Problem Statement
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident is 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?
Verified Solution & Marking Scheme
Define the events and prior probabilities
Let $$E_1$$: Scooter driver, $$E_2$$: Car driver, $$E_3$$: Truck driver. Let $$A$$: Insured person meets with an accident.\nTotal drivers = $$2000 + 4000 + 6000 = 12000$$.\n\n$$P(E_1) = \frac{2000}{12000} = \frac{1}{6}, \quad P(E_2) = \frac{4000}{12000} = \frac{1}{3}, \quad P(E_3) = \frac{6000}{12000} = \frac{1}{2}$$
Conditional probabilities
$$P(A|E_1) = 0.01 = \frac{1}{100}, \quad P(A|E_2) = 0.03 = \frac{3}{100}, \quad P(A|E_3) = 0.15 = \frac{15}{100}$$
Apply Bayes Theorem
$$P(E_1|A) = \frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$$\n\n$$= \frac{\frac{1}{6} \times \frac{1}{100}}{\frac{1}{6} \times \frac{1}{100} + \frac{1}{3} \times \frac{3}{100} + \frac{1}{2} \times \frac{15}{100}} = \frac{\frac{1}{6}}{\frac{1}{6} + 1 + \frac{15}{2}} = \frac{\frac{1}{6}}{\frac{1 + 6 + 45}{6}} = \frac{1}{52}$$