Find the probability that $(a-c)^2+(b-d)^2<1/4$ I was trying to solve for $P((a-c)^2+(b-d)^2<1/4)$ where $a,b,c,d$ are independent and uniformly distributed on $[0,1]$. One thing that I found online was $a-c$ has a triangular distribution.
I think that you have to find the probability cumulative function, but I don't know how to do that. Though I am willing to look through some resources which could point to how to solve these kinds of problems. I briefly looked through a probability and statistics textbook, but couldn't find what I needed.

The quantity $(a-c)^2 + (b-d)^2$ is the square of the distance between the coordinates $(a,b)$ and $(c,d)$, where we clearly have both points independent and identically distributed uniformly in the unit square $[0,1]^2$.  Thus, your probability is geometrically equivalent to asking for the probability that two points, picked uniformly at random in the unit square, will be less than $1/2$ units apart.  How might you reason from this point forward?
