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 \(x^2+y^2+z^2\le 1\)?

supremecheetah May 21, 2023

#1**0 **

The probability that a randomly selected point in space (x, y, z) satisfies the condition $x^2 + y^2 + z^2 \leq 1$ can be found by calculating the volume of the region in the given space that satisfies this condition, and then dividing it by the total volume of the given space.

The given space is defined by the constraints -1 ≤ x ≤ 1, -1 ≤ y ≤ 1, and -1 ≤ z ≤ 1. It forms a cube with sides of length 2 centered at the origin.

The condition $x^2 + y^2 + z^2 \leq 1$ represents a sphere with a radius of 1 centered at the origin.

To find the volume of the region that satisfies the condition, we need to calculate the volume of the sphere and divide it by the volume of the cube.

The volume of the sphere is given by the formula: V_sphere = (4/3) * π * r^3, where r is the radius of the sphere.

In this case, r = 1, so V_sphere = (4/3) * π * 1^3 = (4/3) * π.

The volume of the cube is given by the formula: V_cube = length^3 = 2^3 = 8.

Therefore, the probability that a randomly selected point satisfies the condition is:

P = V_sphere / V_cube = [(4/3) * π] / 8

Simplifying, we have:

P = (π/6) / 8 = π/48

Hence, the probability that a randomly selected point (x, y, z) in the given space satisfies the condition x^2 + y^2 + z^2 ≤ 1 is π/48.

Guest May 21, 2023