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I'm Stuck with this question. It is exercise no. 3.30 from the book "Probability & Statistics for Computer Scientists" :

Eric from Exercise 3.29 continues driving. After three years, he still has no traffic accidents. Now, what is the conditional probability that he is a high-risk driver?

And Exercise 3.29 is:

An insurance company divides its customers into 2 groups. Twenty percent of customers are in the high-risk group, and eighty percent are in the low-risk group. The high-risk customers make an average of 1 accident per year while the low-risk customers make an average of 0.1 accidents per year. Eric had no accidents last year. What is the probability that he is a high-risk driver?

I have successfully solved 3.29 by Poisson Distribution and got the answer 0.09225. But I can't figure out how to do 3.30 and what to do with that "After three years" phrase... Please Explain like I'm 5.

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This is exactly the same as the problem you've already solved, but instead of asking "what is the probability he is a high-risk driver given that he had no accidents last year?" we are asking "what is the probability he is a high-risk driver given that he had no accidents in the last four years?" (the year from 3.29 plus three further years of driving).

So instead of $P(\text{no accidents}\mid\text{high-risk})=P(\text{Poi}(1)=0)=e^{-1}$ for a one-year period, now we have $P(\text{no accidents}\mid\text{high-risk})=P(\text{Poi}(4)=0)=e^{-4}$ for a four-year period, and likewise the low-risk drivers have probability $e^{-0.4}$ instead of $e^{-0.1}$.

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  • $\begingroup$ But that would change the parameter of poisson which shouldn't change...As I think? $\endgroup$ – User5 Feb 5 '18 at 11:24
  • $\begingroup$ But wait do you want to say that if the period of consideration changes, parameter also changes likewise? $\endgroup$ – User5 Feb 5 '18 at 11:26
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    $\begingroup$ Yes, if the number of accidents in a year is distributed $\text{Poi}(\lambda)$ then the number in four years is distributed $\text{Poi}(4\lambda)$. This is because of an important fact about Poisson variables that if $X$ and $Y$ are independent Poissons with parameters $\lambda$ and $\mu$ then $X+Y$ is Poisson with parameter $\lambda+\mu$. $\endgroup$ – Especially Lime Feb 5 '18 at 11:40
  • $\begingroup$ Perfect....got it thanks :) $\endgroup$ – User5 Feb 5 '18 at 11:47

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