A sphere is circumscribed over a cube. Find the probability that a point lies inside the sphere, lies outside the cube.

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Solution

Required probability $= \frac{f a v o r a b l e v a l u m e}{t o t a l v o l u m e}$
Clearly if edge length of cube is a radius of sphere will be $\frac{a \sqrt{3}}{2}$
Thus, volume of sphere $= \frac{4}{3} π {(\frac{a \sqrt{3}}{2})}^{3} = \frac{π a^{3} \sqrt{3}}{2}$
Hence $P = 1 - \frac{1}{π \frac{\sqrt{3}}{2}} = 1 - \frac{2}{π \sqrt{3}}$

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A sphere is circumscribed over a cube. Find the probability that a point lies inside the sphere, lies outside the cube.

Required probability=favorablevalumetotalvolumeClearly if edge length of cube is a radius of sphere will be a√32Thus, volume of sphere=43π(a√32)3=πa3√32Hence P=1−1π√32=1−2π√3