+0  
 

Best Answer 

 #1
avatar+110227 
+1

 

Well it looks like you have got a sphere, radius 1 inside a cube, sidelength 2

 

The volume of the cube is 8 unit cubed

the volume of the sphere is      \(\frac{4}{3}\pi r^3 = \frac{4}{3}\pi\)

 

So the probability that you choose a point inside the cube that is also inside the sphere   \(\frac{4\pi}{3}\div 8=\frac{4\pi}{24}=\frac{\pi}{6}\approx0.52\)

 

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Always check for reasonableness.

I expected the prob to be a bit higher than that but I guess it sounds feasible.

 Feb 24, 2020
edited by Melody  Feb 24, 2020
 #1
avatar+110227 
+1
Best Answer

 

Well it looks like you have got a sphere, radius 1 inside a cube, sidelength 2

 

The volume of the cube is 8 unit cubed

the volume of the sphere is      \(\frac{4}{3}\pi r^3 = \frac{4}{3}\pi\)

 

So the probability that you choose a point inside the cube that is also inside the sphere   \(\frac{4\pi}{3}\div 8=\frac{4\pi}{24}=\frac{\pi}{6}\approx0.52\)

 

----------------------

 

Always check for reasonableness.

I expected the prob to be a bit higher than that but I guess it sounds feasible.

Melody Feb 24, 2020
edited by Melody  Feb 24, 2020
 #2
avatar
+1

Thanks Melody!

Guest Feb 24, 2020
 #3
avatar+110227 
+1

You are welcome    laugh

Note: my answer was changed a bit.

Melody  Feb 24, 2020

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