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Question: Answered & Verified by Expert
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)=10 P(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 is
MathematicsProbabilityJEE Main
Options:
  • A 3673
  • B 4779
  • C 7893
  • D 7583
Solution:
1754 Upvotes Verified Answer
The correct answer is: 7893
PT1=20100

PT2=80100

Let, PDT2=x (where, D represents defective units)

PDT1=10x

PD=7100 (given)

PT1 PDT1+PT2 PDT2=7100

20100×10x+80100×x=7100

x=140

PDT2=140 

PD-T2=3940Probability of not defective, given that it is produced in plant T2

PDT1=1040   PD-T1=3040

Now using Bayes' theorem

Probability of computer from T2 given that it is not defective :

PT2D-=80100×394020100×3040+80100×3940=7893

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