Question

a point in space $(x,y,z)$ is randomly selected so that $-1\le x \le 1$,$-1\le y \le 1$,$-1\le z \le 1$. what is the probability that $x^2 y^2 z^2\le 1$?

Answer 1

Probability that $x^2+y^2+z^2\leq 1$ is 0.523

Given that a point (x, y, z) is in space which is randomly selected as

$-1\le x \le 1 , -1\le y \le 1 , -1\le z \le 1$

These three equations represents a cube in space.

Length of each side of the cube = 1 - -1 = 2

Now  $x^2+y^2+z^2\leq 1$  represents a sphere with center (0, 0, 0) and of radius less than or equal to 1.

So we can imagine that this sphere is inside the cube.

So the probability that   $x^2+y^2+z^2\leq 1$  = Volume of the sphere/Volume of the cube

Volume of sphere with radius 1 = $\frac{4}{3} \pi r^3=\frac{4}{3} \pi \times 1 = \frac{4}{3}\times 3.14 = 4.186$ cubic unit

Volume of the cube of length 2 = $2^3 = 8$ cubic units

So the probability that  $x^2+y^2+z^2\leq 1$  = 4.186/8 = 0.523

Learn more about probability at brainly.com/question/25870256

#SPJ4