I Prefer That You Wait

In the following diagram, it is represented on the vertical axis when the person $A$ can arrive to the particular place depending on the hour (time) the person $B$ arrives to that particular place which it is represented on the horizontal axis.Furthemore, on the horizontal axis is represented when the person $B$ can arrive to the particular place depending on the time that the person $A$ arrives. Assuming that the square has area $1$ then the probability that $A$ and $B$ meet each other is represented by colored red and yellow areas, which are $\frac{5}{18},\space \frac{7}{32}$ . And therefore, the probability for $A$ meets $B$ is $\frac{5}{18} + \frac{7}{32} = \frac{143}{288} \Rightarrow 143 + 288 = 431$

It is easier to calculate the complement, i.e. the likelihood of A and B missing each other. This can happen in two ways: either A is more than 20 minutes later than B, or B is more than 15 minutes later than A.

These situations can be represented in a $(t_A, t_B)$ -diagram (see Guillermo Templado's solution), where they have the shape of right triangles. The probabilities correspond to the areas: $p_{A>B} = \frac12\cdot \frac23\cdot \frac23 = \frac29;$ $p_{B>A} = \frac12\cdot \frac34\cdot \frac34 = \frac9{32}.$ Alternatively, using calculus: $p_{A>B} = \int_{0}^{2/3} dt_B \int_{t_B+1/3}^1 dt_A = \int_{0}^{2/3} \left(\tfrac23 - t_B\right)dt_B = \left.\tfrac23t_B-\tfrac12t_B^2\right|_0^{2/3} = \frac29;$ $p_{B>A} = \int_{0}^{3/4} dt_A \int_{t_A+1/4}^1 dt_B = \int_{0}^{3/4} \left(\tfrac34 - t_A\right)dt_A = \left.\tfrac34t_A-\tfrac12t_A^2\right|_0^{3/4} = \frac9{32}.$ The total is $p_{miss} = \frac29 + \frac9{32} = \frac{64 + 81}{288} = \frac{145}{288};$ Finally, taking the complement, $p_{meet} = 1 - \frac{145}{288} = \frac{143}{288},$ making the answer to be submitted $\boxed{431}$ .