Question 1203815
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Three machines turn out all the products in a factory, with the first machine 
producing 10% of the products, the second machine 15%, and the third machine 75%. 
The first machine produces defective products 6% of the time, the second machine 15% of the time 
and the third machine 8% of the time. What is the probability that a non-defective product 
came from the second machine? (Round your answer to four decimal places.
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It is about calculating a conditional probability.

This conditional probability is the fraction, whose denominator is the probability
to get a non-defective product from all 3 machines, while the numerator
is the probability to get a non-defective product from second machine.


By knowing it, we write the formula immediately


    P = {{{(0.15*(1-0.15))/(0.1*(1-0.06) + 0.15*(1-0.15) + 0.75*(1-0.08))}}} = 0.1399  (rounded as requested).



<U>ANSWER</U>.  P = 0.1399  (rounded as requested).
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Solved, with complete explanations.