Probability of Transferring a White Ball Given a White Ball is Drawn from Urn 2

Consider two urns, urn 1 containing two white balls and one black ball, and urn 2 containing one white ball and five black balls. One ball is drawn at random from urn 1 and placed in urn 2. A ball is then drawn from urn 2, and it happens to be white. What is the probability that the transferred ball was white?

To solve this problem, we can use Bayes' theorem. Bayes' theorem states that the probability of an event A given event B is equal to the probability of B given A multiplied by the probability of A, divided by the probability of B. In this case, event A is the transfer of a white ball and event B is the drawing of a white ball from urn 2.

First, let's find the probability of drawing a white ball from urn 2, P(B). We have:

P(B) = P(white transferred from urn 1) * P(white from urn 2) + P(black transferred from urn 1) * P(white from urn 2)

P(B) = (2/3) * (1/6) + (1/3) * (1/6)

P(B) = 2/18 + 1/18

P(B) = 3/18

Next, let's find the probability of transferring a white ball given a white ball was drawn from urn 2, P(A | B). Using Bayes' theorem, we have:

P(A | B) = P(B | A) * P(A) / P(B)

P(B | A) is the probability of drawing a white ball from urn 2 given a white ball was transferred from urn 1. This is equal to 1/6. P(A) is the probability of transferring a white ball from urn 1, which is 2/3.

P(A | B) = 1/6 * 2/3 / (3/18)

P(A | B) = 4/3

So, the probability that the transferred ball was white, given that a white ball was drawn from urn 2, is 4/3.

In conclusion, to find the probability of transferring a white ball given a white ball was drawn from urn 2, we can use Bayes' theorem. By calculating the probability of drawing a white ball from urn 2 and the probability of drawing a white ball from urn 2 given a white ball was transferred from urn 1, we can find the overall probability of a white ball being transferred. In this case, the probability of transferring a white ball given a white ball was drawn from urn 2 is 4/3.