Probability that a Red ball comes from Bag I ($p$):

$p(B_1/R) = \frac{p(B_1) \cdot p(R/B_1)}{p(R)}$

Substituting the values,

$p(B_1/R) = \frac{\frac{1}{3} \times \frac{3}{10}}{\frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10} + \frac{1}{3} \times \frac{5}{10}}$

Simplify to find $p$:

$=\frac{\frac{1}{10}}{\frac{1}{3}(1.2)} = \frac{1}{4}$

So, $p = \frac{1}{4}$.

Probability that a Green ball comes from Bag III ($q$):

$p(B_3/G) = \frac{p(B_3) \cdot p(G/B_3)}{p(G)}$

Substitute the values,

$p(B_3/G) = \frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{5}{10} + \frac{1}{3} \times \frac{3}{10} + \frac{1}{3} \times \frac{4}{10}}$

Simplify to find $q$:

$=\frac{\frac{2}{15}}{\frac{1}{3} \times 1.2} = \frac{1}{3}$

So, $q = \frac{1}{3}$.

Calculation of $(\frac{1}{p} + \frac{1}{q})$:

$\frac{1}{p} = 4$, $\frac{1}{q} = 3$

Therefore,

$(\frac{1}{p} + \frac{1}{q}) = 4 + 3 = 7$