An actuary has done an analysis of all policies that cover two cars. 70% of the policies are of type A for both cars, and 30% of the policies are of type B for both cars. The number of claims on different cars across all policies are mutually independent. The distributions of the number of claims on a car are given in the following table.

Number of Claims	Type A	Type B
0	40%	25%
1	30%	25%
2	20%	25%
3	10%	25%

Calculate the probability that exactly one of the four policies has the same number of claims on both covered cars.

• 0.104
• 0.250
• 0.285
• 0.417
• 0.739

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: D

If a policy is of Type A, the probability that the two claims are equal is

(0.4)(0.4) + (0.3)(0.3) + (0.2)(0.2) + (0.1)(0.1) = 0.16 + 0.09 + 0.04 + 0.01 = 0.30.

If a policy is of Type B, the probability that the two claims are equal is 4(0.25)(0.25) = 0.25. Therefore, the probability that a randomly selected policy has equal claims is

0.70(0.30) + 0.30(0.25) = 0.285.

If four policies are selected, the desired probability is the probability that a binomial random variable with n = 4 and p = 0.285 is 1. This is

4(0.285)(1 – 0.285)3 = 0.417.

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.