JEE Advance - Mathematics (2016 - Paper 1 Offline - No. 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 $${T_1}$$)
$$ = 10P$$ (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
$$ = 10P$$ (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
$${{36} \over {73}}$$
$${{47} \over {79}}$$
$${{78} \over {93}}$$
$${{75} \over {83}}$$
Explanation
Let $P_1$ be the defective computers that are produced from plant $T_1$ and $P_2$ be that from plant $T_2$.
The total percentage of the defective computers produced is $7 \%$.
Now, $P_1=10 P$ and $P_2=P$.
The computers produced that are defective:
$$ \begin{gathered} \frac{20}{100} \times P_1+\frac{80}{100} \times P_2=\frac{7}{100} \\\\ 20 P_1+80 P_2=7 \\\\ 200 P+80 P=7 \\\\ P=\frac{7}{280}=\frac{1}{40} \end{gathered} $$
Now, the probability of the defective products is calculated as follows:
$$ \frac{20}{100} P_1+\frac{80}{100} P_2=\frac{20}{100} \times \frac{1}{4}+\frac{80}{100} \times \frac{1}{40}=\frac{28}{400}=\frac{7}{100} $$
The probability of producing NOT defective computers is
$$ 1-\frac{7}{100}=\frac{93}{100} $$
The probability that plant $T_2$ produces NOT defective computers is calculated as follows:
$$ \frac{(80 / 100) \times(39 / 40)}{(93 / 100)}=\frac{78}{93} $$
The total percentage of the defective computers produced is $7 \%$.
Now, $P_1=10 P$ and $P_2=P$.
The computers produced that are defective:
$$ \begin{gathered} \frac{20}{100} \times P_1+\frac{80}{100} \times P_2=\frac{7}{100} \\\\ 20 P_1+80 P_2=7 \\\\ 200 P+80 P=7 \\\\ P=\frac{7}{280}=\frac{1}{40} \end{gathered} $$
Now, the probability of the defective products is calculated as follows:
$$ \frac{20}{100} P_1+\frac{80}{100} P_2=\frac{20}{100} \times \frac{1}{4}+\frac{80}{100} \times \frac{1}{40}=\frac{28}{400}=\frac{7}{100} $$
The probability of producing NOT defective computers is
$$ 1-\frac{7}{100}=\frac{93}{100} $$
The probability that plant $T_2$ produces NOT defective computers is calculated as follows:
$$ \frac{(80 / 100) \times(39 / 40)}{(93 / 100)}=\frac{78}{93} $$
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