The problem can be solved using the Poisson distribution.

Here, n = 1000 (the number of policy holders) and p = 0.01% or 0.0001 (the probability of a policy holder being involved in an accident). So, lambda = 1000 * 0.0001 = 0.1.

Finally the Answer Will be **0.9048+0.0905+0.0045≈1**

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an insurance company found that only 0.01% of the population is involved in a certain type of accident each year. if its 1000 policy holders were randomly selected. what is the probability that not more than two of its clients are involved in such an accident next year? (given e^-0.1=0.9048)