it's pretty. Geometrical probability

Assume that the cube has a length of $x$ . Therefore, its volume is $x^3$ . Using the Pythagorean Theorem, one can see that the diagonal of one side of the cube is $\sqrt{2}x$ and the diagonal from one vertex of the cube to the opposite vertex is $\sqrt{3}x$ . This is equivalent to the diameter of the sphere.

The radius of the sphere is $\frac{\sqrt{3}}{2}x$ . The volume of a sphere is $\frac{4}{3}{\pi}{r^3}$ . Using this formula, the volume of the sphere in the problem is $\frac{\sqrt{3}}{2}{\pi}{x^3}$ .

To find the probability of a point being inside the cube given that we know it is inside the sphere, we simply need to divide the volume of the sphere: $\frac{\sqrt{3}}{2}{\pi}(x^3)$ by the volume of the cube: $x^3$ or $\frac{x^3}{\frac{\sqrt{3}}{2}{\pi}{x^3}} = \frac{1}{\frac{\sqrt{3}}{2}}{\pi}$

We now know the probability of a point being inside the cube given that we know it is inside the sphere. However, we want to find the probability of the point not being in the cube, given that it is inside the sphere. This would be the total probability minus the answer we found, or ...

${1}-\frac{1}{\frac{\sqrt{3}}{2}{\pi}}$ .

$\lambda = \frac{\sqrt{3}}{2}{\pi}$ . This can be approximated to $2.721$ on a calculator.