Question

Three similar urns \(A,B,C\) contain \(2\) red and \(3\) white balls; \(3\) red and \(2\) white balls; \(1\) red and \(4\) white balls, respectively. If a ball is selected at random from one of the urns is found to be red, then the probability that it is drawn from urn \(C\) is ?

• A. \(\tfrac{1}{6}\)
• B. \(\tfrac{1}{3}\)
• C. \(\tfrac{1}{2}\)
• D. \(\tfrac{2}{9}\)

Hint:

Use Bayes'Theorem when selecting from multiple sources.
- Probability(\(\text{red}\)) is the total from each urn weighed by the chance to pick that urn.

Answer 1

The Correct Option is A

Step 1: Probability of picking each urn and drawing a red ball.
- Probability of choosing any one urn is \(\tfrac13\).
- From urn \(A\): probability of red is \(\tfrac{2}{5}\).
- From urn \(B\): probability of red is \(\tfrac{3}{5}\).
- From urn \(C\): probability of red is \(\tfrac{1}{5}\). 

Step 2: Bayes'Theorem.
Overall probability of drawing a red ball: \[ P(\text{red}) = \tfrac13\cdot\tfrac{2}{5} + \tfrac13\cdot\tfrac{3}{5} + \tfrac13\cdot\tfrac{1}{5} = \tfrac{2}{15} + \tfrac{3}{15} + \tfrac{1}{15} = \tfrac{6}{15} = \tfrac{2}{5}. \] We want \(P(\text{from }C \mid \text{red})\): \[ = \frac{P(\text{choose }C)\times P(\text{red}\mid C)}{P(\text{red})} = \frac{\tfrac13 \cdot \tfrac{1}{5}}{\tfrac{2}{5}} = \frac{\tfrac{1}{15}}{\tfrac{2}{5}} = \frac{1}{15}\,\times \frac{5}{2} = \frac{1}{6}. \] Hence \(\boxed{\tfrac{1}{6}}\).