+118608 
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.
+118608
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.
