SOLUTION: A bag contains 7 yellow balls,3 red balls and 2 blue balls.A ball is chosen at random from the bag and not replaced.A second ball is then chosen.Determine the probability that the

Question 1142487: A bag contains 7 yellow balls,3 red balls and 2 blue balls.A ball is chosen at random from the bag and not replaced.A second ball is then chosen.Determine the probability that the two balls chosen,one red and the other is blue
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Answer by math_helper(2461) (Show Source): You can put this solution on YOUR website!




The outcomes and their probabilities are:



YY  (7/12)(6/11) = 42/132  

YR  (7/12)(3/11) = 21/132  

YB  (7/12)(2/11) = 14/132

 

RY  (3/12)(7/11) = 21/132

RR               = 6/132

RB               = 6/132



BY               = 14/132

BR               = 6/132

BB               = 2/132

                -----------

          Total  = 132/132





Adding the two entries RB and BR:  (6+6)/132 = 12/132 =  

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