A discrete random variable takes values in a countable set, usually the non-negative integers. Its probability function $p(x)=P(X=x)$ is non-negative and sums to 1, and $F(x)=P(X\le x)$ is a step function. Exam P uses six named discrete families. Each one describes a counting situation, and most exam questions reduce to recognising the situation, choosing the parameters, and applying one of the formulas below.

The six distributions

Distribution	Counts	$p(x)$	Mean	Variance
Binomial$(n,p)$	successes in $n$ independent trials	$\binom{n}{x}p^x q^{n-x}$, $x=0,\dots,n$	$np$	$npq$
Geometric, failures version	failures before the first success	$pq^x$, $x=0,1,2,\dots$	$q/p$	$q/p^2$
Geometric, trials version	trials up to and including the first success	$pq^{x-1}$, $x=1,2,\dots$	$1/p$	$q/p^2$
Negative binomial$(r,p)$	failures before the $r$th success	$\binom{x+r-1}{x}p^r q^x$, $x=0,1,\dots$	$rq/p$	$rq/p^2$
Hypergeometric	marked items in a sample of $n$ drawn without replacement from $N$ items, $K$ marked	$\dfrac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}$	$n\frac{K}{N}$	$n\frac{K}{N}\left(1-\frac{K}{N}\right)\frac{N-n}{N-1}$
Poisson$(\lambda)$	events in a fixed period at rate $\lambda$	$e^{-\lambda}\lambda^x/x!$, $x=0,1,\dots$	$\lambda$	$\lambda$
Discrete uniform on $\{1,\dots,m\}$	an equally likely choice among $m$ values	$1/m$	$(m+1)/2$	$(m^2-1)/12$

Here $q=1-p$. For a discrete uniform on $\{a,\dots,b\}$ with $m=b-a+1$ values, the mean is $(a+b)/2$ and the variance is still $(m^2-1)/12$.

Geometric counting conventions and memorylessness

The two geometric versions differ by 1: if $N$ counts failures before the first success and $T$ counts trials, then $T=N+1$. They have the same variance but means that differ by 1. Read the question for the phrase that fixes the convention ("before the first", "including the first", "on the $k$th trial"). The quickest route to tail probabilities is the survival function: $P(N\ge k)=q^k$ and $P(T\gt k)=q^k$, because both events say the first $k$ trials all failed.

The geometric is the only discrete distribution that is memoryless: $P(N\ge k+j\mid N\ge k)=P(N\ge j)$. Given $k$ failures so far, the number of further failures has the original geometric distribution. So $E[N\mid N\ge k]=k+E[N]$ and $\operatorname{Var}(N\mid N\ge k)=\operatorname{Var}(N)$.

Negative binomial through the binomial

The trial on which the $r$th success occurs exceeds $n$ exactly when the first $n$ trials contain fewer than $r$ successes. So $P(T_r\gt n)=P(\text{Bin}(n,p)\le r-1)$. This replaces a long sum of negative binomial terms with a short binomial sum. The same idea works without replacement: the $r$th marked item is found within the first $n$ draws exactly when a hypergeometric sample of $n$ contains at least $r$ marked items.

Poisson facts used on the exam

Ratio recursion: $p(k)=p(k-1)\cdot\lambda/k$. It gives successive terms quickly and shows that the mode is $\lfloor\lambda\rfloor$; when $\lambda$ is an integer, both $\lambda-1$ and $\lambda$ are modes.
Sums: independent Poisson variables with means $\lambda_1,\dots,\lambda_k$ add to a Poisson with mean $\lambda_1+\dots+\lambda_k$. A rate per week scales to a rate per month.
Thinning: if $N\sim$ Poisson$(\lambda)$ and each event is independently of type A with probability $p$, the numbers of type A and non-A events are independent Poissons with means $\lambda p$ and $\lambda(1-p)$.
Approximation: a binomial with large $n$ and small $p$ is close to Poisson with $\lambda=np$. Use it when the question asks for it or when $n$ is in the thousands.

The binomial mode is $\lfloor (n+1)p\rfloor$. Hypergeometric probabilities are close to binomial ones with $p=K/N$ only when $n$ is small relative to $N$; on the exam, sampling without replacement from a small group always requires the hypergeometric.

Worked example 1

A claims adjuster reviews files one at a time. Each file independently needs a site visit with probability 0.25. Calculate the probability that more than 6 files must be reviewed to find 2 that need site visits.

More than 6 files are needed exactly when the first 6 files contain at most 1 site visit. With $X\sim\text{Bin}(6,0.25)$:

$$P(X\le 1)=0.75^6+6(0.25)(0.75)^5=0.17798+0.35596=0.5339.$$

The direct route sums the negative binomial probabilities for 7, 8, 9, … files, which is an infinite sum.

Worked example 2

The number of hail claims in a month is Poisson with mean 4. Each claim is independently for roof damage with probability 0.6. Given that exactly 3 roof claims occur in a month, calculate the expected total number of hail claims that month.

By thinning, roof claims $R\sim$ Poisson$(2.4)$ and other claims $O\sim$ Poisson$(1.6)$ are independent. So

$$E[R+O\mid R=3]=3+E[O]=3+1.6=4.6.$$

The answer 4 ignores the information, and $3/0.6=5$ treats the observed count as if it fixed the total.

Worked example 3

A group of 12 policies contains 4 with late payments. An auditor samples 3 policies without replacement. Calculate the probability that the sample contains at least 1 late-payment policy.

$$1-\frac{\binom{8}{3}}{\binom{12}{3}}=1-\frac{56}{220}=0.7455.$$

The binomial answer $1-(2/3)^3=0.7037$ is wrong here because the draws are dependent.