Urn A contains four white balls and five black balls. Urn B contains seven white balls and six 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 black?
To find the probability that the transferred ball was black given that the second ball drawn was black, we can use Bayes' Theorem.
Let's define the following events:
A: The transferred ball was black.
B: The second ball drawn was black.
We need to find P(A|B), which represents the probability of event A (transferred ball was black) given event B (second ball drawn was black).
We can start by finding P(A), the probability that the transferred ball was black.
In Urn A, there are a total of 4 white balls and 5 black balls, so the probability of drawing a black ball from Urn A is P(A) = 5 / (4 + 5) = 5/9.
Now, we need to find P(B|A), the probability of drawing a black ball from Urn B given that the transferred ball was black.
After transferring a black ball from Urn A to Urn B, Urn B will have 7 white balls and 7 black balls (original 6 black balls + transferred black ball). So, the probability of drawing a black ball from Urn B is P(B|A) = 7 / (7 + 7) = 7/14 = 1/2.
Next, we need to find P(B), the probability of drawing a black ball from Urn B.
Before the transfer, Urn B had 7 white balls and 6 black balls. So, the probability of drawing a black ball from Urn B without any knowledge of the transfer is P(B) = 6 / (7 + 6) = 6/13.
Finally, we can use Bayes' Theorem to find P(A|B):
P(A|B) = (P(B|A) * P(A)) / P(B)
= (1/2 * 5/9) / (6/13)
= 5/18 * 13/6
= 65/108
Therefore, the probability that the transferred ball was black given that the second ball drawn was black is 65/108.