Example 4. A cube is inscribed in a sphere. A point is randomly fixed inside the sphere. Find the probability that the point will fall inside the cube.
Solution. Let's introduce the following notations: event $A$ - "the point falls into the cube"; $R$ - radius of the sphere, $a$ - edge of the cube, $V$ - volume of the sphere, $V_{1}$ - volume of the inscribed cube.

As is known, $V=\frac{4}{3} \pi R^{3}$; since $V_{1}=a^{3}$ and $a=2 R / \sqrt{3}$, then $V_{1}=\frac{8}{3 \sqrt{3}} R^{3}$. According to formula (1.5.2), taking $V_{g}=V_{1}$, $V_{G}=V$, we get

$$
P(A)=\frac{V_{1}}{V}=\frac{(8 / 3 \sqrt{3}) R^{3}}{(4 / 3) \pi R^{3}}=\frac{2}{\pi \sqrt{3}} \approx 0.368
$$