A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of the computers produced in the factory turn out to be defective. It is known that P(computer turns out to be defective given that it is produced in plant T2) = 10 × P(computer turns out to be defective given that it is produced in plant T1). A computer produced in the factory is randomly selected and it does not turn out to be defective, then the probability that it is produced in plant T2 is:

A computer producing factory has only two plants T1 and T2. Plant T1 p

1.	A computer producing factory has only two plants T1 and T2. Plant T1 produces 20% and plant T2 produces 80% of the total computers produced. 7% of the computers produced in the factory turn out to be defective. It is known that P(computer turns out to be defective given that it is produced in plant T2) = 10 × P(computer turns out to be defective given that it is produced in plant T1). A computer produced in the factory is randomly selected and it does not turn out to be defective, then the probability that it is produced in plant T2 is:
A.	\(\frac {36}{73}\)
B.	\(\frac {47}{79}\)
C.	\(\frac {78}{93}\)
D.	\(\frac {75}{83}\)
Answer» D. \(\frac {75}{83}\)

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