imc

The rectangle has a width of $4$ and a height of $1$. The area of this rectangle is $4\cdot 1=4$. The line $x=y$ intersects the rectangle at $(0,0)$ and $(1,1)$. The area which $x<y$ is the right isosceles triangle with side length $1$ that has vertices at $(0,0)$, $(1,1)$, and $(0,1)$. The area of this triangle is $\frac{1}{2}\cdot 1^2=\frac{1}{2}$ Therefore, the probability that $x<y$ is $\frac{\frac{1}{2}}{4}=\frac{1}{8}\Rightarrow \boxed{\frac{1}{8}}$