**Problem 33A**

The number of claims in a calendar year for an insured has a Poisson distribution with mean 1. When a claim occurs, the individual claim amount, regardless of how many claims the insured will have in the calendar year, is either 1 or 2, with probabilities 0.6 and 0.4, respectively.

When multiple claims occur for this insured, the individual claim amounts are independent.

In the next calendar year, what is the probability that the total claim amount for this insured will be 5?

**Problem 33B**

The number of claims in a calendar year for an insured has a Poisson distribution with mean 1.2. When a claim occurs, the individual claim amount, regardless of how many claims the insured will have in the calendar year, is either 2 or 4, with equal probabilities.

When multiple claims occur for this insured, the individual claim amounts are independent.

In the next calendar year, what is the probability that the total claim amount for this insured will be 10?

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**Answers**

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Could you please provide solution for this question? Thank you so much.

Sincerely

In 33A, there’s 3 ways of reaching 5:

1. 1+2+2 (3 claims)

2. 1+1+1+2 (4 claims)

and

3. 1+1+1+1+1 (5 claims)

1 can be arranged in 3 ways, 2 in 4 and 3 has only 1 arrangement.

Then the probability is:

3exp(-1)/3!*0.6*0.4^2 + 4exp(-1)/4!*0.4*0.6^3 + exp(-1)/5!*0.6^5

which is basically adding up 1, 2 and 3.