Question

A factory production line is manufacturing bolts using three machines, A, B and C. Of the total output, machine A is responsible for $25\%$, machine B for $35\%$ and machine C for the rest. It is known from previous experience with the machines that $5\%$ of the output from machine A is defective, 4% from machine B and $2\%$ from machine C. A bolt is chosen at random from the production line and found to be defective. What is the probability that it came from machine A.
A.$0.632$
B.$0.362$
C.$0.487$
D.None of these

Answer 1

Here, we will use the concept of probability. First, we will write all the probabilities given in the question. Then we will apply the Bayes’ theorem to get the probability of selecting a bolt, which is chosen at random from the production line and found to be defective is from machine, A. Probability is the branch of mathematics which gives the possibility of the event occurrence

Complete step-by-step answer:
Let A be bolt manufactured from machine A, B be bolt manufactured from machine B, C be bolt manufactured from machine C and D be bolt is defective.
Now we will write all the probabilities given in the question with the proper manner.
It is given that machine A is responsible for $25\%$, machine B for $35\%$ and machine C for the rest of the manufacturing. Therefore, we get
Probability of bolt manufactured from machine A is $= 25\% = 0.25$
Probability of bolt manufactured from machine B is $= 35\% = 0.35$
Probability of bolt manufactured from machine C is $= 1 - \left( {0.25 + 0.35} \right) = 0.40$
Now it also given that 5% of the output from machine A is defective, 4% from machine B and 2% from machine C. Therefore, we get
Probability of bolt manufactured from machine A is defective :
$\Rightarrow$ $P\left( {\dfrac{D}{A}} \right) = 5\% = 0.05$
Probability of bolt manufactured from machine B is defective :
$\Rightarrow$ $P\left( {\dfrac{D}{B}} \right) = 4\% = 0.04$
Probability of bolt manufactured from machine C is defective:
$\Rightarrow$ $P\left( {\dfrac{D}{C}} \right) = 2\% = 0.02$
Now by using the Bayes’ theorem we will find the value of the probability of selecting a bolt which is chosen at random from the production line and found to be defective is from machine A. Therefore, we get
$P\left( {\dfrac{A}{D}} \right) = \dfrac{{\left( {P\left( A \right) \cdot P\left( {\dfrac{D}{A}} \right)} \right)}}{{\left( {P\left( A \right) \cdot P\left( {\dfrac{D}{A}} \right)} \right) + \left( {P\left( B \right) \cdot P\left( {\dfrac{D}{B}} \right)} \right) + \left( {P\left( C \right) \cdot P\left( {\dfrac{D}{C}} \right)} \right)}}$
Simplifying the expression, we get
$\Rightarrow P\left( {\dfrac{A}{D}} \right) = \dfrac{{\left( {0.25 \times 0.05} \right)}}{{\left( {0.25 \times 0.05} \right) + \left( {0.35 \times 0.04} \right) + \left( {0.40 \times 0.02} \right)}}$
$\Rightarrow P\left( {\dfrac{A}{D}} \right) = \dfrac{{0.0125}}{{0.0345}} = 0.362$
Hence the probability of selecting a bolt which is chosen at random from the production line and found to be defective from machine A is $0.362$.
So, option B is the correct option.

Note: Here we should note that we have to apply the Baye’s theorem carefully. Baye’s theorem formula is generally used to calculate a probability when we know other certain probabilities also. Probability is equal to the ratio of the number of favorable outcomes to the total number of outcomes. Probability generally lies between the values of 0 to 1. Similarly in the above question, we got the probability in the range of 0 and 1. Probability can never be above the value of 1 or less than 0. So if we got the value of probability greater than 1 then we need to check out the calculation to remove the error for calculating the probability.