Probability that the closest integer to (y/x) being equal to 2.

We are interested when $\frac{3}{2} < \frac{y}{x} < \frac{5}{2} \Rightarrow \frac{3x}{2} < y < \frac{5x}{2}$ . Plotting these two lines over the square region defined by $(0,0); (1,0); (1,1); (0,1)$ will yield the feasible triangular region defined by the vertices: $(0,0); (\frac{2}{5},1); (\frac{2}{3},1)$ . Our required probability is equal to:

$P = \frac{A_{triangle}}{A_{square}}$

where $A_{triangle} = \frac{1}{2} \cdot \begin{vmatrix} 1 & 0 & 0 \\ 1 & 2/3 & 1 \\ 1 & 2/5 & 1 \end{vmatrix} = \frac{1}{2} (\frac{2}{3} - \frac{2}{5}) = \frac{2}{15}.$

Hence, $P = \frac{2/15}{1} = \frac{2}{15} \Rightarrow 2 + 15 = \boxed{17}.$