Question: Given three IID random variables $X,Y,Z$ uniformly distributed on $[0,1].$ Calculate $\mathbb{P}(X+Y>Z).$
My attempt: Let $f_{XYZ}$ be the joint density function of $X,Y,Z.$ by IID, $f_{XYZ} = f_X\cdot f_Y \cdot f_Z = 1$ where $f_X,f_Y,f_Z$ are probability density function of $X,Y,Z$ respectively. It follows that $$\begin{align*} \mathbb{P}(X+Y>Z) & = \mathbb{E}[ 1_{\{X+Y>Z \}} ] \\ & = \int_{-\infty}^\infty \int_{-\infty}^\infty \int_{-\infty}^\infty 1_{\{X+Y>Z \}}(x,y,z) f_{XYZ}(x,y,z) dzdxdy \\ & = \int_0^1 \int_0^1 \int_0^{x+y} dzdxdy \\ & = \int_0^1 \int_0^1 (x+y) dxdy \\ & = \int_0^1 0.5 + y dy \\ & = 0.5 + 0.5 \\ & = 1 \end{align*}.$$ However, suggested answer is $\frac{5}{6}.$ I am not sure where is my mistake.