Question

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:

• A. $\tfrac{35}{68}$
• B. $\tfrac{7}{38}$
• C. $\tfrac{14}{37}$
• D. $\tfrac{34}{43}$

Hint:

Whenever a condition is given (like "ball is red"), apply Bayes' theorem: $P(B|R)=\dfrac{P(B)\,P(R|B)}{P(R)}$.

Answer 1

The Correct Option is A

Step 1: Define events.
Let $A$ = ball drawn from Bag A, $B$ = ball drawn from Bag B. Each bag is equally likely $\Rightarrow P(A)=P(B)=\tfrac{1}{2}$.

Step 2: Find probability of Red from each bag.
- From Bag A: $P(R|A)=\tfrac{3}{7}$.
- From Bag B: $P(R|B)=\tfrac{5}{11}$.

Step 3: Total probability of Red.
\[ P(R) = P(A)P(R|A)+P(B)P(R|B) = \tfrac{1}{2}\cdot\tfrac{3}{7}+\tfrac{1}{2}\cdot\tfrac{5}{11} \] \[ = \tfrac{3}{14}+\tfrac{5}{22} = \tfrac{33}{154}+\tfrac{35}{154}=\tfrac{68}{154}=\tfrac{34}{77}. \]

Step 4: Apply Bayes' theorem.
\[ P(B|R)=\frac{P(B)\cdot P(R|B)}{P(R)}=\frac{\tfrac{1}{2}\cdot\tfrac{5}{11}}{\tfrac{34}{77}} \] \[ = \frac{5}{22}\cdot\frac{77}{34}=\frac{385}{748}=\tfrac{35}{68}. \]

Step 5: Conclusion.
Thus, the probability that the Red ball came from Bag $B$ is $\tfrac{35}{68}$.