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In a factory that produces machine parts, 90% of the produced parts are good quality and 10% are of defective quality. However, a fault in the testing machine incorrectly identifies good-quality parts as defective 5% of the time and defective parts as good quality 2% of the time. What is the probability that a randomly selected part is defective, given that the testing machine identified it as defective? 

Option: 1

0.65


Option: 2

0.84


Option: 3

0.96


Option: 4

0.89


Answers (1)

best_answer

Let A be the event that the part is defective, and B be the event that the testing machine identified the part as defective. 

From the given information, we have: 

P(A) = 0.10 (10% of the parts are defective) 

P(B|A) = 0.95 (the machine correctly identifies 95% of the defective parts as defective) 

P(B|A') = 0.02 (the machine incorrectly identifies 2% of the good parts as defective) 

We want to find P(A|B), the probability that the part is defective given that the machine identified it as defective. 

Using Bayes' theorem, we have:

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

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

P(B)=P(B / A) \times P(A)+P\left(B / A^{\prime}\right) \times P\left(A^{\prime}\right)

Where P(A') = 1 - P(A) = 0.90 (90% of the parts are good quality) 

Substituting the values, we get:

\\{P(B)=0.95\times 0.10+\ 0.02\times 0.90}\\ \\\Rightarrow {P(B)=0.113}

Substituting into Bayes' theorem, we get:

\begin{aligned} & P(A / B)=\frac{0.95 \times 0.10}{0.113} \\ \\& \Rightarrow P(A / B)=0.84 \end{aligned}

Therefore, the probability that a randomly selected part is defective given that the testing machine identified it as defective is 0.84.

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shivangi.shekhar

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