$X$ and $Y$ are independent uniformly distributed on $[a,b]$ random variables.
How to calculate $$ P (Y<X \, | \, X>\frac{a+b}{2})? $$ My calculations: $$ P (Y<X \, | \, X>\frac{a+b}{2})=\frac{P(X>Y,X>\frac{a+b}{2})}{P(X>\frac{a+b}{2})}= \frac{P(X>\max\{Y,\frac{a+b}{2}\})}{P(X>\frac{a+b}{2})}. $$ It is obviously that the denominator $$ P(X>\frac{a+b}{2})=\frac{1}{2}. $$ But how to calculate the numerator?
On
Cm7F7Bb's graphical method is effective (and simple), but if you wish to do it via integration, a change of variables makes it easy too: $$\small\begin{align}\mathsf P(X>Y\mid X>\tfrac{(a+b)}2) ~=~& \mathsf P(S>T\mid S>\tfrac 12) &~:~& S=\tfrac{(X-a)}{(b-a)}, T=\tfrac{(Y-a)}{(b-a)}\\[0.5ex] ~=~& \int_{1/2}^1\int_0^s ~\mathsf d t ~\mathsf d s\Big/\tfrac 12 &~\because~&S,T\overset{\rm iid}{\sim}\mathcal U(0,1)\\[0.5ex]~=~& \dfrac 34\end{align}$$
I think there is a nice way to find this probability graphically. Unfortunately, my computer drawing skills are terrible. But the picture is as follows.
The rectangle represents all possible outcomes. The vertical line in the middle splits the outcomes into two parts: $X<(a+b)/2$ on the left and $X>(a+b)/2$ on the right. The diagonal line splits the outcomes again into two parts: $X>Y$ below the line and $X<Y$ above the line. Hence, $$ P(Y<X\mid X>(a+b)/2)=\frac34. $$