14.35 A regular tetrahedron has two spheres, one inscribed and one circumscribed. Additionally, there are four spheres between each face of the tetrahedron and the circumscribed sphere, each tangent to the center of the face. Point $P$ is a point inside the circumscribed sphere. The probability that $P$ falls within one of the five smaller spheres is closest to
(A) 0 .
(B) 0.1 .
(C) 0.2 .
(D) 0.3 .
(E) 0.4 .
(50th American High School Mathematics Examination, 1999)
［Solution］Let the edge length of the regular tetrahedron be $\mathrm{a}$. It is easy to know that the radius of its inscribed sphere is
$$
R_{1}=\frac{\sqrt{6}}{12} a,
$$

and the radius of its circumscribed sphere is $R_{2}=\frac{\sqrt{6}}{4} a$.
If the radius of the sphere that is tangent to the circumscribed sphere and the faces of the tetrahedron is $r$, then
$$
r=\frac{R_{2}-R_{1}}{2}=\frac{\sqrt{6}}{12} a .
$$

Thus, the required probability is
$$
\begin{array}{l} 
\frac{\frac{4}{3} \pi R_{1}^{3}+4 \cdot \frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R_{2}^{3}}=\frac{5\left(\frac{1}{12}\right)^{3}}{\left(\frac{1}{4}\right)^{3}} \\
=\frac{5}{27} \approx 0.19, \\
\text { Therefore, the answer is (B). }
\end{array}
$$

Therefore, the answer is $(B)$.
Note If the ratio of the distance from the center of the regular tetrahedron to its faces and vertices is 1:3, it can be considered simpler.