#1**+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.

Melody Feb 24, 2020

#1**+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