6. Johan has two bags. The first bag contains 2 red and 7 white balls, while the second contains 4 red and 5 white balls. One of the bags is selected randomly, and a ball is drawn from this bag. If the ball drawn is white, what is the probability that it is drawn from the second bag? A. \( \frac{5}{12} \) B. \( \frac{1}{2} \) C. \( \frac{5}{9} \) D. \( \frac{7}{9} \) E. \( \frac{70}{81} \)

To solve this problem, we need to apply Bayes' Theorem. Let's denote: - \( B_1 \): the event that the first bag is chosen, - \( B_2 \): the event that the second bag is chosen, - \( W \): the event that a white ball is drawn. We can find the probability of drawing a white ball from each bag and then use Bayes' theorem to find the probability that the white ball came from the second bag given that a white ball was drawn. 1. **Calculate the total probability of drawing a white ball**: - Probability of choosing Bag 1: \( P(B_1) = \frac{1}{2} \) - Probability of choosing Bag 2: \( P(B_2) = \frac{1}{2} \) - Probability of drawing a white ball from Bag 1: \( P(W | B_1) = \frac{7}{9} \) (since there are 7 white balls out of a total of 2 red + 7 white = 9 balls) - Probability of drawing a white ball from Bag 2: \( P(W | B_2) = \frac{5}{9} \) (since there are 5 white balls out of a total of 4 red + 5 white = 9 balls) Using the law of total probability: \[ P(W) = P(W | B_1) P(B_1) + P(W | B_2) P(B_2) = \left(\frac{7}{9} \cdot \frac{1}{2}\right) + \left(\frac{5}{9} \cdot \frac{1}{2}\right) = \frac{7}{18} + \frac{5}{18} = \frac{12}{18} = \frac{2}{3} \] 2. **Now apply Bayes' Theorem**: \[ P(B_2 | W) = \frac{P(W | B_2) P(B_2)}{P(W)} = \frac{\left(\frac{5}{9}\right) \left(\frac{1}{2}\right)}{\frac{2}{3}} = \frac{\frac{5}{18}}{\frac{2}{3}} = \frac{5}{18} \cdot \frac{3}{2} = \frac{5 \times 3}{18 \times 2} = \frac{15}{36} = \frac{5}{12} \] Thus, the probability that the drawn white ball is from the second bag is \( \frac{5}{12} \). The correct answer is **A. \( \frac{5}{12} \)**.