Question

Coloured balls are distributed in three containers according to the following table: \[\begin{array}{|c|c|c|c|}\hline \text{Container} & \text{Black} & \text{White} & \text{Red} \\ \hline I & 3 & 4 & 5 \\ \hline II & 2 & 2 & 2 \\ \hline III & 1 & 2 & 3 \\ \hline \end{array}\] A ball is drawn out from a container randomly chosen. If the ball is black, then find the probability that the ball is drawn from Container-III.

Hint:

In problems involving conditional probability, Bayes' Theorem is useful to calculate the probability of an event given some other event.

Answer 1

Let the event \( A \) be that the ball drawn is black and event \( B \) be that the ball is drawn from Container-III. We need to find \( P(B|A) \), the probability that the ball is from Container-III given that it is black. By Bayes' Theorem: \[ P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)}. \] Now, calculate the individual probabilities: 1. \( P(A|B) \) is the probability of drawing a black ball from Container-III. This is the ratio of black balls in Container-III to the total number of balls in Container-III: \[ P(A|B) = \frac{1}{1 + 2 + 3} = \frac{1}{6}. \] 2. \( P(B) \) is the probability that Container-III is chosen. Since there are 3 containers, the probability of choosing any one container is \( \frac{1}{3} \). 3. \( P(A) \) is the total probability of drawing a black ball from any container. This can be calculated by adding the probabilities of drawing a black ball from each container, weighted by the probability of choosing each container: \[ P(A) = \frac{1}{3} \cdot \frac{3}{3+4+5} + \frac{1}{3} \cdot \frac{2}{2+2+2} + \frac{1}{3} \cdot \frac{1}{1+2+3}. \] Simplifying: \[ P(A) = \frac{1}{3} \cdot \frac{3}{12} + \frac{1}{3} \cdot \frac{2}{6} + \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{3} \cdot \left( \frac{1}{4} + \frac{1}{3} + \frac{1}{6} \right). \] Finding a common denominator: \[ P(A) = \frac{1}{3} \cdot \frac{11}{12} = \frac{11}{36}. \] Now, we can substitute into Bayes' Theorem: \[ P(B|A) = \frac{\frac{1}{6} \cdot \frac{1}{3}}{\frac{11}{36}} = \frac{\frac{1}{18}}{\frac{11}{36}} = \frac{2}{11}. \] Conclusion: The probability that the ball is drawn from Container-III given that the ball is black is \[ \boxed{\frac{2}{11}}. \]