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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! smiley

 Mar 13, 2021
 #1
avatar+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
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Sorry, but I don't get why you are doing this. Could you please elaborate? Thanks!!! laugh

Guest Mar 13, 2021
 #3
avatar+112966 
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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?  

That will be your answer.

 

Just as Alan has already said.

 Mar 13, 2021
 #4
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Oh ok! That makes much more sense! smiley

 

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

Guest Mar 13, 2021

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