What is $\Pr(Y\in[\pi,X+\pi]\mid X)$ if $X \sim U(0,\pi)$ and $Y \sim U(0,2\pi)$? Let $X \sim U(0,\pi)$ and $Y \sim U(0,2\pi)$ be two uniform independent distributions. What is $\Pr(\left.Y\in[\pi,X+\pi]\right|X)$? Intuitively I know  that the result is $\frac{1}{4}$ but how can I formally derive the density function in order to compute the integral?

It depends on the joint distribution of $X$ and $Y$. If almost surely, $Y=2X$, then $\Pr(Y\in [\pi, \pi+X]|X) = \mathbb I(X>\pi/2)$.
If $X$ and $Y$ are independent, $\Pr(Y\in [\pi, \pi+X]|X) = X/(2\pi)$. Indeed, the unconditional probability in this last case is 1/4.  I can't find, off the top of my head, a situation when the conditional probability is 1/4.
