Question
A computer producing factory has only two plants $${T_1}$$ and $${T_2}.$$ Plant $${T_1}$$ produces 20% and plant $${T_2}$$ 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 $${T_1}$$)
= 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 defective. Then the probability that it is produced in plant $${T_2}$$ is
A.
$$\frac{{36}}{{73}}$$
B.
$$\frac{{47}}{{79}}$$
C.
$$\frac{{78}}{{93}}$$
D.
$$\frac{{75}}{{83}}$$
Answer :
$$\frac{{78}}{{93}}$$
Solution :
$$\eqalign{
& P\left( {{T_1}} \right) = \frac{{20}}{{100}},P\left( {{T_2}} \right) = \frac{{80}}{{100}},P\left( D \right) = \frac{7}{{100}} \cr
& {\text{Let }}P\left( {\frac{D}{{{T_2}}}} \right) = x,\,{\text{then }}P\left( {\frac{D}{{{T_1}}}} \right) = 10x \cr
& {\text{Also }}P\left( D \right) = P\left( {{T_1}} \right)P\left( {\frac{D}{{{T_1}}}} \right) + P\left( {{T_2}} \right)P\left( {\frac{D}{{{T_2}}}} \right) \cr
& \Rightarrow \,\,\frac{7}{{100}} = \frac{{20}}{{100}} \times 10x + \frac{{80}}{{100}} \times x \cr
& \Rightarrow \,\,\frac{7}{{280}} = x\,\,{\text{or }}x = \frac{1}{{40}} \cr
& P\left( {\frac{D}{{{T_1}}}} \right) = \frac{{10}}{{40}}\,\,{\text{and }}P\left( {\frac{D}{{{T_2}}}} \right) = \frac{1}{{40}} \cr
& \Rightarrow \,\,P\left( {\frac{{\overline D }}{{{T_1}}}} \right) = \frac{{30}}{{40}}\,\,{\text{and}}{\text{ }}P\left( {\frac{{\overline D }}{{{T_2}}}} \right) = \frac{{39}}{{40}} \cr
& P\left( {\frac{{{T_2}}}{{\overline D }}} \right) = \frac{{P\left( {\frac{{\overline D }}{{{T_2}}}} \right)P\left( {{T_2}} \right)}}{{P\left( {\frac{{\overline D }}{{{T_1}}}} \right)P\left( {{T_1}} \right) + P\left( {\frac{{\overline D }}{{{T_2}}}} \right)P\left( {{T_2}} \right)}} \cr
& = \frac{{\frac{{80}}{{100}} \times \frac{{39}}{{40}}}}{{\frac{{20}}{{100}} \times \frac{{30}}{{40}} + \frac{{80}}{{100}} \times \frac{{39}}{{40}}}} \cr
& = \frac{{156}}{{186}} \cr
& = \frac{{26}}{{31}} \cr
& {\text{Also }}\frac{{78}}{{93}} = \frac{{26}}{{31}} \cr} $$