6. (1999 American High School Mathematics Exam) A regular tetrahedron has two spheres, one inscribed and one circumscribed. Additionally, there are four smaller 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
6. C. Reason: Let the edge length of the regular tetrahedron be \( a \). It is known 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
\[
r = \frac{R_{2} - R_{1}}{2} = \frac{\sqrt{6}}{12} a \text{.}
\]

Thus, the required probability is
\[
\frac{\frac{4}{3} \pi R_{1}^{3} + 4 \cdot \frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi R_{2}^{3}} = \frac{5 \left( \frac{1}{12} \right)^{3}}{\left( \frac{1}{4} \right)^{3}} = \frac{5}{27} \approx 0.19.
\]