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| Question 1056099:  Suppose three machines A, B, and C produce similar engine components. Machine A produces 45% of the
 total components, Machines produces 30%, and machine C produces 25%. For the usual production
 schedule, 6% of the components produced by Machine A do not meet established specifications; for
 Machines B and Machine C the figures are 4% and 3% respectively. One component is selected at random
 from from the total output and is found to be defective. What is the probability that the component selected
 was produced by machine A ?
 Answer by Boreal(15235)
      (Show Source): 
You can put this solution on YOUR website! Look at 2000 components Machine A produces 900 and 54 (6%) are defective  846 not--900
 Machine B produces 600 and 24 (4%) are defective, 576 not--600
 Machine C produces 500 and 15 (3%) are defective, 485 not--500
 Totals===========2000----93-------------------1907 not-2000
 Given that one component is defective, the probability it is from A is 54/93 (18/31)=58.0%
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