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0
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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 $x^2+y^2+z^2\le 1$?

TYSM!

Mar 13, 2021

#1
+32062
+4

This constraints represent a sphere of radius 1 inside a cube of side length 2.

Volume of cube  $$Vc=2^3=8$$

Volume of sphere $$Vs=\frac{4}{3}\pi 1^3=\frac{4}{3}\pi$$

Probability $$p=\frac{Vs}{Vc}$$

I'll let you finish!

Mar 13, 2021
#2
+1

Sorry, but I don't get why you are doing this. Could you please elaborate? Thanks!!!

Guest Mar 13, 2021
#3
+112966
+2

Try and imagine it.

$$-1\le x\le1\\ -1\le y\le1\\ -1\le z\le1\\$$

this encloses a cube.  The centre is (0,0,0)  what is the side length,  What is the area?

Now

$$x^2+y^2+z^2\le 1$$

is the formula for a sphere with centre (0,0,0) and radius  ?    What is the area?

The sphere is inside the cube.  What fraction of the cube space does the sphere take up?

Just as Alan has already said.

Mar 13, 2021
#4
+1

Oh ok! That makes much more sense!

Also thanks for not just telling the answer :D I should of specified that earlier I wanted hints not answers

Guest Mar 13, 2021