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 $P\left(E\right)$ 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
Event $T_1$= The computer is produced in plant $1$
Event $T_2$= The computer is produced in plant $2$
Event $D$= The computer produced turns out to be defective

Given that
$P\left(T_1\right)=0.2,P\left(T_2\right)=0.8,P\left(D\right)=0.07\&P\left(D/T_1\right)=10P\left(D/T_2\right)$

Let $P\left(D/T_2\right)=p\Rightarrow P\left(D/T_1\right)=10p$

$P\left(D\right)=P\left(D/T_1\right).P\left(T_1\right)+P\left(D/T_2\right).P\left(T_2\right)$
$\Rightarrow 0.07=10p\left(0.2\right)+p\left(0.8\right)$
$\Rightarrow p=\frac{1}{40}$

$P\left(T_2/\overline{D}\right)=\frac{P\left(\overline{D}/T_2\right).P\left(T_2\right)}{P\left(\overline{D}/T_1\right).P\left(T_1\right)+P\left(\overline{D}/T_2\right).P\left(T_2\right)}$

$\Rightarrow P\left(T_2/\overline{D}\right)=\frac{0.8(1-\frac{1}{40})}{0.2(1-\frac{10}{40})+0.8(1-\frac{1}{40})}$
$\Rightarrow P\left(T_2/\overline{D}\right)=\frac{78}{93}$