Sum of three independent random variables
The sum of a given number of i.i.d. continuous uniform variables, on the support $[0,1]$ follows the Irwin-Hall distribution.
Since you are considering variables on the interval $[0,2]$, it is just a matter of re-scaling.
Idea:
- $X \sim {\rm Unif}([0,2])$
- In the required event, $0 \le Y \le 2 - X$.
- Determine the value of $Z$ similarily: $0 \le Z \le 2 - X - Y$.
$$\begin{aligned} & P(X+Y+Z \le 2) \\ &= \int_0^2 \int_0^{2-x} \int_0^{2-x-y} \left(\frac12\right)^3 \ dz \, dy \, dx \\ &= \left(\frac12\right)^3\int_0^2 \int_0^{2-x} (2-x-y) \, dy \, dx \\ &= \left(\frac12\right)^3\int_0^2 \int_0^{2-x} y \, dy \, dx \quad (\int_0^a f(x) dx = \int_0^a f(a-x) dx) \\ &= \left(\frac12\right)^3\int_0^2 \frac{(2-x)^2}{2} \, dx \\ &= \dots \text{(same trick)} \\ &= \left(\frac12\right)^3 \cdot \frac{2^3}{3!} = \frac16 \end{aligned}$$
Edit in response to OP's comment:
Consider $S = \{ (x,y,z) \in \Bbb{R}_+^3 \mid x + y + z \le 2\}$ and $S' = \{ (x,y,z) \in \Bbb{R}_+^3 \mid x \in [0,2], y \in [0,2-x], z \in [0,2-x-y]\}$. It suffices to show that $S = S'$. $S' \subseteq S$ is clear, and is left as exercise. To finish the argument, let $(x,y,z) \in S$. Then verify the criteria which define $S'$ componentwise.
- $x \in [0,2]$ is clear
- $y \in [0,2-x-z] \subseteq [0,2-x]$ is a bit tricky
- $z \in [0,2-x-y]$ is also clear
Hence the upper and lower limits of the above iterated integral is justified.
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.)