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  • Suppose a factory has two machines, Machine A and Machine B, which produce bolts. Machine A produces 60% of the bolts, while Machine B produces the remaining 40%. Machine A has a defect rate of 2%,

Suppose a factory has two machines, Machine A and Machine B, which produce bolts. Machine A produces 60% of the bolts, while Machine B produces the remaining 40%. Machine A has a defect rate of 2%, while Machine B has a defect rate of 5%. A bolt is randomly chosen and is found to be defective. What is the probability that it was produced by Machine A?

Option: 1

0.375


Option: 2

0.398


Option: 3

0.368


Option: 4

0.558


Answers (1)

best_answer

Let A be the event that the bolt is produced by Machine A, and D be the event that the bolt is defective. 

Then we need to find P(A | D), the probability that the bolt was produced by Machine A given that it is defective. 

We can use Bayes' theorem to calculate this probability:

P(A / D)=\frac{P(D / A) \times P(A)}{P(D)}

Where P(D | A) is the probability of the bolt being defective given that it was produced by Machine A, P(A) is the prior probability of a bolt being produced by Machine A, and P(D) is the probability of a bolt being defective. 

We can calculate these probabilities as follows: 

P(D | A) = 0.02 (Machine A's defect rate) 

P(D | B) = 0.05 (Machine B's defect rate) 

P(A) = 0.6 (Machine A produces 60% of the bolts) 

P(B) = 0.4 (Machine B produces 40% of the bolts) 

We can calculate P(D) using the law of total probability:

{P(D)=P(D/A)\times P(A)+P(D/B)\times P(B)}

Substituting the values, we get:

\\\Rightarrow {P(D)=0.02\times 0.6+\ 0.05\times 0.4}\\ \\\Rightarrow {P(D)=0.032}

Substituting these values into Bayes' theorem, we get:

\begin{aligned} & P(A / D)=\frac{0.02 \times 0.6}{0.032} \\ & \Rightarrow P(A / D)=0.375 \end{aligned}

Therefore, the probability that the defective bolt was produced by Machine A is 0.375.

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Riya

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