# Easiest way to recognize a discrete random variable distribution

I'm having quite a large issue recognizing which distribution to use in a given task.

I know the mathematical definitions for each of them, but I'm unable to apply them in real problems. Is there like an easy way or recognizing which one to use to solve the problem?

Like for e.g.

For poisson distribution, the "event" usually happens on a given timeframe( like daily, weeklly etc. ) or we're given some kind of an average value.

For hypergeometric distribution, usually we need to pick k objects of a set with n objects with some kind of a special property.

The ones I'm having trouble with are binomal, negative binomal and geometric.

Let's see the following problem:

We have a lab. consisting of 20 computers. 5 students need to take an exam. The probability that the computers have the exam software installed is 1/3.

a) What's the probability that there will be enough computers for the students in a single lab?

b) The professor turns the computers on one by one and checks if the software is installed. Describe the random variable "amount of turned on computers until a working one is found" (this is geometric distribution, number of failures before success happens)

c) The professor turns the computers on one by one, checks if the software is installed and puts a student on it if it's working. What's the probability that all the students will take the exam in the same lab? (this looks to me like negative binomal? Not sure.)

d) If the professor knows that the software is installed on 7 computers, what's the probability that he will find those 7 computers if he picks the first half of the turned on computers? (hypergeometric? 10 computers picked, 7 have special property?)

Tl;dr What's the easiest way to recognize which random variable distribution needs to be used in a task?

Thanks!

• a) is Binomial as you want 5 or more successes in 20 trials. – math_noob Oct 29 '15 at 2:13
• c) is same as a). d) is hypergeometric, you recognized that correctly. b) would've been a geometric if trials were infinite. – math_noob Oct 29 '15 at 2:23
• So b) is a negative binomial with a parameter 1? – kjanko Oct 29 '15 at 12:03
• Negative Binomial with parameter 1 is just geometric. As I said earlier, this is not geometric because the trials are not infinite. – math_noob Oct 29 '15 at 13:53
• So what would be the solution? Because in my mind, the task goes like ( q q q q q q p ) – kjanko Oct 29 '15 at 17:30

## 2 Answers

Binomial distribution mainly applies to what are called bernoulli trials with a finite constraint. Binomial has a finite amount of outcomes. For example in binomial there is a finite amount of trials(i.e 20 trials). So whenever you are given a finite constraint(i.e total of 20 trials) you will generally be using a Binomial. The idea of Binomial distribution can be extended to an infinite amount of trials through the Poisson distribution(you should read about how the Poisson distribution can be the limit of a binomial distribution).

The geometric distribution has to do with the idea of waiting time. These problems have an infinite amount of outcomes. An example would be the probability of hitting the bullseye on a dartboard for the first time on your 10th dart. It could potentially take you forever to hit the bullseye, so the amount of trials is infinite(it could take you $10^{1000}$ tries to hit the bullseye). So if the variable represents the amount of tries required to pass for the first time the varaible is geometrically distributed.

• So whenever we are asked to find the amount of attempts to reach a certain amount of success-es we use binomial? What abou negative binomial? – kjanko Oct 29 '15 at 12:01

A Bernoulli random variable is the result of a single success-or-failure trial.   The relevant parameter is the success rate: $p$.

$$X\sim\mathcal{Bern}(p) \iff \mathsf P(X{=}1)= p, \mathsf P(X{=}0)=1-p$$

A Binomial random variable is the count of successes amid a certain number of iid Bernoulli trials.   The relevant parameters are the number of trials, $n$ and the Bernoulli success rate $p$.

$$X\sim\mathcal{Bin}(n, p) \iff \mathsf P(X=k) = {^n{\rm C}_k}\,p^k\,(1-p)^{n-k}\;\big[k\in\{0,1,..., n\}\big]$$

A Hypergeometric random variable is the count of favoured items in a sample drawn without replacement from a certain sized population containing a certain amount of favoured items.   The relevant parameters are population size $N$, favoured population size $K$, and sample size $n$.

$$X\sim\mathcal{Hyp}(n, N, K) \iff \mathsf P(X{=}k) = \dfrac{{^{K}{\rm C}_{k}}{^{N-K}{\rm C}_{n-k}}}{^{N}{\rm C}_{n}}\;\big[k\in\{0,...,n\}\big]$$

A Geometric random variable is the count of potentially infinite Bernoulli trials until a success.   The parameter is the Bernoulli success rate, $p$.   Some texts consider it the count of failures before the first success.   I've taken to subscripting to specify which.

$$X\sim\mathcal{Geo}_1(p) \iff \mathsf P(X{=}k) = (1-p)^{k-1}p\;\big[k\in\{1,...\infty\}\big] \\ Y\sim\mathcal{Geo}_0(p) \iff \mathsf P(Y{=}k) = (1-p)^k p\;\big[k\in\{0,...\infty\}\big]$$

A Negative Binomial random variable is the count of successful Bernoulli trials until a certain number of failures have occurred.   The parameter is the success rate $p$ and terminating number of failures $r$.   This is related to a Geometric random variable as $\mathcal{NB}(1,p) \simeq \mathcal{Geo}_0(1-p)$ or $\mathcal{Geo}_0(p) \simeq \mathcal{NB}(1, 1-p)$

$$X\sim\mathcal{NB}(r, p) \iff \mathsf P(X{=}k) = {^{k+r-1}{\rm C}_{k}}p^k(1-p)^r$$

Similarly if you have a series of independent and identically distributed geometric${}_0$ random variables, then their sum is negatively binomially distributed: $$\{X_k\}_{k\in\{1,.,n\}}\mathop{\sim}^\text{iid}\mathcal{Geo}_0(p) \implies \sum_{k=1}^n X_k \sim\mathcal{NB}(n, 1-p)$$

A Poisson random variable is the count of (potentially infinite) discrete events occurring within a given interval (of time or space) when the events occur independently across the interval and with an average rate.   The relevant parameters are the rate of occurrence $\lambda$ and the interval size $t$.   Sometimes the product, a dimensionless parameter, is used $\lambda_t = \lambda\; t$.

$$X\sim\mathcal{Pois}(\lambda, t) \iff \mathsf P(X{=}k) = \frac{{(\lambda\,t)}^k \mathsf e^{-\lambda\,t}}{k!}\;\big[k\in\{0,...,\infty\}\big]$$

• As I said in the question, I do know the mathematical definitions of the distributions but I'm having a hard time recognizing them in a given task :( – kjanko Oct 29 '15 at 12:01