From a research is known that average 1% of drivers are causing a car accident every year.One safe company has 1000 drivers that are been on them.What is the possibility that the company should pay for more than 2 accidents of their customers in the next year. What i did , i thought i have 1% on average so to find
λ =0.01 *1000=10 so i have λ
f(x)=e^-λ * λ^x/x!= e^-10 * 10^2/2!
I have done Poisson ,am i right?
As $$X=\text{number of accidents per year}$$ is bounded by $1000$ we can't have a Poisson here at all, strictly speaking. Definitely $X$ is binomial with $n=1000$ and $p=0.01$. The task ist to calculate $$P(X>2)=1-P(X\leq2)=0.997320568006$$ via binomial distribution.
As $p$ is small, one may approximate the binomial distribution by a Poisson. In this case $$P(X>2)=1-P(X\leq2)=0.997320604284$$ via Poisson.