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Question
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
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Solution
Let E1: be the scooter driver.
E2: The car is a driver.
E3: Is a truck driver.
E: The insured driver gets into an accident
Driver = 2000 + 4000 + 6000 = 12000
P(E1) = `2000/12000 = 1/6`,
P(E2) = `4000/12000 = 1/3`,
P(E3) = `6000/12000 = 1/2`
P(E|E1) − 0.01, P(E|E2) = 0.03, P(E|E3) = 0.15
The probability that the accident driver is a scooter driver = P(E1|E)
By Bayes' theorem
P(E1|E) = `(P(E_1) xx P(E|E_1))/(P(E_1) xx P(E|E_1) + P(E_2) xx P(E|E_2) + P(E_3) xx P(E|E_3)`
= `(1/6 xx (0.01))/(0.01/6 + 0.03/3 + 0.15/2)`
= `0.01/(0.01 + 0.06 + 0.45)`
= `0.01/0.52`
= `1/52`
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