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 October 13th, 2013, 07:30 AM #1 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Probability of Black Ball Taken Given two boxes. The first box contains 4 red balls and 3 black balls. The second one contains 3 white balls and 5 black balls. If a ball is taken from the first box and placed in the second box (unseen), what is the probability that a ball taken from the second box is a black one? How to do it? I know that the probability of a black ball taken from the first box is 3/7, however there's also probability that the ball taken from the first box is the red one. So, how should I do it?
 October 13th, 2013, 08:16 AM #2 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Probability of Black Ball Taken There is probability of black ball on first pick and black ball on second pick. There is also probability of red ball on first pick and black ball on second pick. Other cases are not to contribute to the probability. Your anwer is therefore $\ \dfrac{3}{7} \times \dfrac{6}{9} \ + \ \dfrac{4}{7} \times \dfrac{5}{9}$.
 October 15th, 2013, 03:14 AM #3 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Re: Probability of Black Ball Taken OK, thank you very much for your help.

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