A factory has three machines A, B and C, which produce 100, 200 and 300 items of a particular type daily. The machines produce 2%, 3% and 5% defective items respectively. One day when the production was over, an item was picked up randomly and it was found to be defective. Find the probability that it was produced by machine A.

Hint: In this problem, let us find the probabilities of each machine with the total number of produced items. Find the probability of the item being produced being a defective by all the machines using the conditional probability formula, $P\left( X|Y \right)=\dfrac{P\left( X\cap Y \right)}{P(Y)}$ and then find the probability of the defective produced by the given machine by using Baye’s theorem, $P\left( {{E}_{1}}|D \right)=\dfrac{P\left( D|{{E}_{1}} \right)\cdot P\left( {{E}_{1}} \right)}{P\left( D \right)}$.

Complete step-by-step answer:
Here let us denote machines A, B and C by ${{E}_{1}},\,{{E}_{2}}$and ${{E}_{3}}$, also denote the defective items to be D. Firstly, we will find the probability of the selected item produced by all the three machines
Total number of productions = 100 + 200 + 300
                                                     = 600
Now, probability of machine ${{E}_{1}}$,
$\begin{align}
  & P\left( {{E}_{1}} \right)=\dfrac{100}{600} \\
 & =\dfrac{1}{6}
\end{align}$
Probability of machine ${{E}_{2}}$,
$\begin{align}
  & P\left( {{E}_{2}} \right)=\dfrac{200}{600} \\
 & =\dfrac{2}{6}
\end{align}$
Probability of machine ${{E}_{3}}$,
$\begin{align}
  & P\left( {{E}_{3}} \right)=\dfrac{300}{600} \\
 & =\dfrac{3}{6}
\end{align}$
Now, we know according to conditional probability theorem, $P\left( X|Y \right)=\dfrac{P\left( X\cap Y \right)}{P(Y)}$
Here, let us find the probability of the item being produced being a defective by machine ${{E}_{1}}$,
\[\begin{align}
  & P\left( D|{{E}_{1}} \right)=\dfrac{P\left( D\cap {{E}_{1}} \right)}{P\left( {{E}_{1}} \right)} \\
 & =\dfrac{2}{100}
\end{align}\]
Similarly find the probability of the item being produced being defective by machines ${{E}_{2}}$and ${{E}_{3}}$.
\[\begin{align}
  & P\left( D|{{E}_{2}} \right)=\dfrac{P\left( D\cap {{E}_{2}} \right)}{P\left( {{E}_{2}} \right)} \\
 & =\dfrac{3}{100}
\end{align}\]
 \[\begin{align}
  & P\left( D|{{E}_{3}} \right)=\dfrac{P\left( D\cap {{E}_{3}} \right)}{P\left( {{E}_{3}} \right)} \\
 & =\dfrac{5}{100}
\end{align}\]
By using Baye’s theorem, we know that
$P\left( Y|X \right)=\dfrac{P\left( X|Y \right)\cdot P\left( Y \right)}{P\left( X \right)}$
To find the picked defective item which can be from machine ${{E}_{1}}$, the required probability is
$P\left( {{E}_{1}}|D \right)=\dfrac{P\left( D|{{E}_{1}} \right)\cdot P\left( {{E}_{1}} \right)}{P\left( D \right)}$, here $P(D)$is the total probability of the defective items from all the three machines. Therefore, we get
$\begin{align}
  & P\left( {{E}_{1}}|D \right)=\dfrac{P\left( D|{{E}_{1}} \right)\cdot P\left( {{E}_{1}} \right)}{P\left( {{E}_{1}} \right)\cdot P\left( \dfrac{D}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)\cdot P\left( \dfrac{D}{{{E}_{2}}} \right)+P\left( {{E}_{3}} \right)\cdot P\left( \dfrac{D}{{{E}_{3}}} \right)} \\
 & =\dfrac{\dfrac{2}{100}\cdot \dfrac{1}{6}}{\left( \dfrac{1}{6} \right)\cdot \left( \dfrac{2}{100} \right)+\left( \dfrac{2}{6} \right)\cdot \left( \dfrac{3}{100} \right)+\left( \dfrac{3}{6} \right)\cdot \left( \dfrac{5}{100} \right)} \\
 & =\dfrac{\dfrac{2}{600}}{\dfrac{23}{600}} \\
 & =\dfrac{2}{23}
\end{align}$
Hence the probability for the item picked produced by machine ${{E}_{1}}$to be defective is $\dfrac{2}{23}$.

Note: Here, the question depends on the Bayes theorem and conditional probability theorem. Also, do not cancel out the fractions and keep them the way they are which makes the calculations easy. We denote the three machines three different annotations to be less confused.