Bag A contains 3 Red and 4 Black balls while Bag B contains 5 Red and 6 Black balls. One ball is drawn at random from one of the bags and is found to be red. Then, the probability that it was drawn from Bag B is
Step-by-step Solution:
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)