Question 44

Mathematics Probability Hard

A computer producing factory has only two plants T<sub>1</sub> and T<sub>2</sub>. Plant T<sub>1</sub> produces 20% and plant T<sub>2</sub> 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 T<sub>1</sub>) = 10 &times; P(computer turns out to be defective given that it is produced in plant T<sub>2</sub>). 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 T<sub>2</sub> is:

(A) <span class="math-tex">\(\frac {36}{73}\)</span>
(B) <span class="math-tex">\(\frac {47}{79}\)</span>
(C) <span class="math-tex">\(\frac {78}{93}\)</span>
(D) <span class="math-tex">\(\frac {75}{83}\)</span>
View Dynamic Solution & Explanation
Correct Solution: Option C

Step-by-step Solution:

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