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Law of addition; satisfactory trainee Bayes

Short answer

For any events A,B, P(A ∪ B) = P(A)+P(B)−P(A∩B); if mutually exclusive, P(A∪B)=P(A)+P(B). Here P(acceptable) = 0.665, so P(satisfactory | acceptable) = 0.56 / 0.665 ≈ 0.842.

State the law of addition of probability. In a training, the 70% of persons achieved a rating of Satisfactory. Of those rated as Satisfactory, 80% had Acceptable Scores on the personality test. Of those rated as Unsatisfactory, 35% had Acceptable Scores. Find the probability that an applicant would be a Satisfactory trainee given the Acceptable scores on personality test.

Law of addition of probability

For any two events A and B:

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

If A and B are mutually exclusive, P(A ∩ B) = 0, so P(A ∪ B) = P(A) + P(B).

Trainee problem

Let S be "rated Satisfactory" and T be "Acceptable score on the personality test".

P(S) = 0.70,   P(S′) = 0.30
P(T|S) = 0.80, P(T|S′) = 0.35

Probability of an acceptable score (total probability):

P(T) = P(S) P(T|S) + P(S′) P(T|S′)
     = (0.70)(0.80) + (0.30)(0.35)
     = 0.560 + 0.105
     = 0.665

Probability of a Satisfactory trainee given an acceptable score (Bayes' theorem):

P(S|T) = P(S) P(T|S) / P(T) = 0.560 / 0.665 = 0.842

Answer: P(Satisfactory | Acceptable score) ≈ 0.842.

✓ Answer checked by Claude Sonnet 5.5

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