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 $T_1)$ = 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

The correct option is C

$\frac{78}{93}$

Let x = P (computer turns out to be defective, given that it is produced in plant $T_2$)
$\Rightarrow$ $x=P\left(D|T_2\right)$ ...(i)
Where, D = Defective computer
$\therefore P$ (computer turns out to be defective given that it is produced in plant $T_1$) = 10x
i.e $P\left(D|T_1\right)=10x$ ...(ii)
Also, $P\left(T_1\right)=\frac{20}{100}andP\left(T_2\right)=\frac{80}{100}$
Given, P (defective computer)$=\frac{7}{100}$
i.e. $P\left(D\right)=\frac{7}{100}$
Using law of total probability,
$P\left(D\right)=P\left(T_1\right).P\left(D|T_1\right)+P\left(T_2\right).P\left(D|T_2\right)\therefore \frac{7}{100}=\left(\frac{20}{100}\right).10x+\left(\frac{80}{100}\right).x\Rightarrow 7=\left(280\right)x\Rightarrow x=\frac{1}{40}...\left(iii\right)$
$\therefore P\left(D|T_2\right)=\frac{1}{40}andP\left(D|T_1\right)=\frac{10}{40}\Rightarrow P\left(\overline{D}|T_2\right)=1-\frac{1}{40}=\frac{39}{40}andP\left(\overline{D}|T_1\right)=1-\frac{10}{40}=\frac{30}{40}...\left(iv\right)$
Using Bayes' theorem,
$P\left(T_2|\overline{D}\right)=\frac{P(T_2\cap \overline{D})}{P(T_1\cap \overline{D})+P(T_2\cap \overline{D})}=\frac{P\left(T_2\right).P\left(\overline{D}|T_2\right)}{P\left(T_1\right).P\left(\overline{D}|T_1\right)+P\left(T_2\right).P\left(\overline{D}|T_2\right)}\frac{\frac{80}{100}.\frac{39}{40}}{\frac{20}{100}.\frac{30}{40}+\frac{80}{100}.\frac{39}{40}}=\frac{78}{93}$