# Layman's terms to describe common discrete distributions

I am currently taking a probability and statistics course. We recently started studying discrete random variables, specifically discrete distributions lately. I want to ensure I understand what each of these distributions enables one to compute. Are the following descriptions accurate?

$$X\sim\textit{Bernoulli}\left(p\right)$$, where $$p$$ is the probability of a success. The distribution solves the probability of a success in one trial.

$$X\sim\textit{Geometric}\left(p\right)$$, where $$p$$ is the probability of a success. The distribution solves the number of trials until a success.

$$X\sim\textit{Binomial}\left(n,p\right)$$, where $$n$$ is the number of trials and $$p$$ is the probability of a success. The distribution solves the number of successes in $$n$$ trials.

$$X\sim\textit{Pascal}\left(m,p\right)$$, where $$m$$ is the number of successes and $$p$$ is the probability of a success. The distribution solves the number of trials until $$m$$ successes.

$$X\sim\textit{Poisson}\left(\lambda\right)$$. I am having some trouble with this one. Could someone please explain?

Thank you!

• Distributions don't solve anything; better wording would be something like "$X \sim \text{Binomial}(n,p)$. $X$ denotes the number of successes in $n$ independent trials of an experiment in which the probability of success on each trial is $p$." Jul 9, 2021 at 21:42
• Thank you for that clarification. I do understand this. I think I stated my descriptions in terms of what their expected values would calculate. Jul 10, 2021 at 18:02

All correct, though I think the usual convention is that the Pascal distribution models the number of $$failures$$, rather than $$trials$$. Of course these just differ by the constant $$m$$, so what you said is essentially the same - just make sure you use the right formula in each case:

$${\rm Pr}(X=k)={k-1 \choose m-1} (1-p)^{k-m}p^m,$$ gives the probability that the $$m^{{th}}$$ success occurs on the $$k^{th}$$ trial.

$${\rm Pr}(X=k)={m+k-1 \choose m-1} (1-p)^{k}p^m,$$ gives the probability that the $$m^{{th}}$$ success occurs after precisely $$k$$ failures.

As for the Poisson distribution, it is hard to describe its motivation in purely discrete terms, even though it is a discrete distribution.

The idea is that some process is going on continuously for a period of time $$T$$, and $$\lambda$$ is the success rate. So in any period of time $${\rm d}t$$, you expect $$\lambda\frac{{\rm d}t}T$$ successes.

Then the Poisson distribution models the number of successes in the entire period.

• Thank you. I never thought of Poisson to describe that the "$m^{\text{th}}$ success occurs on the $k^{\text{th}}$ trial," but that obviously aligns with my description: "number of trials until $m$ successes." "Until $m$ successes" implies the $m^{\text{th}}$ success occurs on the last trial. Jul 10, 2021 at 17:54
• Further question: For Pascal, the expected number of trials until the $m^{\text{th}}$ success occurs is calculated with $\mathrm{E}\left[X\right]=\frac{m}{p}$. What about the expected number number of trials until the until the $m^{\text{th}}$ success occurs after precisely $k$ failures? Jul 10, 2021 at 18:03
• If the $m^{th}$ success occurs after exactly $k$ failures, then it will occur on the $m+k^{th}$ trial.
– tkf
Jul 10, 2021 at 19:31

As the other answer says Poisson distribution is counts that happen in a certain interval of time or space. It could be the number of algae filtered on a filtration system, or the number of errors printed in a page of a book. Whereas the other distributions you listed all have to do with trials of successes and failures, Poisson doesn't really involve this. However, it is true that a binomial distribution with large n is approximately Poisson. The mean $$\lambda$$ is the average number of counts you would expect for one occurrence of the Poisson random variable.

• Thank you very much. Jul 10, 2021 at 17:55