Two circles of radius $1$: the center of circle $A$ is chosen uniformly from the segment joining $(0,0)$ to $(3,0)$, and the center of circle $B$ uniformly and independently from $(0,1)$ to $(3,1)$. What is the probability that $A,B$ intersect?

1. Let the centers be $(x,0)$ and $(y,1)$ with $x,y$ uniform on $[0,L]$. The circles intersect iff the distance between the centers is less than $2$: $(x-y)^2+1<4$, i.e. $|x-y|<\sqrt3$. In the $L\times L$ square of pairs $(x,y)$, the complement $|x-y|\ge\sqrt3$ is two right isosceles triangles with legs $L-\sqrt3$, so the probability is $1-\left(\dfrac{L-\sqrt3}L\right)^2$ (for $L>\sqrt3$).
2. With $L=3$: probability $=1-\left(\dfrac{3-\sqrt3}3\right)^2=\dfrac{6\sqrt3-3}9=\dfrac{2\sqrt3-1}3$.
3. The answer is $\boxed{\dfrac{2\sqrt3-1}3}$.