What is the probability that $X$ is located closer to $Y$ than to $0$?

Solution 1:

The idea is correct but you need to complete the proof. Consider the set of all points $(X,Y)$ satisfying your conditions. All the points must satisfy the conditions $0\le X\le R$, $R\le Y\le 1$ where $R$ is $L/2$ a fixed number between $0, 1$. So the set of all points you consider is a rectangle $U$ of height $1-R$ and width $R$, area $A=R-R^2$.

The subset you are interested in is the set $S$ of points $(X,Y)$ such that $X>Y-X$ or $Y<2X$. So $S$ consists of all points from $U$ under the line $Y=2X$. The area $A_S$ of $S$ depends on $R$ and can be easily found. Indeed, let, say $R=2/3$. Then $S$ is a trapezium with height $1-R=1/3$, and bases $R-R/2=1/3$ and $R-1/2=1/6$, so $A_S=1/2(1/3+1/6)1/3=1/12$. Then the probability is $A_S/A$ (in the above case when $R=2/3$, the probability is $(1/12)/(2/3-4/9)=9/24=3/8$).