Urn A contains eight white and four black balls. Urn B contains seven white and three black balls. A ball is drawn from urn A and then transferred to urn B. A ball is then drawn from urn B. What is the probability that the transferred ball was black given that the second ball drawn was white?

Question

Quizplus · Accepted Answer

The probability that the transferred ball was black given that the second ball drawn was white can be calculated using Bayes' theorem. Let B represent the event that the transferred ball is black, and W represent the event that the second ball drawn is white. We want to find P(B|W), the probability that the transferred ball is black given that the second ball drawn is white.First, calculate the probability of drawing a white ball from urn B after transferring a black ball from urn A (P(W|B)). If a black ball is transferred, urn B will have 7 white and 4 black balls. So, P(W|B) = 7/(7+4) = 7/11.Next, calculate the probability of drawing a white ball from urn B after transferring a white ball from urn A (P(W|B')). If a white ball is transferred, urn B will have 8 white and 3 black balls. So, P(W|B') = 8/(8+3) = 8/11.Now, calculate the prior probabilities of transferring a black ball (P(B)) and transferring a white ball (P(B')). Since urn A initially contains 8 white and 4 black balls, P(B) = 4/12 = 1/3, and P(B') = 8/12 = 2/3.Finally, use Bayes' theorem to find P(B|W):$$P(B|W) = \frac{P(W|B)P(B)}{P(W|B)P(B) + P(W|B')P(B')} = \frac{(7/11) \times (1/3)}{(7/11) \times (1/3) + (8/11) \times (2/3)} = \frac{7}{7 + 16} = \frac{7}{23} \approx 0.3.$$Therefore, the correct answer is 0.3.

Urn a Contains Eight White and Four Black Balls