# Poisson Distributions: Defining the poisson random variable

Accidents occur in a factory at a rate of 2 per week: $(X \sim Po(2)).$

What is the probability that the time to the first accident is greater than 2 weeks?

What is the probability that the time to the first accident is less than 2 days?

What I've tried so far:

There are 4 accidents every 2 weeks, so $X \sim Po(4)$. I think the question is meaning to ask what the probability is that there will be $0$ accidents in the first 2 weeks, so the poisson rv would be $0$, making the equation --

$$P(X = 0) = \frac{e^{-4}4^0}{0!} = 0.0183$$

for the second one, there are $0.5714$ accidents in $2/7$ of a week (2 days), so $X \sim Po(0.5174)$, but I think the question is asking what the probability of one accident on the first day will be, so that would make $0.2857$ accidents in one day, and the poisson rv would be $1$, so $X \sim Po(0.2857)$, making the equation --

$$P(X = 1) = \frac{e^{-0.2857} 0.2857^1}{1!} = 0.2147$$

Is my thinking correct? should I be looking at these problems from a different perspective?

• $\mathbb{P}(\text{time of first accident}<2)=1-\mathbb{P}(X=0)$ You can also think these probability as an exponential random variables. Jun 5, 2014 at 20:02

## 1 Answer

A Poisson random variable is frequently used to model the probability of observing a count or frequency; i.e., "how many times did something happen?" It is a discrete random variable--we don't say "we observed exactly 3.4 car accidents in three days," or "there were 4.7 bad apples in a bucket."

An exponential random variable is used to model the probability of a failure time, or event time; i.e., "how long did it take for something to happen?" It is a continuous random variable because we can measure time in very fine increments that do not have to be integers; e.g., "the phone failed after 3.26 years of use."

When a Poisson process is used to model the occurrence of an event of interest, it happens that the frequency distribution is Poisson, but the interarrival times are exponentially distributed; that is, the random number of events that occur in a given period of time follows a Poisson distribution, but the random time that elapses between one observed event and the next is exponentially distributed.

Therefore, the time $T$ that elapses until you observe the first factory accident is exponentially distributed with rate parameter $\lambda = 2$ accidents per week. The survival function of $T$ is $S_T(t) = e^{-\lambda t}$, so we want $$\Pr[T > 2] = S_T(2) = e^{-2(2)} = e^{-4}.$$ The other part of your question is addressed similarly, except you need to convert $2$ days into an equivalent (fractional) number of weeks.

• Thanks for the response. Our class hasn't touched exponential random variables yet; funny that our professor expects us to understand them to answer these questions correctly. Jun 5, 2014 at 20:09
• You can see that the way you answered the first question is correct and agrees with the solution I provided above. However, consider that for the second part of the question, simply calculating $\Pr[X = 1]$ does not take into account the possibility that more than one accident can be observed within 2 days. So the answer you got is not correct; you want $\Pr[X \ge 1]$ because the probability that the first accident occurs in 2 days also includes the probability of many accidents occurring in that time, not just exactly one. Jun 5, 2014 at 20:15
• Yeah, I did consider the "many accidents occuring in that time" angle. It's what tripped me up originally, but because of my ignorance of exponential random variables, I assumed it was the answer our prof was looking for. Thanks again! your explanation really helped. Jun 5, 2014 at 20:22
• @heropup can you maybe help me with this question: math.stackexchange.com/questions/2598781/… Thank you Jan 9, 2018 at 20:43