Probability of 3 variables

Relevant wiki: 3-dimensional Geometric Probability

The region which satisfies $x,y,z$ in interval $[-1,1]$ is a cube with side length $2$ , center $(0,0,0)$ . The volume of this region is $8$ . $\begin{aligned} x^2+y^2+z^2+2y-2z+1&\leq 0 \\ x^2+(y+1)^2+(z-1)^2&\leq 1^2 \end{aligned}$ The region which satisfies $x^2+y^2+z^2+2y-2z+1\leq 0$ is a sphere with radius $1$ , center $(0,-1,1)$ . But $x,y,z$ lies on interval $[-1,1]$ , the part of sphere which lies outside of cube don't exist. Hence, the required region is $\frac{1}{4}$ of sphere mentioned above.(Only one quarter of sphere lies inside the cube, because the center of sphere lies on the midpoint of side of cube. And its radius is half of side length of cube.) Its volume is $\frac{1}{4}×\frac{4}{3}\pi=\frac{\pi}{4}$ .

The probability is $\frac{V_{sphere}}{V_{cube}}$ . $\begin{aligned} \frac{V_{sphere}}{V_{cube}}&= \frac{\frac{\pi}{3}}{8} \\ &=\frac{\pi}{24} \\ &=\large 0.131 \end{aligned}$

The implicit equation can be written as $(x-0)^2-(y+1)^2+(z-1)^2 \leq 0$ , namely, the unit sphere centered in $(0,-1,1)$ . The set of all possible $(x,y,z)$ describe a cube in the 3D plane, where the vertices are given by the $2^3$ combination of $[-1,1]$ . By analytical geometry consideration, the set of all point $(X,Y,Z)$ such that $(X,Y,Z)\in C \cup S$ , where $S$ is the sphere and $C$ the cube is

$\{(X,Y,Z) \in \mathbb{R}^3 : \begin{cases} -1 \leq X \leq 1\\ -1 \leq Y \leq \sqrt{1-X^2}-1 \\ 1-\sqrt{1-X^2-(Y+1)^2} \leq Z \leq 1 \end{cases}\}$

The probability density function of each random variable is $\frac{1}{2}$ , ( $\int_{-1}^{1} \frac{1}{2} di=1 , i\in\{X,Y,Z\}$ ), so

$\displaystyle \mathbb{P}[(X,Y,Z)\in C \cup S] = \int_{-1}^{1} \int_{-1}^{\sqrt{1-x^2}-1}\int_{1-\sqrt{1-x^2-(y+1)^2}}^{1} \frac{1}{8} dzdydx = 0.131$