Probability 4 Question 3
3. 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: 3. (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.
$$ \begin{gathered} P \frac{D}{T _1}=10 x \\ P\left(T _1\right)=\frac{20}{100} \text { and } P\left(T _2\right)=\frac{80}{100} \end{gathered} $$
Also,
Given, $P$ (defective computer) $=\frac{7}{100}$
$$ \text { i.e. } \quad P(D)=\frac{7}{100} $$
Using law of total probability,
$$ \begin{aligned} & & P(D)=9\left(T _1\right) \cdot P \quad \frac{D}{T _1}+P\left(T _2\right) \cdot P \frac{D}{T _2} \\ & \therefore & \frac{7}{100}=\frac{20}{100} \cdot 10 x+\frac{80}{100} \cdot x \\ \Rightarrow & \quad 7 & =(280) x \Rightarrow \quad x=\frac{1}{40} \\ & P & \frac{D}{T _2}=\frac{1}{40} \text { and } \quad P \frac{D}{T _1}=\frac{10}{40} \\ \Rightarrow & P & \frac{\bar{D}}{T _2}=1-\frac{1}{40}=\frac{39}{40} \text { and } P \frac{\bar{D}}{T _1}=1-\frac{10}{40}=\frac{30}{40} \end{aligned} $$
Using Baye’s theorem,
$$ \begin{aligned} P \frac{T _2}{\bar{D}} & =\frac{P\left(T _2 \cap \bar{D}\right)}{P\left(T _1 \cap \bar{D}\right)+P\left(T _2 \cap \bar{D}\right)} \\ & =\frac{P\left(T _2\right) \cdot P \frac{\bar{D}}{T _2}}{P\left(T _1\right) \cdot P \frac{\bar{D}}{T _1}+P\left(T _2\right) \cdot P \frac{\bar{D}}{T _2}} \\ & =\frac{\frac{80}{100} \cdot \frac{39}{40}}{\frac{20}{100} \cdot \frac{30}{40}+\frac{80}{100} \cdot \frac{39}{40}}=\frac{78}{93} \end{aligned} $$
