Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Get accurate and detailed answers to your questions from a dedicated community of experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
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
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
