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Sagot :
Using conditional probability, it is found that there is a 0.6262 = 62.62% probability that the part is indeed defective.
Conditional Probability
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
- P(B|A) is the probability of event B happening, given that A happened.
- [tex]P(A \cap B)[/tex] is the probability of both A and B happening.
- P(A) is the probability of A happening.
In this problem:
Event A: Positive test result.
Event B: Defective.
For the probability of a positive test result, we consider two scenarios.
- 99.5% of 1%(defective).
- 0.6% of 100 - 1 = 99%(false positive).
Hence:
[tex]P(A) = 0.995(0.01) + 0.006(0.99) = 0.01589[/tex]
The probability of a positive test result and being defective is:
[tex]P(A \cap B) = 0.995(0.01)[/tex]
The conditional probability is:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.995(0.01)}{0.01589} = 0.6262[/tex]
0.6262 = 62.62% probability that the part is indeed defective.
A similar problem is given at https://brainly.com/question/14398287
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