Discrete random variables and distributions

Updated 18 Oct 2026

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Random variables
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  Random variable -> a function that maps each outcome of an experiment to a number
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  Discrete -> takes countable values, such as $0,1,2,\ldots$ ; described by a probability mass function with $\sum_xP(X=x)=1$ .
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    $x$ is the specific value, $X$ is random variable where its value is determined from a result of a random event
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  Continuous -> takes values across an interval; probabilities come from an area under a density, and a single value has probability 0
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  Building a discrete distribution -> list each possible value with its probability, then check that the probabilities sum to 1.
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    for a spinner example the values $2,3,4$ have probabilities $0.0625,0.375,0.5625$ .
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Mean, variance, and expected value
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  Mean -> $\mu=E(X)=\sum_xxP(X=x)$ .
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    Spinner example -> $2(0.0625)+3(0.375)+4(0.5625)=3.5$ .
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    the mean is a theoretical long-run average and need not be a possible outcome of one trial.
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  Variance -> $\sigma^2=\sum_x(x-\mu)^2P(X=x)$ .
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    Spinner example -> $0.140625+0.09375+0.140625=0.375$ . The slide reports about $0.376$ because of rounding.
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  Standard deviation -> $\sigma=\sqrt{\sigma^2}$ , so about $0.612$ for the spinner.
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    variance is in squared units; standard deviation uses the variable's own units.
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  Expected value -> $E(X)=\mu$ , the same quantity as the mean for a discrete variable.
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  Lottery example -> 500 tickets at $1 each; one prize of$ 100 and one of $50$ .
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    Net gain -> $99$ with probability $1/500$ , $49$ with probability $1/500$ , and $-1$ with probability $498/500$ .
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    Expected gain -> $99\left(\frac1{500}\right)+49\left(\frac1{500}\right)-1\left(\frac{498}{500}\right)=- \\$ 0.70$.
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    this is an average loss of $0.70$ per ticket in the long run, not a guaranteed loss for every buyer.
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Binomial distribution
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  four conditions
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    fixed number of trial $n$
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    trials are independent
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    each trial has exactly two outcomes of interest, success or failure
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    success probability $p$ is the same on every trial
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  $X$ counts successes in $n$ trials, with $q = 1-p$
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  "Success" is just a label for the coutned outcome and need not be a desirable one
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  Formula -> $P(X=x)=\binom{n}{x}p^xq^{n-x}$ .
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    $\binom{n}{x}=\frac{n!}{x!(n-x)!}$ is the number of ways to place the successes among the trials.
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    the formula holds only when all four conditions hold
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  Qualifying example -> draw a card, record whether it is an ace, replace, repeat 8 times: $n=8$ , $p=4/52=1/13$ , $X=0\ldots8$ .
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  Non-qualifying example -> rolling a die 10 times and recording the face is not binomial when the recorded value is the die number, because six outcomes are of interest rather than two.
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  Cumulative probability -> $P(X\leq k)$ is the sum of $P(X=x)$ for $x=0,\ldots,k$ ; the complement rule gives the upper tail.
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  Mean, variance, standard deviation -> $\mu=np$ , $\sigma^2=npq$ , $\sigma=\sqrt{npq}$ .
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    Skipped-breakfast example -> 1 in 5 students skips breakfast, 10 students sampled: $\mu=10(0.2)=2$ and $\sigma^2=10(0.2)(0.8)=1.6$ .
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Poisson distribution
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  Use -> counts how many times an event occurs within a fixed interval, where the interval may be time, area, volume, length, or pages
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  Assumptions
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    count of events in an interval is what is being computed
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    the rate is roughly constant across equal-sized intervals
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    counts in different intervals are independent
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    over a very small interval, the probability is proportional to interval length
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  Typical contexts -> customer arrivals per hour, phone calls per day, typos per page, late employees per day, power outages per year
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  Formula -> $P(X=x)=\frac{e^{-\mu}\mu^x}{x!}$ , where $\mu$ is the average count over the interval in question.
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    if a rate $\lambda$ is given per unit and the interval length is $t$ , use $\mu=\lambda t$ .
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    match $\mu$ to the interval in the question: a rate per 100 pages must be rescaled before answering a question about 400 pages
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Hypergeometric distribution
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  Use -> sampling without replacement from a finite population where each item is either a success or a failure
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  Condition
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    the population has size $N$
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    it contains $M$ successes and $N-M$ failures
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    a sample of size $n$ is drawn without replacement, and every sample of that size is equally likley
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  Formula -> $P(X=x)=\frac{\binom{M}{x}\binom{N-M}{n-x}}{\binom{N}{n}}$ .
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    numerator counts ways to take $x$ successes and $n-x$ failures; denominator counts all samples of size $n$
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    Valid range -> $\max(0,n-N+M)\leq x\leq\min(n,M)$ .
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  Card example -> 52 cards with 13 spades, 5 drawn without replacement; $X$ counts spades with $N=52$ , $M=13$ , $n=5$ .
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Binomial versus hypergeometric
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  Sampling -> binomial assumes independent trials with replacement, or a large population approximating it; hypergeometric samples without replacement from a finite population.
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  Use binomial -> when the population is large relative to the sample, or items are returned after each draw.
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  Use hypergeometric -> when the sample is a noticeable fraction of a small population.
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Choosing a distribution
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  decide what is being counted.
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  ask whether draws are with or without replacement.
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  note wording that implies replacement, such as “returned” or “put back”.
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  write the parameters clearly, matching the interval or population.
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  sanity-check the result: a probability in $[0,1]$ and a mean that matches the situation.
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Discrete random variables and distributions

Random variables

Mean, variance, and expected value

Binomial distribution

Poisson distribution

Hypergeometric distribution

Binomial versus hypergeometric

Choosing a distribution

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