Question

A computer producing factory has only two plants ${\text{T}}_1$ and ${\text{T}}_2$ . Plant ${\text{T}}_1$ produces $20\%$ and plant ${\text{T}}_2$ 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 $T_1$ ) $=10P($ 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)
$P\left(T_1\right)P\left(\frac{D}{T_1}\right)+P\left(T_2\right)P\left(\frac{D}{T_2}\right)=\frac{7}{100}$
$\frac{20}{100}\times 10x+\frac{80}{100}\times x=\frac{7}{100}$
$x=\frac{1}{40}$
$P\left(\frac{D}{T_2}\right)=\frac{1}{40}$
$\Rightarrow P\left(\frac{\stackrel{-}{D}}{T_2}\right)=\frac{39}{40}=$ Probability of not defective, given that it is produced in plant $T_2$
$P\left(\frac{D}{T_1}\right)=\frac{10}{40}\Rightarrow P\left(\frac{\stackrel{-}{D}}{T_1}\right)=\frac{30}{40}$
Now using Bayes' theorem
Probability of computer from $T_2$ given that it is not defective :
$P\left(\frac{T_2}{\stackrel{-}{D}}\right)=\frac{\frac{80}{100}\times \frac{39}{40}}{\frac{20}{100}\times \frac{30}{40}+\frac{80}{100}\times \frac{39}{40}}=\frac{78}{93}$