As far as I understand the question your sketch isn't correct. I've made a

**sketch **but it doesn't take into account that we are dealing with a 3D object.

Probably (

) the probability could be calculated by:

$\displaystyle p(P\ in\ spheres) = \dfrac{\text{sum of the 5 volumes}}{\text{volume of the circumscribing sphere}}$

Let a denote the length of one edge of the tetrahedron, then

$\displaystyle R_c = \frac23 \cdot h = a\cdot \frac2{9} \cdot \sqrt{6}$

$\displaystyle R_i = \frac12 \cdot R_c$

$\displaystyle r = \frac12 \cdot R_i$ That means the radii are proportional like: $\displaystyle r : R_i : R_c = 1 : 2 : 4$ Can you take it from here?