probability that a randomly chosen point within a cube lies within an inscribed cone

1. Understand the Problem: We need to find the probability that a randomly selected point inside a cube also lies inside a cone inscribed within that cube. This probability is the ratio of the volume of the cone to the volume of the cube.

1. Define Variables: Let the side length of the cube be $s$. Since the cone is inscribed within the cube, the diameter of the base of the cone will be equal to the side length of the cube, $s$. Therefore, the radius of the cone's base is $r = \frac{s}{2}$. The height of the cone will also be equal to the side length of the cube, $h = s$.

1. Calculate the Volume of the Cube: The volume of a cube with side length $s$ is given by the formula $V_{cube} = s^3$.

4. Calculate the Volume of the Cone: The volume of a cone is given by the formula $V_{cone} = \frac{1}{3}\pi r^2 h$. Substituting the values $r = \frac{s}{2}$ and $h = s$ into the formula, we get:
$$V_{cone} = \frac{1}{3}\pi \left(\frac{s}{2}\right)^2 s$$
$$V_{cone} = \frac{1}{3}\pi \left(\frac{s^2}{4}\right) s$$
$$V_{cone} = \frac{1}{12}\pi s^3$$

5. Calculate the Probability: The probability that a randomly chosen point within the cube lies within the cone is the ratio of the volume of the cone to the volume of the cube:
$$P = \frac{V_{cone}}{V_{cube}}$$
$$P = \frac{\frac{1}{12}\pi s^3}{s^3}$$
$$P = \frac{\pi}{12}$$

1. Simplify the Result: The $s^3$ terms cancel out, leaving the probability as $\frac{\pi}{12}$.