Probability Ques 113
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 $\left.T _1\right)=10 P$ (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
(2016 Adv.)
(a) $\frac{36}{73}$
(b) $\frac{47}{79}$
(c) $\frac{78}{93}$
(d) $\frac{75}{83}$
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Answer:
Correct Answer: 113.(c)
Solution:
- Let $x=P$ (computer turns out to be defective, given that it is produced in plant $T _2$ )
$ \Rightarrow \quad x=P (\frac{D}{T _2}) $
where, $D=$ Defective computer
$\therefore P$ (computer turns out to be defective given that is produced in plant $\left.T _1\right)=10 x$
i.e.
$P (\frac{D}{T _1})=10 x $
$P\left(T _1\right)=\frac{20}{100} \text { and } P\left(T _2\right)=\frac{80}{100}$
Also,
Given, $P$ (defective computer) $=\frac{7}{100}$
$ \text { i.e. } \quad P(D)=\frac{7}{100} $
Using law of total probability,
$P(D)=9\left(T _1\right) \cdot P \quad (\frac{D}{T _1})+P\left(T _2\right) \cdot P (\frac{D}{T _2}) $
$ \therefore \frac{7}{100}=(\frac{20}{100}) \cdot 10 x+(\frac{80}{100}) \cdot x $
$\Rightarrow \quad 7 =(280) x \Rightarrow \quad x=\frac{1}{40} $
$P (\frac{D}{T _2})=\frac{1}{40} \text { and } \quad P (\frac{D}{T _1})=\frac{10}{40} $
$\Rightarrow \quad P (\frac{\bar{D}}{T _2})=1-\frac{1}{40}=\frac{39}{40} \text { and } P \frac{\bar{D}}{T _1}=1-\frac{10}{40}=\frac{30}{40}$
Using Baye’s theorem,
$P \frac{T _2}{\bar{D}} =\frac{P\left(T _2 \cap \bar{D}\right)}{P\left(T _1 \cap \bar{D}\right)+P\left(T _2 \cap \bar{D}\right)} $
$=\frac{P\left(T _2\right) \cdot P (\frac{\bar{D}}{T _2})}{P\left(T _1\right) \cdot P (\frac{\bar{D}}{T _1})+P\left(T _2\right) \cdot P (\frac{\bar{D}}{T _2})} $
$ =\frac{\frac{80}{100} \cdot \frac{39}{40}}{\frac{20}{100} \cdot \frac{30}{40}+\frac{80}{100} \cdot \frac{39}{40}}=\frac{78}{93}$
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