Suppose a pool of insurance policyholders can be classified into groups 1, 2, and 3. For...

Suppose a pool of insurance policyholders can be classified into groups 1, 2, and 3. For a person from group i, the number of accidents that the person will have in a year follows a Poisson distribution with parameter i, for i=1, 2, 3. Suppose 20% of the policyholders are in group 1, 40% in group 2 and 40% in group 3.
1. For a policyholder in group 3, what is the probability that the person will have at least one accident in a year?
2. Find the probability that a randomly selected person from the pool will have exactly two accidents in a year.

Expert Solution

a) For policyholder in group 3, the probability that the person will have at least one accident in a year is computed here as:

Therefore 0.9502 is the required probability here.

b) We are given here that:
P( group 1) = 0.2, P(group 2) = P(group 3) = 0.4

The probability that the person will have at least one accident in a year for different groups is computed here as:

Using law of total probability, we get here:
P(X = 2) = P(X = 2| group-1)P(group 1) + P(X = 2| group-2)P(group 2) + P(X = 2| group-3)P(group 3)
P(X = 2) = 0.1839*0.2 + 0.2707*0.4 + 0.2240*0.4 = 0.2347

Therefore 0.2347 is the required probability here.

orchestra answered 2 years ago

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Suppose a pool of insurance policyholders can be classified into groups 1, 2, and 3. For...

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