Question

Assume that a factory has two machines A and B. Past records show that machine A produces 60% of the items of output and machine B produces 40% of the items. Further, 2% of the items produced by machine A were defective and only 1% produced by machine B were defective. If a defective item is drawn at random, what is the probability that it was produced by machine A?

Answer 1

We solve this problem by first assuming that the items produced by machine A and B are ${{E}_{1}}$ and ${{E}_{2}}$ respectively. Then we can find the probabilities of those events from given information. Then we assume that the event that the item is defective is $D$. Then we can write the values of $P\left( {}^{D}/{}_{{{E}_{1}}} \right)$ and $P\left( {}^{D}/{}_{{{E}_{2}}} \right)$ from the given information. Then we consider the formula for Bayes Theorem, $P\left( A \right)=P\left( {{B}_{1}} \right)P\left( {}^{A}/{}_{{{B}_{1}}} \right)+P\left( {{B}_{2}} \right)P\left( {}^{A}/{}_{{{B}_{2}}} \right)$ and find the value of $P\left( D \right)$. Then we find the required probability, that is probability that item is from A given it is defective, that is $P\left( {}^{{{E}_{1}}}/{}_{D} \right)$ by using the formula for probability $P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( B\cap A \right)}{P\left( A \right)}$ and modifying it as $P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( {}^{A}/{}_{B} \right)\times P\left( B \right)}{P\left( A \right)}$ and substituting the values.

Complete step by step answer:
Let ${{E}_{1}}$ be the event for producing items produced by machine A.
Let ${{E}_{2}}$ be the event for producing items produced by machine B.
According to the question, we have been given the probabilities of items produced by machine A and machine B.
Therefore, the probability of items produced by machine A is
$\Rightarrow P\left( {{E}_{1}} \right)=\dfrac{60}{100}=\dfrac{3}{5}$
The probability of items produced by machine B is
$\Rightarrow P\left( {{E}_{2}} \right)=\dfrac{40}{100}=\dfrac{2}{5}$
Let the event of producing defective pieces be D.
So, the probability of producing defective pieces can be given as $P\left( D \right)$.
We are given the probabilities of defective pieces provided it is produced by machine A and B separately.
Therefore, we can say that the probability of defective pieces by machine A, that is probability that machine produced by A is defective is
$\Rightarrow P\left( {}^{D}/{}_{{{E}_{1}}} \right)=\dfrac{2}{100}$
The probability of defective pieces by machine B, that is probability that item produced by B is defective
$\Rightarrow P\left( {}^{D}/{}_{{{E}_{2}}} \right)=\dfrac{1}{100}$
We need to find the probability that a defective item is made by machine A.
It can be shown by $P\left( {}^{{{E}_{1}}}/{}_{D} \right)$.
Now let us consider the formula for Bayes Theorem.
$P\left( A \right)=P\left( {{B}_{1}} \right)P\left( {}^{A}/{}_{{{B}_{1}}} \right)+P\left( {{B}_{2}} \right)P\left( {}^{A}/{}_{{{B}_{2}}} \right)$
Using this we can write the probability of defective machines as,
$P\left( D \right)=P\left( {{E}_{1}} \right)P\left( {}^{D}/{}_{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)P\left( {}^{D}/{}_{{{E}_{2}}} \right)$
Substituting all the values we get,
$P\left( D \right)=\dfrac{2}{5}\times \dfrac{1}{100}+\dfrac{3}{5}\times \dfrac{2}{100}$
Solving this further, we get,
$P\left( D \right)=\dfrac{2}{125}$ .
Now let us consider the formula for conditional probability.
$P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( B\cap A \right)}{P\left( A \right)}$
Which can be modified again as,
$P\left( {}^{B}/{}_{A} \right)=\dfrac{P\left( {}^{A}/{}_{B} \right)\times P\left( B \right)}{P\left( A \right)}$
Using this we get,
$\Rightarrow P\left( {}^{{{E}_{1}}}/{}_{D} \right)=\dfrac{P\left( {}^{D}/{}_{{{E}_{1}}} \right)\times P\left( {{E}_{1}} \right)}{P\left( D \right)}$
Substituting all the values we get,
$P\left( {}^{{{E}_{1}}}/{}_{D} \right)=\dfrac{\dfrac{3}{5}\times \dfrac{2}{100}}{\dfrac{2}{125}}$
Solving this equation, we get,
$P\left( {}^{{{E}_{1}}}/{}_{D} \right)=\dfrac{3}{4}$
Hence, the probability that the defective item is produced by machine A is $\dfrac{3}{4}$.

Hence the answer is $\dfrac{3}{4}$.

Note: In this problem, we need to understand the method to find the probability by the condition on it. We need to multiply the probabilities of the event and the condition acting upon it. Also, we can cross-check by seeing the probabilities of an event is never greater than 1. Also, the percentage can be written as the number divided by 100.