Discussion Forum



1)

A computer producing factory has only two plants   $T_{1}$   and $T_{2}$. Plant  $T_{1}$ produces 20% and $T_{2}$ 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:

Option C

Explanation:

Let x=P   (computer turns out to be defective, given that it is produced in plant $T_{2}$)

$\Rightarrow$        $x=P(\frac{D}{T_{2}})$           .............(i)

where  , D= Defective computer

$\therefore$ P (computer  turns out to be defective given that is produced  in plant  $T_{1}$)=10x

 i.e,     $P(\frac{D}{T_{1}})=10x$            .......(ii)

Also,   $P(T_{1})=\frac{20}{100}$  and  $P(T_{2})=\frac{80}{100}$

Given, P  (defective computer )=  $\frac{7}{100}$

Using law of total probability,

 $P(D)=9(T_{1}).P(\frac{D}{T_{1}})+ P(T_{2}).P(\frac{D}{T_{2}})$

$\therefore$       $\frac{7}{100}=(\frac{20}{100}).10x+(\frac{80}{100}).x$

$\Rightarrow$     7= (280)x

$\Rightarrow$     $x=\frac{1}{40}$          ..........(iii)

$\therefore$    $P(\frac{D}{T_{2}})=\frac{1}{40}$  and   $P(\frac{D}{T_{1}})=\frac{10}{40}$

$\Rightarrow$    $P(\frac{\overline{D}}{T_{2}})=1-\frac{1}{40}=\frac{39}{40}$

and    $P(\frac{\overline{D}}{T_{1}})=1-\frac{10}{40}=\frac{30}{40}$                       .........(iv)

 Using Baye's theorem,

 $P(\frac{T_{2}}{D})=\frac{P(T_{2}\cap\overline{D})}{P(T_{1}\cap\overline{D})+P(T_{2}\cap\overline{D})}$

 $=\frac{P(T_{2}).P(\frac{\overline{D}}{T_{2}})}{P(T_{1}).P(\frac{\overline{D}}{T_{1}})+P(T_{2}).P(\frac{\overline{D}}{T_{2}})}$

 $=\frac{\frac{80}{100}.\frac{39}{40}}{\frac{20}{100}.\frac{30}{40}+\frac{80}{100}.\frac{39}{40}}=\frac{78}{93}$