SOLUTION: One bag contains 4 white balls and 3 black balls, and a second bag contains 3 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in the second

Question 1172368: One bag contains 4 white balls and 3 black balls, and a second bag contains 3 white
balls and 5 black balls. One ball is drawn from the first bag and placed unseen in
the second bag. What is the probability that a ball now drawn from the second
bag is black?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

There are two cases to analyze: the ball drawn from the first bag can be either white or black.

Case 1: first ball white

P(white from first bag) = 4/7.
There are now 4 white and 5 black in the second bag; P(black from second bag) = 5/9.
P(white then black) = (4/7)(5/9) = 20/63.

Case 2: first ball black

P(black from first bag) = 3/7.
There are now 3 white and 6 black in the second bag; P(black from second bag) = 6/9.
P(black then black) = (3/7)(6/9) 18/63.

ANSWER: P black from second bag = 20/63+18/63 = 38/63.

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