document.write( "Question 1115761: . A company has three machines A, B and C which all produce the same two parts, X and Y. of all the parts produced, machine A produces 60%, machine B produces 30%, and machine C produces the rest. 40% of the parts made by machine A are part X, 50% of the parts made by machine B are part X, and 70% of the parts made by machine C are part X. A part produced by this company is randomly sampled and is determined to be an X part. With the knowledge that it is an X part, find the probabilities that the part came from machine A, B or C. \n" ); document.write( "
Algebra.Com's Answer #730593 by ikleyn(52778)\"\" \"About 
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document.write( "Let aX = the number of parts X produced by machine A;\r\n" );
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document.write( "    aY = the number of parts Y produced by machine A;\r\n" );
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document.write( "    bX = the number of parts X produced by machine B;\r\n" );
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document.write( "    bY = the number of parts Y produced by machine B;\r\n" );
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document.write( "    cX = the number of parts X produced by machine C;\r\n" );
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document.write( "    cY = the number of parts Y produced by machine C.\r\n" );
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document.write( "Let P = aX + aY + bX + bY + cX + cY is the FULL NUMBER of all produced Parts.   \r\n" );
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document.write( "Then the probability that the part X came from machine A  is equal to  \"%28aX%29%2F%28aX+%2B+bX+%2B+cX%29\",   and we have\r\n" );
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document.write( "    aX = (0.6*P)*0.4 = 0.24*P\r\n" );
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document.write( "    bX = (0.3*P)*0.5 = 0.15*P\r\n" );
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document.write( "    cX = (0.1*P)*0.7 = 0.07*P.\r\n" );
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document.write( "Therefore,  the probability that the part X came from machine A = \"%28aX%29%2F%28aX+%2B+bX+%2B+cX%29\" = \"%280.24%2AP%29%2F%280.24%2AP+%2B+0.15%2AP+%2B+0.07P%29\" = \"0.24%2F%280.24%2B0.15%2B0.07%29\" = 0.5217 = 52.17%.\r\n" );
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