Respuesta :
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
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
What is Bayes' theorem?
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:
[tex]P(A|B) = \dfrac{P(B|A)P(A)}{P(B)}[/tex]
(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
- P(A) = 20% = 0.2
- P(A') = 1- P(A) = 0.8
- P(B|A) = 1 - 0.95 = 0.05
- P(B|A') = 1 - 0.7 = 0.3
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,
[tex]P(A|B) = \dfrac{P(A|B)P(A)}{P(B)} = \dfrac{P(B|A))P(A)}{P(B|A)P(A) + P(B|A')P(A')}\\\\P(A|B) = \dfrac{0.05 \times 0.2}{0.05 \times 0.2 + 0.3 \times 0.8 } = \dfrac{0.001}{0.241} \approx 0.004[/tex]
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:
https://brainly.com/question/13318017
