Question

A computer producing factory has only two plants $\textit{T}_{1}$ and $\textit{T}_{2}$ . Plant $\textit{T}_{1}$ produces $20\%$ and plant $\textit{T}_{2}$ 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 1

Answer: (C)

$P\left(T_{1}\right)=\frac{20}{100}$
$P\left(T_{2}\right)=\frac{80}{100}$
Let, $P\left(\frac{D}{T_{2}}\right)=x$ (where, $D$ represents defective units)
$P\left(\frac{D}{T_{1}}\right)=10x$
$P\left(D\right)=\frac{7}{100}$ (given)
$P\left(T_{1}\right) \, P\left(\frac{D}{T_{1}}\right)+P\left(T_{2}\right) \, P\left(\frac{D}{T_{2}}\right)=\frac{7}{100}$
$\frac{20}{100}\times 10x+\frac{80}{100}\times x=\frac{7}{100}$
$x=\frac{1}{40}$
$P\left(\frac{D}{T_{2}}\right)=\frac{1}{40} \, $
$\Rightarrow P\left(\frac{\overset{-}{D}}{T_{2}}\right)=\frac{39}{40}=$ Probability of not defective, given that it is produced in plant $T_{2}$
$P\left(\frac{D}{T_{1}}\right)=\frac{10}{40} \, \Rightarrow \, \, P\left(\frac{\overset{-}{D}}{T_{1}}\right)=\frac{30}{40}$
Now using Bayes' theorem
Probability of computer from $T_{2}$ given that it is not defective :
$P\left(\frac{T_{2}}{\overset{-}{D}}\right)=\frac{\frac{80}{100} \times \frac{39}{40}}{\frac{20}{100} \times \frac{30}{40} + \frac{80}{100} \times \frac{39}{40}}=\frac{78}{93}$