A point $(x,y)$ is randomly picked from inside the rectangle with  vertices $(0,0)$, $(4,0)$, $(4,1)$, and $(0,1)$. What is the probability that $x<y$? 
$\mathrm{(A) \ } \frac{1}{8}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{3}{8}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{3}{4}$
The rectangle has a width of $4$ and a height of $1$. 
The area of this rectangle is $4\cdot1=4$. 
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 $\frac{1}{2}\cdot1^{2}=\frac{1}{2}$
Therefore, the probability that $x<y$ is $\frac{\frac{1}{2}}{4}=\frac{1}{8} \Rightarrow \boxed{\mathrm{(A)}\ \frac{1}{8}}$