# Poisson and Exponential Distribution- Is the following question correct?

Accidents occur with a Poisson distribution at an average of 4 per week. i.e., $\lambda= 4$.

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

2. 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.

• What exactly is the question? One answer to what appears to be the question is "Yes, it is necessary to be clear that Part 2 is based on exponential distribution" Another answer is "No, it is not necessary to be clear that Part 2 is based on exponential distributions" since all that you are being asked is "What is the probability that there are no arrivals in two weeks given that there are, on average, $8$ arrivals in two weeks?" which can be answered purely as the probability that a Poisson random variable with mean $8$ has value $0$ without any mention of exponential distributions. – Dilip Sarwate Dec 8 '13 at 21:03
• Are you attempting to write a question ("Is it necessary when writing this question to be clear..."), or answer one ("Is it necessary when answering this question to be clear...")? – Glen_b -Reinstate Monica Dec 8 '13 at 21:39
• Is this a question you are asking students? If yes, have they learned that inter-arrival times in a poisson process are exponential? If they haven't, do you expect them to prove this in the question? Questions to consider. – bdeonovic Dec 8 '13 at 23:37
• @Benjamin: Yes this is for students and they have learned inter-arrival times in a Poisson process are exponential. I have mentioned related outcome too in the Note accompanying question. – Madhu Dec 9 '13 at 9:13
• @Gleb_b: I am attempting to write a question. – Madhu Dec 9 '13 at 9:13

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.

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!

• @ Dilip: Thanks for the discussion. I was trying to write a question and was having doubt about second part as the sentence is starting with Poisson distribution. – Madhu Dec 10 '13 at 10:07

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.