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# I am confused...

#1
+108679
+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
+108679
+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   $$\frac{4\pi}{3}\div 8=\frac{4\pi}{24}=\frac{\pi}{6}\approx0.52$$

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

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
+1

Thanks Melody!

Guest Feb 24, 2020
#3
+108679
+1

You are welcome

Note: my answer was changed a bit.

Melody  Feb 24, 2020