Question

A sphere is inscribed inside a cube. What is that probability that a point that is inside the cube is also inside the sphere?

• A. $\frac{6}{\pi }$
• B. $\frac{\pi }{12}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{4}$
• E. $\frac{2}{\pi }$

Answer 1

The correct option is A $\frac{6}{\pi }$

A Sphere is inscribed inside a cube. Let the length of one side of the cube be ${}^{\prime }{a}^{\prime }$. So, the diameter of sphere is also ${}^{\prime }{a}^{\prime }.$

$\therefore$ Radius of sphere $=a/2$

$\Rightarrow$ Volume of sphere $=\frac{4}{3}\pi r^3=\frac{4}{3}\pi {\left(a/2\right)}^3=\frac{\pi a^3}{6}$

A point inside the sphere will be a point inside the cube also.

$\therefore \frac{Volumeofcube}{Volumeofsphere}=\frac{a^3}{\pi a^3/6}=\frac{6}{\pi }$

Hence, the answer is $\frac{6}{\pi }.$