Question

A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of the cars and plant Y manufactures 30%. At plant X, 80% of the cars are rated of standard quality and at plant Y, 90% are rated of standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plant X.

Answer 1

Hint: In this question, we have to find the probability that, if car picked is of standard quality then it has come from plant X. For this, we will use the Bayes theorem according to which $P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{P\left( {{E}_{1}} \right)\cdot P\left( \dfrac{A}{{{E}_{1}}} \right)}{P\left( {{E}_{1}} \right)\cdot P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{A}{{{E}_{2}}} \right)}$.
Where $P\left( {{E}_{1}} \right)$ is the probability of event ${{E}_{1}}$ happening , $P\left( \dfrac{A}{{{E}_{1}}} \right)$ is the probability of selecting A from event ${{E}_{1}}$, $P\left( \dfrac{{{E}_{1}}}{A} \right)$ is the probability event ${{E}_{1}}$ happening under condition of A and similarly for $P\left( {{E}_{2}} \right)$ and $P\left( \dfrac{A}{{{E}_{2}}} \right)$.

Complete step-by-step answer:
Here we are given that a car manufacturing factory has two plants X and Y.
Let us suppose that ${{E}_{1}}$ represent the event such that cars manufactured from plant X is drawn, ${{E}_{2}}$ represent the event such that car manufactured from plant Y is drawn and A represent that the car drawn is of standard quality.
We need to find probability that if the selected car is of standard quality, then it is drawn from plant X which means we need to find $P\left( \dfrac{{{E}_{1}}}{A} \right)$.
We know that Bayes theorem is given as, $P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{P\left( {{E}_{1}} \right)\cdot P\left( \dfrac{A}{{{E}_{1}}} \right)}{P\left( {{E}_{1}} \right)\cdot P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{A}{{{E}_{2}}} \right)}$.
So let us find the value of $P\left( {{E}_{1}} \right)$,$P\left( {{E}_{2}} \right)$,$P\left( \dfrac{A}{{{E}_{1}}} \right)$,$P\left( \dfrac{A}{{{E}_{2}}} \right)$.
For $P\left( {{E}_{1}} \right)$.
It represents the probability of choosing a car from plant X. As we are given that, 70% of the cars are manufactured at plant X. So required probability will be $\dfrac{70}{100}$. Therefore, $P\left( {{E}_{1}} \right)=\dfrac{70}{100}=\dfrac{7}{10}$.
For $P\left( {{E}_{2}} \right)$.
It represents the probability of choosing a car from plant Y. As we are given that, 30% of the cars are manufactured at plant Y. So required probability will be $\dfrac{30}{100}$. Therefore, $P\left( {{E}_{2}} \right)=\dfrac{30}{100}=\dfrac{3}{10}$.
For $P\left( \dfrac{A}{{{E}_{1}}} \right)$.
It represents the probability of picking a standard car from plant X. As we are given that, 80% of the cars are standard in plant X. So required probability will be $\dfrac{80}{100}$. Therefore, $P\left( \dfrac{A}{{{E}_{1}}} \right)=\dfrac{80}{100}=\dfrac{8}{10}$.
For $P\left( \dfrac{A}{{{E}_{2}}} \right)$.
It represents the probability of picking a standard car from plant Y. As we are given that 90% of the cars are standard in plant Y. So the required probability will be $\dfrac{90}{100}$. Therefore, $P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{90}{100}=\dfrac{9}{10}$.
Putting in all the values we get:
\[\begin{align}
  & P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{\dfrac{7}{10}\times \dfrac{8}{10}}{\left( \dfrac{7}{10}\times \dfrac{8}{10} \right)+\left( \dfrac{3}{10}\times \dfrac{9}{10} \right)} \\
 & \Rightarrow \dfrac{\dfrac{56}{100}}{\dfrac{56}{100}+\dfrac{27}{100}} \\
 & \Rightarrow \dfrac{\dfrac{56}{100}}{\dfrac{83}{100}} \\
 & \Rightarrow \dfrac{56}{100}\times \dfrac{83}{100} \\
 & \Rightarrow \dfrac{56}{83} \\
\end{align}\]
Hence our required probability is $\dfrac{56}{83}$.

Note: Students should carefully use Bayes theorem by finding accurate probability. Student can get confused between $P\left( \dfrac{A}{{{E}_{1}}} \right)\text{ and }P\left( \dfrac{{{E}_{1}}}{A} \right)$. Take care while solving fractions. Generally, $P\left( \dfrac{A}{B} \right)$ means probability of event A occurring under the condition of B. Probability should always lie between 0 and 1.