A husband and wife agree to meet at a street corner between 4 and 5 o'clock to go shopping together. The one who arrives first will await the other for 15 minutes, and then leave. What is the probability that the two meet within the given time interval, assuming that they can arrive at any time with the same probability? 
Denote by $x$, respectively, $y$, the fraction of the hour when the husband, respectively, wife, arrive. The configuration space is the square $$ D=\{(x, y) \mid 0 \leq x \leq 1,0 \leq y \leq 1\} . $$In order for the two people to meet, their arrival time must lie inside the region$$ D_{f}=\left\{(x, y)|| x-y \mid \leq \frac{1}{4}\right\} . $$The desired probability is the ratio of the area of this region to the area of the square.The complement of the region consists of two isosceles right triangles with legs equal to $\frac{3}{4}$, and hence of areas $\frac{1}{2}\left(\frac{3}{4}\right)^{2}$. We obtain for the desired probability$$ 1-2 \cdot \frac{1}{2} \cdot\left(\frac{3}{4}\right)^{2}=\frac{7}{16} \approx 0.44 . $$