In a bolt factory, three machines A, B and C manufacture 25, 35 and 40 percent of the total bolts respectively. Out of the total bolts manufactured by the machines 5, 4 and 2 percent are defective from machine A, B and C respectively. A bolt drawn at random is found to be defective. Find the probability that it was manufactured by (i) Machine A or C (ii) Machine B.

Hint: We will be using the concepts of probability to solve the problem. We will first find the probability of a bolt manufactured by the machines A, B, C respectively then we will find the probability of a bolt manufactured by a machine is defective and then will be using Bayes theorem to find the answer.

Complete step-by-step answer:

We have been given that;
Bolts manufactured from machine A = 25%
Bolts manufactured from machine B = 35%
Bolts manufactured from machine C = 40%
So, probability of bolt manufactured by machine $A=P\left( A \right)=\dfrac{25}{100}=0.25$
Probability of bolt manufactured by machine $B=P\left( B \right)=\dfrac{35}{100}=0.35$
Probability of bolt manufactured by machine $C=P\left( C \right)=\dfrac{40}{100}=0.40$
Now, probability of a defective bolt manufactured by machine $A=P\left( D\left| A \right. \right)=\dfrac{5}{100}=0.05$
Probability of a defective bolt manufactured by machine $B=P\left( D\left| B \right. \right)=\dfrac{4}{100}=0.04$
Probability of a defective bolt manufactured by machine $C=P\left( D\left| C \right. \right)=\dfrac{2}{100}=0.02$

Now, we know that according to Bayes theorem;
$P\left( A\left| B \right. \right)=\dfrac{P\left( B\left| A \right. \right).P\left( A \right)}{P\left( B \right)}$
Where;
$A,B=$ events
$P\left( A\left| B \right. \right)=$ probability of B given A is true
$P\left( A \right),P\left( B \right)=$ The independent probability of A and B
So,
\[P\left( B\left| D \right. \right)=\dfrac{P\left( B \right).P\left( D\left| B \right. \right)}{P\left( A \right).P\left( D\left| A \right. \right)+P\left( B \right).P\left( D\left| B \right. \right)+P\left( C \right).P\left( D\left| C \right. \right)}\]
Where D means bolt is defective
$\begin{align}
  & P\left( B\left| D \right. \right)=\dfrac{0.35\times 0.04}{0.25\times 0.05+0.35\times 0.04+0.4\times 0.02} \\
 & =\dfrac{0.014}{0.0125+0.014+0.008} \\
 & =\dfrac{0.014}{0.0345} \\
 & =\dfrac{140}{345} \\
 & =\dfrac{28}{69} \\
\end{align}$
Therefore, probability that it was manufactured by machine B is $\dfrac{28}{69}$
Now, for is part we have to find the probability that it was manufactured by machine A or C. It is same as the probability that is not manufactured by B. Therefore, probability it was manufactured by machine A or C;
$=1-P\left( B\left| D \right. \right)$
$\begin{align}
  & =1-\dfrac{28}{69} \\
 & =\dfrac{69-28}{69} \\
 & =\dfrac{41}{69} \\
\end{align}$

Note: These types of questions are formula based so remembering the formula of Baye’s theorem will be helpful, also it becomes easier to solve (i) part by solving (ii) first.