
Stochastic processes #2
On the average, Mr. Z drinks and drives once in 4 years. He knows that
According to the laws of his state, the third time when he is caught drinking and driving results in the loss of his driver's license.
Poisson process is the correct model for such ``rare events'' as drinking and driving.
What is the probability that Mr. Z will keep his driver's license for at least 10 years?

Let $\displaystyle N(t)=$ the number of times Mr. Z gets pulled over by year $\displaystyle t.$
Then $\displaystyle N(t)$ is a poisson process with rate $\displaystyle 1/4.$
Therefore $\displaystyle N(t)=poi(t\times 1/4)$
Now, $\displaystyle P(N(1)<3)= P(N(1)=1)+P(N(1)=2)$
$\displaystyle \Rightarrow P(N(1)<3)= e^{1/4}\frac{(1/4)^1}{1!} + e^{1/4} \frac{(1/4)^2}{2!}$
By independent increments,
$\displaystyle \Rightarrow P(N(10)<3)= e^{1/4\times 10}\frac{(1/4\times 10)^1}{1!} + e^{1/4\times 10} \frac{(1/4\times 10)^2}{2!}$
I've been studying for my stochastic process test on monday (Rock).