Assume we have two circles, one with radius 2 and one with radius 3, and we randomly choose a dot in both circles. What's the probability that the dot in the first circle is closer to the center than the one in the second circle?
I thought to approach this with the distribution density functions: for the first circle, lets call it X and for the second circle Y: $$ f_x = 1/(4\pi) , f_Y = 1/(9\pi) $$ and then using convolution. Thats where I got stuck, any ideas?
Denote by $R$ and $S$ the distances of the two dots from the centers of the two disks. Then $$P[R\leq r]={r^2\pi\over 4\pi}\quad(0\leq r\leq 2),\qquad P[S\leq s]={s^2\pi\over 9\pi}\quad(0\leq s\leq 3)\ .$$ It follows that the corresponding probability densities are given by $$f_R(r)={dP[R\leq r]\over dr}={r\over2}\quad(0\leq r\leq 2),\qquad f_S(s)={2s\over9}\quad(0\leq s\leq 3)\ .$$ (Your densities were wrong!) Therefore the probability $P[R\leq S]$ is given by $$P[R\leq S]=\int_0^2 \int_r^3 f_R(r)\>f_S(s) \>ds\>dr=\int_0^2 \int_r^3 {rs\over9}\>ds\>dr={7\over9}\ .$$ Here the domain of integration was the part of the $(r,s)$-rectangle $[0,2]\times[0,3]$ where $r\leq s$.