Bag1 = 3 back, 4 white balls
Bag2 = 4 black, 3 white balls
Probability of selecting bag 1 \(=\frac{2}{6}=\frac{1}{3}\)
Probability of selecting bag 2 \(=1-\frac{1}{3}=\frac{2}{3}\)
Probability of choosing a black ball from bag 1 \(=\frac{3}{7}\)
Probability of choosing a black ball from bag 2 \(=\frac{4}{7}\)
Required probability \(=\frac{1}{3}\times\frac{3}{7}+\frac{2}{3}\times\frac{4}{7}\)
\(=\frac{11}{21}\).So the correct option is (A)