SOLUTION: An urn contains 14 balls; 6 of them are white, and the others are black. Another urn contains 9 balls; 3 are white, and 6 are black. A ball is drawn at random from the first urn
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Question 360608: An urn contains 14 balls; 6 of them are white, and the others are black. Another urn contains 9 balls; 3 are white, and 6 are black. A ball is drawn at random from the first urn and is placed in the second urn. Then, a ball is drawn at random from the second urn. If this ball is white, find the probability that the ball drawn from the first urn was black.
**Well, I know that I have to use the Law of Total Probability and the Bayes'Rule but I don't understand how.
So much thanks Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! Let b1 = event that 1st ball drawn is black,
w1 = event that 1st ball drawn is white,
b2 = event that 2nd ball drawn is black,
w2 = event that 2nd ball drawn is white.
Then P(b1|w2) = P(b1 and w2)/P(w2)= P(b1 and w2)/[P(b1 and w2)+P(w1 and w2)]
=P(w2|b1)*P(b1)/[P(w2|b1)*P(b1) + P(w2|w1)*P(w1)]
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