$A computer producing factory has only two plants T_{1} and T_{2} \cdot plant T_{1} produces 20 %$ $and plant T_{2} produces 80 % of the total computers produced. 7 % of computers roduced$
$in the factory turn out to be defective. It is known that P (computer turns out to be$
$defective given that it is produced in plant T_{1}) = 10 P (computer turns out to be defective$
$given that it is produced in plant T_{2}) where P (E) denotes the probability of an event E .$ $A computer produced in the factory is randomly selected and it does not turn out to be$ $d e f e c t i v e, T h e n t h e p r o b a b i l i t y t h a t i t i s p r o d u c e d i n p l a n t i s$

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$\frac{47}{79}$

$\frac{75}{83}$

$\frac{36}{73}$

$\frac{78}{93}$

answer is C.

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Detailed Solution

$Given P (T_{1}) = \frac{2}{10}, P (T_{2}) = \frac{8}{10},$

$\begin{array}{l} P (D) = \frac{7}{100} \Rightarrow P (\bar{D}) = \frac{93}{100} \\ G i v e n P (D / T_{1}) = 10 P (D / T_{2}) \\ N o w P (D) = P (D / T_{1}) P (T_{1}) + P (D / T_{2}) P (T_{2}) \\ \Rightarrow \frac{7}{100} = 10 P (D / T_{2}) \times \frac{2}{10} + P (D / T_{2}) \times \frac{8}{10} \\ \Rightarrow P (D / T_{2}) = \frac{1}{40} \Rightarrow P (D / T_{1}) = \frac{1}{4} \\ \Rightarrow P (\bar{D}) = P (\bar{D} / T_{1}) P (T_{1}) + P (\bar{D} / T_{2}) P (T_{2}) \\ = (1 - P (D / T_{1})) P (T_{1}) + (1 - P (D / T_{2})) P (T_{2}) \\ = \frac{3}{4} \times \frac{2}{10} + \frac{39}{40} \times \frac{8}{10} = \frac{372}{400} \end{array}$

$P (T_{2} / \bar{D}) = \frac{P (\bar{D} / T_{2}) P (T_{2})}{P (\bar{D})} using BAYER THEOREM$

$= \frac{\frac{39}{40} \times \frac{8}{10}}{\frac{372}{400}} = \frac{39 \times 2}{93} = \frac{78}{93}$

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