Meeting for lunch

Idea. The pair of arrival times is a uniform point in the unit square, so the probability is an area. The friends miss each other in two corner triangles, which are easy to measure.

Step 1. With $X$ and $Y$ the arrival times in hours after 12:00, the point $(X,Y)$ is uniform in the unit square. They meet exactly when the later one arrives within 15 minutes of the earlier one: $|X-Y|\le \frac{1}{4}$.

Step 2. They miss when $Y>X+\frac{1}{4}$ or $X>Y+\frac{1}{4}$. Each of these is a right triangle with legs $\frac{3}{4}$, of area $\frac{1}{2}\cdot {\left(\frac{3}{4}\right)}^2=\frac{9}{32}$.

Step 3. $$\mathbb{P}(\text{meet})=1-2\cdot \frac{9}{32}=1-\frac{9}{16}=\frac{7}{16}.$$

In general. If each waits a fraction $w$ of the hour, they meet with probability $1-(1-w{)}^2$: for 30 minutes this is $\frac{3}{4}$.