Find probability from uniform distribution Let $X$, $Y$ be two independent random variables from $U(0,1)$. Then find $P[Y>(X-1/2)^2]$.
I initially  tried drawing the figure but that seemed complicated. I then took expectation on both sides and got $P[E(Y)>V(X)]$. Am I right?

This ends up being the area above the curve
\begin{equation}
Y=(X−\frac{1}{2})^2
\end{equation}

This can be found by integration
\begin{equation}
P[Y>(X−\frac{1}{2})^2] = \int_{0}^{1}\int_{(X−\frac{1}{2})^2}^{1} 1\times1 \,dydx
\end{equation}
\begin{equation}
= \int_{0}^{1}{1-(X−\frac{1}{2})^2} \,dx
\end{equation}
\begin{equation}
=  \Big[X-\frac{1}{3}\times(X−\frac{1}{2})^3\Big]_0^1  
\end{equation}
\begin{equation}
=  \Big(1-\frac{1}{3}\times(1−\frac{1}{2})^3\Big) - \Big(0-\frac{1}{3}\times(0−\frac{1}{2})^3\Big)
\end{equation}
\begin{equation}
=  \frac{23}{24}-\frac{1}{24}
\end{equation}
\begin{equation}
=  \frac{22}{24}
\end{equation}
