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
(A) 36 (B) 47
73 79
(C) 78 (D) 75
93 83
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Answered by
1
Answer:
Answer:
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Answered by
27
Answer:
- c)78 is the Right Answer.
Hope it will be helpful ✌️
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