At a plant, 20% of all the produced parts are subject to a special electronic inspection. it is known that any produced part which was inspected electronically has no defects with probability 0.95. for a part that was not inspected electronically this probability is only 0.7. a customer receives a part and find defects in it. what is the probability that this part went through an electronic inspection?

You can use Bayes' theorem here to find the needed probability.

The probability that the defective part received by the customer went through an electronic inspection is 0.004

Suppose that there are two events A and B.

Then suppose the conditional probability are:

P(A|B) = probability of occurrence of A given B has already occurred.

P(B|A) = probability of occurrence of B given A has already occurred.

Then, according to Bayes' theorem, we have:

(assuming the P(B) is not 0)

Using that theorem to find the needed probabilities

Let the events be

A = Part produced went through electronic inspection

B = Part produced is defective

Thus, from the given data, we have

Then we have the probability that a defective part went through an electronic inspection is P(A|B) (as it is given that part is defective).

Thus,

Thus,

The probability that the defective part received by the customer went through an electronic inspection is 0.004

Learn more about Bayes' theorem here:

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solution:

P(inspect) = 0.20, P(no inspect) = 0.80

P(defect|inspect) = 0.05

P(defect|no inspect) = 0.30

P(defect) = P(defect|inspect)\timesP(inspect) + P(defect|no inspect)\timesP(no inspect) = 0.05\times0.20 + 0.30\times0.80 = 0.10+0.24 = 0.34

\frac{P(inspect and defect)}{P(defect)}

= 0.10/0.34 = 0.294

step-by-step explanation:

explanation:

step-by-step explanation: