Question

A car insurance company offers three types of insurance: liability, collision and comprehensive. The car insurance company has the following information on its customers insuring only one car: (i) 70% of the cars are insured under the liability. (i) There are twice as many comprehensive policies as collision policies. (ii) A comprehensive insured car is three times as likely to be involved in an accident in a given year as a liability insured car. (iv) A collision insured car is as likely to be involved in an accident in a given year as a comprehensive insured ear. The probability that a car is involved in an aceident in a given year is 1/600. Calculate the probability that a comprehensive insured car is involved in an accident in a given year

Answer 1

The probability that a randomly chosen car is insured under liability = 0.70

Since there are twice as many comprehensive policies as there are collision policies, the probability of a randomly chosen car to have a comprehensive policy =0.20 and the probability of a car to have collision policy = 0.10

Let p be the probability that a liability insured car is involved in an accident in a year.

Then the probability that a comprehensive insured car is involved in an accident in a year = 3p and the probability that a collision insured car is involved in an accident = 3p

The overall probability that a randomly chosen car is involved in an accident = 0.7 x p + 0.2 x 3p + 0.1 x 3p = 1/600

1.6 p = 1/600

p = 1/(1.6x600) = 1/960

Probability that a comprehensive insured car is involved in an accident = 3p = 3 x 1/960 = 3/960 = 1/320 = 0.0031