Answer: To solve the problem, we can use Bayes' theorem. Let D be the event that a device is defective, and let A be the event that the quality assurance check concludes that a device is defective.
We want to find P(A and not D) + P(not A and D), which represents the probability of an incorrect conclusion.
We know that P(D) = 1 - P(not D) = 1 - 0.84 = 0.16, and that P(A | not D) = 0.03 and P(A | D) = 0.97.
Using Bayes' theorem, we can compute:
P(not A | not D) = 1 - P(A | not D) = 1 - 0.03 = 0.97
P(not A | D) = 1 - P(A | D) = 1 - 0.97 = 0.03
Therefore,
P(A and not D) = P(not D) * P(A | not D) = 0.84 * 0.03 = 0.0252
P(not A and D) = P(D) * P(not A | D) = 0.16 * 0.03 = 0.0048
So the probability of an incorrect conclusion is:
P(A and not D) + P(not A and D) = 0.0252 + 0.0048 = 0.03
Therefore, the probability of an incorrect conclusion is 0.03, or 3% (rounded to the nearest tenth of a percent).
Why was this answer deleted prior?