Question:
Bag $A$ contains $3$ white and $4$ black balls. Bag $B$ contains $4$ white and $3$ black balls. Bag $C$ contains $2$ white and $5$ black balls. A bag is randomly selected and then a ball is randomly drawn from that bag. If the ball drawn was found to be white, then the probability that the ball is drawn from bag $C$ is
Bag $A$ contains $3$ white and $4$ black balls. Bag $B$ contains $4$ white and $3$ black balls. Bag $C$ contains $2$ white and $5$ black balls. A bag is randomly selected and then a ball is randomly drawn from that bag. If the ball drawn was found to be white, then the probability that the ball is drawn from bag $C$ is
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Bayes' theorem: Posterior probability = $\dfrac{\text{Prior}\times\text{Likelihood}}{\text{Total Probability}}$.
Updated On: Jun 3, 2026
- $\dfrac{1}{6}$
- $\dfrac{2}{9}$
- $\dfrac{1}{4}$
- $\dfrac{2}{13}$
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Apply Bayes' theorem.
Step 2: Meaning
Let \[ A,B,C \] denote the events of selecting the corresponding bags. Since a bag is chosen randomly, \[ P(A)=P(B)=P(C)=\frac13. \] Also, \[ P(W|A)=\frac37, \qquad P(W|B)=\frac47, \qquad P(W|C)=\frac27. \]
Step 3: Analysis
Total probability of drawing a white ball: \[ P(W) = \frac13\left(\frac37+\frac47+\frac27\right) = \frac13\cdot\frac97 = \frac37. \] By Bayes' theorem, \[ P(C|W) = \frac{P(C)\,P(W|C)}{P(W)}. \] Substituting values, \[ P(C|W) = \frac{\frac13\cdot\frac27}{\frac37} = \frac29. \]
Step 4: Conclusion
Hence the probability that the white ball came from bag $C$ is \[ \frac29. \]
Final Answer: (B)
Apply Bayes' theorem.
Step 2: Meaning
Let \[ A,B,C \] denote the events of selecting the corresponding bags. Since a bag is chosen randomly, \[ P(A)=P(B)=P(C)=\frac13. \] Also, \[ P(W|A)=\frac37, \qquad P(W|B)=\frac47, \qquad P(W|C)=\frac27. \]
Step 3: Analysis
Total probability of drawing a white ball: \[ P(W) = \frac13\left(\frac37+\frac47+\frac27\right) = \frac13\cdot\frac97 = \frac37. \] By Bayes' theorem, \[ P(C|W) = \frac{P(C)\,P(W|C)}{P(W)}. \] Substituting values, \[ P(C|W) = \frac{\frac13\cdot\frac27}{\frac37} = \frac29. \]
Step 4: Conclusion
Hence the probability that the white ball came from bag $C$ is \[ \frac29. \]
Final Answer: (B)
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