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Mutually exclusive events; insurance accident Bayes

Short answer

Mutually exclusive events cannot happen together (P(A∩B)=0). Independent events have P(A∩B)=P(A)P(B). Using Bayes: P(Civil|accident) = (1/6 × 0.01) / [(1/6×0.01)+(1/3×0.03)+(1/2×0.15)] = 0.001667/0.086667 ≈ 0.0192 (about 1.9%).

Define mutually exclusive and independent events. An insurance company insured 2000 Civil engineers, 4000 Electrical engineers and 6000 Mechanical engineers. The probability of an accident involving Civil engineer, Electrical engineer and Mechanical engineer during their jobs is 0.01, 0.03 and 0.15 respectively. One of the insured engineers meets with an accident. What is the probability that he is civil Engineer?

Mutually exclusive events

Two events that cannot occur together are mutually exclusive: A ∩ B = ∅, so P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B). Example: getting a head and getting a tail in one toss.

Independent events

Events are independent if P(A ∩ B) = P(A) P(B), that is, one event does not affect the probability of the other. Example: a head on a coin and an even number on a die.

Insurance problem (Bayes' theorem)

Let C, E, M be the events that the engineer is Civil, Electrical, Mechanical, and let X be the event of an accident.

Total insured = 2000 + 4000 + 6000 = 12000.

P(C) = 2000/12000 = 1/6    P(X|C) = 0.01
P(E) = 4000/12000 = 1/3    P(X|E) = 0.03
P(M) = 6000/12000 = 1/2    P(X|M) = 0.15

Total probability of an accident:

P(X) = (1/6)(0.01) + (1/3)(0.03) + (1/2)(0.15)
     = 0.001667 + 0.01 + 0.075
     = 0.086667

Bayes' theorem:

P(C|X) = P(C) P(X|C) / P(X)
       = 0.001667 / 0.086667
       = 0.0192

The same result follows from expected counts: accidents are 20 (civil), 120 (electrical), 900 (mechanical), total 1040, and 20/1040 = 0.0192.

Answer: P(Civil | accident) ≈ 0.0192 (about 1.9 %).

✓ Answer checked by Claude Sonnet 5.5

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