Geometric Probability

Rewriting the probability $P$ into a more useful form leads to $|a-b| > \frac{a + b}{2}$ This implies two equations: $\begin{aligned} a - b & > \frac{ a+b}{2} \\ a-b & < -\frac{a + b}{2} \end{aligned}$ Solving in terms of $a$ leads to $b < \frac{a}{3}$ and $b > 3a$ . Utilizing geometric probability we can generalize this problem using the unit square as shown With $b$ on the $y$ -axis and $a$ on the $x$ -axis, the regions defined by the previous inequalities are the ones touching their respective axes. Thus, the area covered by the sum of the regions is just the sum of the areas of the two triangles both with lengths $1$ and $\frac{1}{3}$ . This implies that the total area covered by these equations is $\frac{(1)(\frac{1}{3})}{2} + \frac{(1)(\frac{1}{3})}{2} = \frac{1}{3}$ . Thus, the probability of the absolute difference between $a$ and $b$ being larger than their average is $\frac{1}{3}$ . Therefore, the answer is $x + y = 1 + 3 = \boxed{4}$