Question

Three bags contain a number of red and white balls as follows:
• Bag I: 3 red balls
• Bag II: 2 red balls and 1 white ball
• Bag III: 3 white balls The probability that bag \( i \) will be chosen and a ball is selected from it is \( \frac{i}{6}, i = 1, 2, 3 \). If a white ball is selected, what is the probability that it came from Bag III?

• A. \( \frac{9}{11} \)
• B. \( \frac{2}{11} \)
• C. 0
• D. \( \frac{1}{11} \)

Hint:

Bayes' Theorem is useful for finding conditional probabilities. When dealing with probabilities of multiple events, use the law of total probability to find the overall probability before applying Bayes' Theorem.

Answer 1

The Correct Option is A

Step 1: Understanding the problem.
We are given three bags, each with a certain number of red and white balls, and the probability that each bag will be chosen. We are asked to find the probability that a white ball selected came from Bag III.

Step 2: Use Bayes' Theorem.
To find the probability that the white ball came from Bag III, we use Bayes' Theorem. Bayes' Theorem is given by:
\[ P(\text{Bag III} \mid \text{White Ball}) = \frac{P(\text{White Ball} \mid \text{Bag III}) \cdot P(\text{Bag III})}{P(\text{White Ball})} \]

Step 3: Calculate the individual probabilities.
- \( P(\text{Bag III}) = \frac{3}{6} = \frac{1}{2} \)
- \( P(\text{White Ball} \mid \text{Bag III}) = \frac{3}{3} = 1 \), because all balls in Bag III are white.
- \( P(\text{White Ball}) \) is the total probability of drawing a white ball from any bag, which is the sum of the probabilities of selecting a white ball from each bag:
\[ \begin{aligned} P(\text{White Ball}) &= P(\text{White Ball} \mid \text{Bag I}) \cdot P(\text{Bag I}) &\quad + P(\text{White Ball} \mid \text{Bag II}) \cdot P(\text{Bag II}) &\quad + P(\text{White Ball} \mid \text{Bag III}) \cdot P(\text{Bag III}) \end{aligned} \] Substituting the values:
\[ P(\text{White Ball}) = 0 \cdot \frac{1}{6} + \frac{1}{3} \cdot \frac{2}{6} + 1 \cdot \frac{3}{6} \] \[ P(\text{White Ball}) = 0 + \frac{2}{18} + \frac{3}{6} = \frac{2}{18} + \frac{9}{18} = \frac{11}{18} \]

Step 4: Apply Bayes' Theorem.
Now, we can apply Bayes' Theorem: \[ P(\text{Bag III} \mid \text{White Ball}) = \frac{1 \cdot \frac{1}{2}}{\frac{11}{18}} = \frac{\frac{1}{2}}{\frac{11}{18}} = \frac{1}{2} \cdot \frac{18}{11} = \frac{9}{11} \]

Step 5: Conclusion.
Therefore, the probability that the white ball came from Bag III is \( \frac{9}{11} \), and the correct answer is option (A).