TITLE: Sum of three independent random variables
QUESTION [5 upvotes]: Let $X$, $Y$, and $Z$ be independently and identically distributed variables, each uniformly distributed between $0$ and $2$. What is the probability that $X+Y+Z\leq2$?
All the other answers on similar questions refer to various things like "convolutions" that I've never heard of or use integrals and change the limits from $[-\infty,+\infty]$ to other things etc. so I'm very confused and don't understand any of it. In particular, I thought we could just work out $P(X+Y\leq2-Z)=\frac{1}{8}(2-Z)^2$ and thus do $\int_{0}^{2}\frac{1}{8}(2-Z)^2 \text{d}Z$, and I don't understand why this doesn't give the right answer of $\frac{1}{6}$. Please would someone take the time to write out all the steps with full explanation for someone who has only basic knowledge of probability.

REPLY [2 votes]: For those who prefer a geometric proof:
The random points are uniformly distributed in $X,Y,Z$ space within a cube of side 2.
$X+Y+Z=2$ defines a plane which intersects the cube at the points (2,0,0),(0,2,0) and (0,0,2).  This is an equilateral triangle of side $2\sqrt 2$ and area $2\sqrt{3}$.
The centre of the triangle is at (2/3,2/3,2/3) which is $2/\sqrt{3}$ from the origin.
The points for which $X+Y+Z\leq2$ are those in the triangular pyramid whose base is this triangle and whose apex is the origin. This has volume ${1 \over 3} A h={1 \over 3}{2\sqrt 3}{2 \over \sqrt 3}={4 \over 3}$. 
Dividing by the total volume of 8 gives the probability ${1 \over 6}$ 
(The integrals are hidden in the formulae for the area ${ 1 \over 2} base \times height$ of a triangle and the volume ${1 \over 3} area \times height$  for a triangular pyramid.)