An urn A contains 6 white balls and 7 black balls. And urn B contains 8 white balls and 6 black balls. A person draws a ball at random from one of the two urns. It turns out to be black. What is probability that the ball was drawn from urn A?

(A) 7/14
(B) 49/88
(C) 39/88
(D) None of the above

To find the probability that the black ball was drawn from urn A, we can use Bayes' theorem. Bayes' theorem states that the probability of an event A given that event B has occurred can be calculated using the formula:

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

Let's let A represent the event that the ball was drawn from urn A, and B represent the event that a black ball was drawn.

We need to find P(A|B), which is the probability that the ball was drawn from urn A given that a black ball was drawn.

P(B|A) is the probability of drawing a black ball given that the ball was drawn from urn A. In this case, there are 7 black balls out of a total of 13 balls in urn A, so P(B|A) = 7/13.

P(A) is the probability of drawing from urn A, which is 1/2 since there are 2 urns.

P(B) is the probability of drawing a black ball. To find this, we need to consider the total number of black balls in both urns. There are 7 black balls in urn A and 6 black balls in urn B, so there are 7+6 = 13 black balls in total. The total number of balls is 6+7+8+6 = 27. Therefore, P(B) = 13/27.

Plugging all these values into Bayes' theorem:

P(A|B) = (P(B|A) * P(A)) / P(B)
P(A|B) = (7/13 * 1/2) / (13/27)
P(A|B) = 7/26

So, the probability that the ball was drawn from urn A given that a black ball was drawn is 7/26.

None of the answer choices provided match this probability, so the answer is None of the above.