Don't be late

The following graph plots the time that Ana shows up on the $x$ -axis and the time that Bastian shows up on the $y$ -axis, and shades green the points that Ana and Bastian will meet and red the points that Ana and Bastian will not meet. (For example, the point $(\text{1:15}, \text{1:20})$ is shaded green since Bastian would arrive only $5$ minutes after Ana and they would meet, but the point $(\text{1:15}, \text{1:45})$ is shaded red since Bastian would be $30$ minutes after Ana and they would not meet.)

Then the probability that they will meet is equivalent to the area of the green part of the graph, which is the area of the whole graph minus the area of the red triangles, or $P = 1 - 2(\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{3}{4}) = 1 - \frac{9}{16} = \frac{7}{16}$ . Therefore $a = 7$ and $b = 16$ , and $a + b = \boxed{23}$ .

It doesn't matter who arrives first.

Case 1: The first person (let it be Ana (A)) arrives in the first 45 minutes

Probability of that would be $\frac {45}{60}$ . Now B will get a time slot of exactly 15 minutes to arrive, depending on when A comes. For example, if A came at 1:39, B can come anywhere between 1:40 to 1:55. Hence, B will have to arrive in those 15 min, and that would have a probability of $\frac {15}{60}$ .

Total probability for this case = $\frac {3}{4} × \frac {1}{4}$

Case 2: Ana arrives in the last 15 minutes .

Probability of that would be $\frac {15}{60}$ . Now, since B is coming after A, B can come at anytime between 1:45 to 2:00, and Ana would still be waiting. Hence B's correct arrival would have a probability of 1.

Total probability for this case = $\frac {1}{4} ×1$

Adding up the two probabilities, we get the answer as $\dfrac {7}{16}$ .