Question

in the diagram, the sphere is inscribed within the cube. let r be the radius of the sphere. if a point within the cube is chosen at random, what is the probability that the point lies within the sphere? (hint: the volume of a sphere can be calculated using the formula $v = \frac{4}{3}\pi r^{3}$.)

Explanation:

Step1: Find the volume of the sphere

The formula for the volume of a sphere is \(V_{sphere}=\frac{4}{3}\pi r^{3}\).

Step2: Find the side - length of the cube

Since the sphere is inscribed in the cube, the diameter of the sphere is equal to the side - length of the cube. The diameter \(d = 2r\), so the side - length of the cube \(a=2r\).

Step3: Find the volume of the cube

The formula for the volume of a cube is \(V_{cube}=a^{3}\). Substituting \(a = 2r\) into the formula, we get \(V_{cube}=(2r)^{3}=8r^{3}\).

Step4: Calculate the probability

The probability \(P\) that a randomly chosen point in the cube lies within the sphere is given by the ratio of the volume of the sphere to the volume of the cube.
\(P=\frac{V_{sphere}}{V_{cube}}=\frac{\frac{4}{3}\pi r^{3}}{8r^{3}}\).
Simplify the expression:

$$
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Answer:

\(\frac{\pi}{6}\)