draw the lines with equations $y=x-1/4$ and $y=x+1/4$.
3) We want the probability that the point $(x,y)$ that records the arrival times of A and B lies between the two lines we drew in 2).
4) Since the square has area $1$, and the arrival times are uniform and independent, this probability is the area of the part of the square between $y=x-1/4$ and $y=x+1/4$.
5) Find that area. Note that our region is the square with two (congruent) isosceles right triangles removed. It is easy to find the area of these triangles.
Define 5 PM as $t=0$ and 6 PM as $t=1$. Then $p\left(t\right) = p_A\left(t\right) = p_B\left(t\right) = 1$ for $0 \le t \le 1$ and is zero otherwise.
If person $A$ arrives at time $t_A$, then they will meet each other if $\left|t_B - t_A\right| \le 1/4$. The total probability of them meeting is then $$ \int_0^1 \ dt_A \ p\left(t_A\right) \ P\left(\left|t_B - t_A\right| \le 1/4\right) = \int_0^1 \ dt_A \ P\left(\left|t_B - t_A\right| \le 1/4\right). $$ Now you need to determine $P\left(\left|t_B - t_A\right| \le 1/4\right)$ as a function of $t_A$. It is not always $1/2$. To see this, consider separately the cases $0 \le t_A \le 1/4$, $1/4 \le t_A \le 3/4$, and $3/4 \le t_A \le 1$.