Answer: (c) \(\frac{3}{7}\)
P(Drawing a green ball from bag A) = \(P\big(\frac{G}{A}\big) =\frac{4}{7}\)
P(Drawing a green ball from bag B) = \(P\big(\frac{G}{B}\big) =\frac{3}{7}\)
∴ Required probability \( P\big(\frac{B}{G}\big)\)
= \(\frac{P(B).P\big(\frac{G}{B}\big)}{P(A).P\big(\frac{G}{A}\big)+P(B).P\big(\frac{G}{B}\big)}\)
\(\frac{\frac{1}{2}.\frac{3}{7}}{\frac{1}{2}.\frac{4}{7}+\frac{1}{2}.\frac{3}{7}}\) = \(\frac{3/14}{4/14+3/14} = \frac{3/14}{7/14} = \frac{3}{7}\)