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
two urns contain white balls and yellow balls the first urn contains 10 white balls and 6 yellow balls and the second urn contains 8 white balls and 3 yellow balls a ball is drawn at random from each urn what is the probability that both balls are white
The probability of drawing a white ball from the first urn is 10/16 = 5/8, and the probability of drawing a white ball from the second urn is 8/11.
To find the probability of both events happening, we multiply the probabilities:
(5/8) × (8/11) = 40/88 = 5/11
Therefore, the probability of drawing two white balls is 5/11.
To find the probability that the ball was drawn from urn A, we can use Bayes' theorem.
Let's define the events:
- A: The ball was drawn from urn A.
- B: The ball drawn is black.
We need to find P(A|B), which represents the probability that the ball was drawn from urn A given that it is black.
Bayes' theorem states that:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) represents the probability of drawing a black ball given that the ball was drawn from urn A. In this case, since there are 7 black balls out of 13 in urn A, P(B|A) = 7/13.
P(A) represents the probability of drawing from urn A, which is 1/2 since there are two urns to choose from.
P(B) represents the probability of drawing a black ball. There are 7 black balls in urn A and 6 black balls in urn B, so the total number of black balls is 7 + 6 = 13. Since there is an equal chance of drawing from either urn, P(B) = 13/26.
Now we can substitute these values into Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(A|B) = (7/13 * 1/2) / (13/26)
P(A|B) = 7/26 * 26/13
P(A|B) = 7/13
Therefore, the probability that the ball was drawn from urn A given that it is black is 7/13.
So, the correct option is (A) 7/14.