Poisson and Exponential Distribution- Is the following question correct? Accidents occur with a Poisson distribution at an average of 4 per week. i.e., $\lambda= 4$.


*

*Calculate the probability of more than 5 accidents in any one week.

*What is the probability that at least two weeks will elapse between accident?
Query: Is is necessary to be clear that Part 2 is based on exponential distribution or time is exponentially distributed or something like this?
Note: The question is based on integrating knowledge of the Poisson and exponential distributions.
 A: What is the probability that at least two weeks will elapse between accident? 
This uses the fact that time between Poisson events follows the exponential distribution (probability distribution that describes the time between events in a Poisson process - wikipedia./exponential distribution). When finding the probability of at least two weeks the lower bound is 2, and the distribution has a mean of 0.25. Hope this helps. 
A: If you are trying to write a question, consider asking for a solution to the second question in two different ways and comparing the answers. 


*

*First, using the fact that in a Poisson process with arrival rate 
$\lambda$, the number of arrivals in an interval of length $T$ is a Poisson random variable $N_T$ with parameter $\lambda T$, what is the probability that there are no arrivals in an interval of length $T$? Answer: $P\{N_T = 0\} = e^{-\lambda T}$.

*Second, using the fact that inter-arrival times as well as the first arrival
time $X$ after $t = 0$ is an exponential random variable with mean $\lambda^{-1}$
and thus parameter $\lambda$, what is $P\{X > T\} = P${no arrivals in $(0,T]$}? Answer: $\int_T^\infty f_X(x)\, dx = e^{-\lambda T}$.

*Are the two results the same? Of course they are!
A: Setup:
$\{N(t),t\ge 0\}$ is the counting process and is a Poisson Process (PP) with rate $\lambda = 2$ per week. Let $T$ be the time between events.  We also know $T\sim Expo(\lambda)$.
(1) Use the Poisson distribution 
$\begin{align}P(N(1)>5) &= 1-P(N(1)\le 4) \\
&= 1 - [P(N(1)=0)+P(N(1)=1)+\cdots+P(N(1)=4)] \\ 
&= 1-\left[\frac{\lambda^0 \text{e}^{-\lambda}}{0!}+  \frac{\lambda^1 \text{e}^{-\lambda}}{1!}+  \frac{\lambda^2 \text{e}^{-\lambda}}{2!}+  \frac{\lambda^3 \text{e}^{-\lambda}}{3!}+  \frac{\lambda^4 \text{e}^{-\lambda}}{4!}\right] \end{align}$
(2) Use the CDF of the Exponential Distribution
$\begin{align}P(T>2) =1- P(T\le 2) &= 1 - (1-\text{e}^{-2\lambda})\\ 
&=\text{e}^{-2\lambda}\end{align}$ 
Notice here you can also use the Poisson distribution directly and achieve the same result.  
