Probability $a+b>c$ for $a,b,c$ uniform in $[0,1]$ Let $a,b,c$ be drawn independently and uniformly at random from $[0,1]$. What is the probability that $a+b>c$?
We can represent $(a,b,c)$ as points in a unit cube, but as we want to consider the quantity $a+b$, it's not clear what to do.
 A: If you want to go the geometric route, consider that you're looking for what portion of the unit cube is "below" the plane $a+b-c=0$. Now, what is that plane? Well, it contains $(0,0,0)$, and $(1,0,1)$ and $(0,1,1)$ - it, in a way, cuts off the corner $(0,0,1)$. A cube is drawn below, with the intersection of the desired plane with the faces drawn as dashed lines:

Now, we are interested in the volume "below" the plane. The area above is a tetrahedron, with a right triangle base. The base has area $\frac{1}2$ and the tetrahedron has a height of $1$, so the area of the tetrahedron is $\frac{1}6$. Thus, as the area of the cube is $1$, the area below the plane is $1-\frac{1}6=\frac{5}6$, which is the desired probability.
A: Let's do this directly:  For us to have $c > a+b$ we must have 
$$
\left\{
\begin{align}
0 \leq a < 1 \\
0 \leq b < 1-a \\
a+b < c \leq 1
\end{align}
\right.
$$
where the second condition is the key to the p[roblem:  If $a+b$ adds to more than $1$, you can't meet the condition for any $c$, so we have to carefully avoid counting that region.
The probability integral becomes
$$
\int_{a=0}^1
\int_{b=0}^{1-a}
\int_{c=a+b}^1 dc\, db \, da \\
=\int_{a=0}^1
\int_{b=0}^{1-a}
(1-a-b) \,db \, da \\
=\int_{a=0}^1
 \left( (1-a) - a(1-a) -\frac{1}{2}(1-a)^2 \right)\, da \\
=\frac{1}{2} \int_{a=0}^1 (1-a)^2 \, da = \frac{1}{2} \int_{x=1}^0 x^2 \, (-1)dx
=\frac{1}{6}
$$
