document.write( "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\r
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Algebra.Com's Answer #763194 by math_helper(2461) $\"\"$ $\"About$

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document.write( "The outcomes and their probabilities are:\r
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document.write( "YY  (7/12)(6/11) = 42/132  
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document.write( "YR  (7/12)(3/11) = 21/132  
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document.write( "YB  (7/12)(2/11) = 14/132
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document.write( "RY  (3/12)(7/11) = 21/132
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document.write( "RR               = 6/132
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document.write( "RB               = 6/132\r
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document.write( "BY               = 14/132
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document.write( "BR               = 6/132
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document.write( "BB               = 2/132
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document.write( "                -----------
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document.write( "          Total  = 132/132\r
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document.write( "Adding the two entries RB and BR:  (6+6)/132 = 12/132 =  \n" );
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