Question

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.

Answer 1

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 $A=P\left(A\right)={\displaystyle \frac{25}{100}}=0.25$
Probability of bolt manufactured by machine $B=P\left(B\right)={\displaystyle \frac{35}{100}}=0.35$
Probability of bolt manufactured by machine $C=P\left(C\right)={\displaystyle \frac{40}{100}}=0.40$
Now, probability of a defective bolt manufactured by machine $A=P\left(D|A\right)={\displaystyle \frac{5}{100}}=0.05$
Probability of a defective bolt manufactured by machine $B=P\left(D|B\right)={\displaystyle \frac{4}{100}}=0.04$
Probability of a defective bolt manufactured by machine $C=P\left(D|C\right)={\displaystyle \frac{2}{100}}=0.02$

Now, we know that according to Bayes theorem;
$P\left(A|B\right)={\displaystyle \frac{P\left(B|A\right).P\left(A\right)}{P\left(B\right)}}$
Where;
$A,B=$ events
$P\left(A|B\right)=$ probability of B given A is true
$P\left(A\right),P\left(B\right)=$ The independent probability of A and B
So,
$$P\left(B|D\right)={\displaystyle \frac{P\left(B\right).P\left(D|B\right)}{P\left(A\right).P\left(D|A\right)+P\left(B\right).P\left(D|B\right)+P\left(C\right).P\left(D|C\right)}}$$
Where D means bolt is defective
$\begin{array}{rl}& P\left(B|D\right)={\displaystyle \frac{0.35\times 0.04}{0.25\times 0.05+0.35\times 0.04+0.4\times 0.02}}\\ & ={\displaystyle \frac{0.014}{0.0125+0.014+0.008}}\\ & ={\displaystyle \frac{0.014}{0.0345}}\\ & ={\displaystyle \frac{140}{345}}\\ & ={\displaystyle \frac{28}{69}}\end{array}$
Therefore, probability that it was manufactured by machine B is ${\displaystyle \frac{28}{69}}$
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;
$=1-P\left(B|D\right)$
$\begin{array}{rl}& =1-{\displaystyle \frac{28}{69}}\\ & ={\displaystyle \frac{69-28}{69}}\\ & ={\displaystyle \frac{41}{69}}\end{array}$

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.