Solution (Bayes’ Theorem):

Let \(A\) be the event “chosen bag is A”, \(B\) be “chosen bag is B”, and \(R\) be “drawn ball is red”. Assume each bag is chosen at random with probability \(\tfrac{1}{2}\).

\[ P(R\mid A)=\frac{3}{3+4}=\frac{3}{7}, \qquad P(R\mid B)=\frac{5}{5+6}=\frac{5}{11}, \qquad P(A)=P(B)=\frac{1}{2}. \]

Using Bayes’ theorem: \[ P(B\mid R)=\frac{P(R\mid B)\,P(B)}{P(R\mid A)\,P(A)+P(R\mid B)\,P(B)} =\frac{\tfrac{5}{11}\cdot \tfrac{1}{2}}{\tfrac{3}{7}\cdot \tfrac{1}{2}+\tfrac{5}{11}\cdot \tfrac{1}{2}} =\frac{\tfrac{5}{11}}{\tfrac{3}{7}+\tfrac{5}{11}} =\frac{35/77}{68/77}=\boxed{\frac{35}{68}}. \]

✅ Correct Answer: 35/68 (Option A)