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.


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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