A probability problem by Josh Rowley

Consider $A$ , the region bounded by the equations $y=4, x=4, x+y=1$ and the two axes in the coordinate plane. Then $a+b \geq 1$ iff $(a, b)$ lies in $A$ . It has area $15.5$

Consider $B$ , the region bounded by the lines $y=x+1, y=x-1$ . Then $(a, b)$ is in $B$ iff $a-b \geq 1$ or $b-a \geq 1$

The intersection of $A$ and $B$ has area $15.5-0.5×3×3-0.5×3×3=6.5$ (It is just 15.5 minus the area of two isoceles right angled triangles of base $4-1=3$ )

Hence probability is $\frac {6.5}{15.5}=\frac {13}{31}$ hence the answer is $13+31=44$ .

thanks for your beautiful problem.It was a pleasure solving it.

0<x,y<0 and x+y>1give us a pentagon whit vertex in (1; 0), (0; 1), (0; 4), (4; 0), (4; 4). |x-y|-1<1give us the part of the plane between y-x-1=0 and y-x+1. The ratio between the two areas is 13/31