document.write( "Question 433452: Please help me with this difficult homework problem:\r
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document.write( "A company uses printer cartridges from three manufacturers. Manufacturers A, B, and C supply, respectively, 60%, 30%, and 10% of the company’s cartridges. 1% of cartridges from A, 2% from B, and 3% from C are defective. If a cartridge is randomly chosen and found to be defective, what is the probability that it is from manufacturer C? \n" );
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Algebra.Com's Answer #300466 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! A company uses printer cartridges from three manufacturers. \n" ); document.write( "Manufacturers A, B, and C supply, respectively, 60%, 30%, and 10% of the company’s cartridges. \n" ); document.write( "---------------------------------- \n" ); document.write( "1% of cartridges from A, \n" ); document.write( "2% from B, \n" ); document.write( "3% from C are defective. \n" ); document.write( "---------------------------------- \n" ); document.write( "P(d|A) = P(d and A)/P(A) = P(d and A)/0.6 = 0.01 \n" ); document.write( "---- \n" ); document.write( "P(d|B) = P(d and B)/P(B) = P(d and B)/0.3 = 0.02 \n" ); document.write( "---- \n" ); document.write( "P(d|C) = P(d and C)/P(C) = P(d and C)/0.1 = 0.03 \n" ); document.write( "----\r \n" ); document.write( "\n" ); document.write( "If a cartridge is randomly chosen and found to be defective, what is the probability that it is from manufacturer C? \n" ); document.write( "--- \n" ); document.write( "P(C|d) = P(C and d)/P(d) \n" ); document.write( "--- \n" ); document.write( "= P(C and d)/[P(d and A)+P(d and B)+P(d and A)] \n" ); document.write( "--- \n" ); document.write( "= [0.1*0.03)/[0.6*0.01+0.3*0.02+0.1*0.03] \n" ); document.write( "--- \n" ); document.write( "= 0.003/0.015 \n" ); document.write( "---- \n" ); document.write( "= 0.2 \n" ); document.write( "================ \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "--- \n" ); document.write( " \n" ); document.write( " |