Question

A compute producing factory has only two plants ${{T}_{1}}$ and ${{T}_{2}}.$ Plant ${{T}_{1}}$ produces 20% and the plant ${{T}_{2}}$ produces 80% of the total computes produced. 7% of computers produced in the factory turn out to be defective. It is know that P (computer turns out to be defective given that it produced in the plant ${{T}_{1}}$ ) = 10P (computer turns out to be defective given that it is produced in the 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 the plant ${{T}_{2}}$ is:
$\left( a \right)\dfrac{36}{73}$
$\left( b \right)\dfrac{47}{79}$
$\left( c \right)\dfrac{78}{93}$
$\left( d \right)\dfrac{75}{83}$

Answer 1

Assume ${{E}_{1}}$ as the event that computer is produced in plant ${{T}_{1}}$ and ${{E}_{2}}$ as the event that computer is produced in plant ${{T}_{2}}.$ Now, assume A as the event that computer turns out to be defective and $\overline{A}$ as the event that it does not turn out to be defective. We will use the information given in the question to find out the value of conditional probability $P\left( \dfrac{A}{{{E}_{2}}} \right)$ and $P\left( \dfrac{A}{{{E}_{1}}} \right).$ Then we will use the formula of total probability theorem which is given as $P\left( A \right)=P\left( {{E}_{1}} \right).P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right).P\left( \dfrac{A}{{{E}_{2}}} \right).$ Finally, we will apply Bayes theorem to find the probability $P\left( \dfrac{{{E}_{2}}}{\overline{A}} \right)$ using the formula $P\left( \dfrac{{{E}_{2}}}{\overline{A}} \right)=\dfrac{P\left( {{E}_{2}} \right).P\left( \dfrac{\overline{A}}{{{E}_{2}}} \right)}{P\left( {{E}_{1}} \right).P\left( \dfrac{\overline{A}}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right).P\left( \dfrac{\overline{A}}{{{E}_{2}}} \right)}.$

Complete step-by-step solution:
Here, to solve the above question, let us make the following assumptions. So, we are assuming,
${{E}_{1}}$ as the event that the computer is produced in the plant ${{T}_{1}}.$
${{E}_{2}}$ as the event that the computer is produced in the plant ${{T}_{2}}.$
A is the event that the computer turns out to be defective.
$\overline{A}$ as the event that the computer does not turn out to be defective.
Now, we have been given that plant ${{T}_{1}}$ produced 20% and plant ${{T}_{2}}$ produced 80% of the total computer. Therefore, we have,
$P\left( {{E}_{1}} \right)=\dfrac{20}{100}=\dfrac{1}{5}$
$P\left( {{E}_{2}} \right)=\dfrac{80}{100}=\dfrac{4}{5}$
It is also given to us that 7% computer turns out to be defective, so we have,
$P\left( A \right)=\dfrac{7}{100}$
We have been provided with the conditional probability that P (computer turns out to be defective given that it produced in the plant ${{T}_{1}}$ ) = 10P (computer turns out to be defective given that it is produced in the plant ${{T}_{2}}$ ). So, mathematically we have,
$P\left( \dfrac{A}{{{E}_{1}}} \right)=10P\left( \dfrac{A}{{{E}_{2}}} \right)$
Let us assume $P\left( \dfrac{A}{{{E}_{2}}} \right)=x,$ therefore, $P\left( \dfrac{A}{{{E}_{1}}} \right)=10x.$ Now, applying the total probability theorem, we get,
$P\left( A \right)=P\left( {{E}_{1}} \right).P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right).P\left( \dfrac{A}{{{E}_{2}}} \right)$
$\Rightarrow \dfrac{7}{100}=\dfrac{1}{5}\times 10x+\dfrac{4}{5}\times x$
$\Rightarrow \dfrac{7}{100}=\dfrac{14x}{5}$
$\Rightarrow x=\dfrac{1}{40}$
Hence, $P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{1}{40}$ and $P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{10}{40}=\dfrac{1}{4}.$
Here, we have to find the probability of selecting a computer from a plant ${{T}_{2}}$ with the given condition that it is not defective, i.e. we have to find $P\left( \dfrac{{{E}_{2}}}{\overline{A}} \right).$ Since, we have, $P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{1}{40},$
$\Rightarrow P\left( \dfrac{\overline{A}}{{{E}_{2}}} \right)=1-\dfrac{1}{40}=\dfrac{39}{40}$
$P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{1}{4},$ so, we get,
$\Rightarrow P\left( \dfrac{\overline{A}}{{{E}_{1}}} \right)=1-\dfrac{1}{4}=\dfrac{3}{4}$
Now applying Bayes Theorem, we get,
$P\left( \dfrac{{{E}_{2}}}{\overline{A}} \right)=\dfrac{P\left( {{E}_{2}} \right).P\left( \dfrac{\overline{A}}{{{E}_{2}}} \right)}{P\left( {{E}_{1}} \right).P\left( \dfrac{\overline{A}}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right).P\left( \dfrac{\overline{A}}{{{E}_{2}}} \right)}$
Substituting all the values, we get,
$\Rightarrow P\left( \dfrac{{{E}_{2}}}{\overline{A}} \right)=\dfrac{\dfrac{4}{5}\times \dfrac{39}{40}}{\dfrac{1}{5}\times \dfrac{3}{4}+\dfrac{4}{5}\times \dfrac{39}{40}}$
$\Rightarrow P\left( \dfrac{{{E}_{2}}}{\overline{A}} \right)=\dfrac{78}{93}$
Hence, the option (c) is the correct answer.

Note: One may note that in the solution of the above question, we have used the theorem of conditional probability, total probability, and Baye’s theorem. So, it is very important to remember the statements of these theorems otherwise it will be very difficult for us to solve such a question. It is important to assume the event as some variable alphabet so that we do not have to write the long conditions again and again.