Question

# 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 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 T1)=10P(computer turns out to be defective given that it is produced in plant T2). Where P(E) denotes the probability of an event E. 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 isÂ Â

A
3673
B
4779
C
7893
D
7583

Solution

## The correct option is C 7893Let Event T1= The computer is produced in plant 1  Event T2= The computer is produced in plant 2  Event D= The computer produced turns out to be defective Given that P(T1)=0.2, P(T2)=0.8,P(D)=0.07 & P(D/T1)=10P(D/T2) Let P(D/T2)=p⇒P(D/T1)=10p P(D)=P(D/T1).P(T1)+P(D/T2).P(T2) ⇒0.07=10p(0.2)+p(0.8) ⇒p=140 P(T2/ ¯¯¯¯¯D )=P( ¯¯¯¯D /T2).P(T2)P( ¯¯¯¯D /T1).P(T1)+P( ¯¯¯¯D /T2).P(T2) ⇒P(T2/ ¯¯¯¯¯D )=0.8(1−140)0.2(1−1040)+0.8(1−140) ⇒P(T2/ ¯¯¯¯¯D )=7893

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