Binomial and Bernoulli Distributions

Almost every quantitative statement in physics that begins "how many, out of many tries?" is a binomial statement. A Bernoulli trial is the atom of chance: a single experiment with exactly two outcomes — heads or tails, click or silence, decay or survival, spin up or spin down — carrying one parameter, the success probability $p$ . Stack $n$ such trials, independent and identically weighted, and count the successes: that count is a binomial random variable, and its law $P(k)=\binom{n}{k}p^k q^{n-k}$ is the oldest non-trivial distribution in probability. This article takes the migrated skeleton — the mass function, the mean $np$ , the variance $npq$ , the generating functions, and the Poisson and Gaussian limits — and rebuilds it as deep canon, with every number re-derived and re-checked. One structural fact organizes everything: a binomial variable is literally a sum of $n$ independent Bernoulli indicators, $X=\sum_i X_i$ . From that single decomposition the mean follows by linearity of expectation, the variance by additivity for independent summands, and the generating functions by the multiplicativity of independence — each is one line, not a calculation. The same decomposition is why the binomial sits at a crossroads of limits: hold $p$ fixed and let $n$ grow and the standardized sum flows to a Gaussian (de Moivre–Laplace, the first central limit theorem); hold the mean $np=\lambda$ fixed while $n\to\infty$ and it collapses onto the Poisson law of rare events. Photon counts at fixed detector efficiency, up-spins in a paramagnet, steps of a random walker, bit errors on a channel, "yes" answers in a poll — all are binomial counts, and their $1/\sqrt{n}$ relative fluctuations are why macroscopic averages are sharp.

Key takeaways

A Bernoulli trial is one two-outcome experiment: success with probability $p$ , failure with $q=1-p$ ; its indicator has mean $p$ and variance $pq$ .
The binomial law gives the chance of exactly $k$ successes in $n$ independent identical trials, $P(k)=\binom{n}{k}p^k q^{n-k}$ , normalized by the binomial theorem $(p+q)^n=1$ .
Because $X=\sum_{i=1}^n X_i$ is a sum of independent Bernoulli bricks, its mean is $np$ and its variance is $npq$ — additivity, not algebra.
The generating functions factorize, $G(s)=(q+ps)^n$ and $M(t)=(q+pe^t)^n$ , so moments and sums of binomials are one-liners.
Two limits meet here: $np=\lambda$ fixed, $n\to\infty$ gives Poisson $\lambda^k e^{-\lambda}/k!$ ; $p$ fixed, $n\to\infty$ gives a Gaussian of mean $np$ , variance $npq$ .
Relative fluctuations scale as $\sqrt{q/np}\sim 1/\sqrt{n}$ — the reason a million fair tosses land overwhelmingly near a half-heads fraction.
Key idea: a binomial variable is just a sum of $n$ independent Bernoulli indicators, and mean $np$ , variance $npq$ , the generating function $(q+ps)^n$ , and the Poisson and Gaussian limits all follow from that one additive fact.

The core insight

A binomial variable is just a sum of $n$ independent Bernoulli indicators; everything — mean $np$ , variance $npq$ , the generating function $(q+ps)^n$ , and the Poisson and Gaussian limits — follows from that single additive fact.

Where this lives today

Binomial statistics are the everyday arithmetic of the laboratory. Single-photon detectors register each incident photon independently with a quantum efficiency $\eta$ , so out of $N$ arriving photons the count of clicks is $\mathrm{Bin}(N,\eta)$ with mean $N\eta$ and variance $N\eta(1-\eta)$ ; "binomial thinning" of a Poisson light source stays Poisson, the foundation of shot-noise analysis and of loss budgets in quantum-key-distribution experiments. In statistical mechanics the number of up-spins among $N$ non-interacting two-state moments is binomial, so magnetization fluctuates with variance $Npq$ and relative fluctuation $\propto 1/\sqrt N$ — the very reason a macroscopic magnet has a sharp, effectively deterministic equilibrium. Pollsters quoting a margin of error $\sqrt{p(1-p)/n}$ , geneticists counting alleles under Hardy–Weinberg, and engineers bounding bit-error rates all reach for the same $\binom{n}{k}p^k q^{n-k}$ . And the two limits remain live tools: the Poisson limit governs rare-decay counting in dark-matter searches, while the de Moivre–Laplace Gaussian underwrites the " $5\sigma$ " discovery threshold that announced the Higgs boson.

The physical picture

Flip a biased coin $n$ times and count the heads. Two forces fight inside $\binom{n}{k}p^k q^{n-k}$ : the combinatorial factor $\binom{n}{k}$ , which counts how many arrangements give $k$ heads and is largest at the middle $k=n/2$ , and the weight $p^k q^{n-k}$ , the probability of any one such arrangement, which pulls toward whichever face is favoured. The distribution's peak is their compromise, and it sits at the mean $k\approx np$ . Shift $p$ and the peak slides with it; the first figure shows the fair coin peaking at $np=5$ while a $p=0.2$ coin peaks at $np=2$ .

Figure

Two binomial mass functions over n = 10 trials — the fair coin p = 0.5 in blue peaks at its mean np = 5 and is symmetric about it, while the biased p = 0.2 coin in amber peaks at np = 2. The mode of each tracks its mean, and the widths differ because the variance npq = 2.5 for the fair coin exceeds npq = 1.6 for the biased one

Now widen the width by holding $p$ fixed and letting $n$ grow. The mean climbs as $np$ but the standard deviation grows only as $\sqrt{npq}$ , so the shape, rescaled by that width, freezes into a single universal bell. This is the de Moivre–Laplace theorem, and it is not an analogy but an exact asymptotic: the binomial bars fall under a Gaussian envelope of mean $np$ and variance $npq$ . The second figure is the punchline a physicist actually uses — one hundred fair tosses, mean $50$ , standard deviation exactly $5$ , with about $68\%$ of the mass inside one standard deviation of the mean.

Figure

The de Moivre–Laplace limit for n = 100 fair tosses — the exact binomial mass (blue bars) is captured almost perfectly by the Gaussian envelope of mean np = 50 and standard deviation √(npq) = 5 shown in green. The peak height is 1/√(2π·25) = 0.0798, and about 68% of all outcomes fall in the one-sigma band 45 ≤ k ≤ 55

The core equations

The Bernoulli distribution is the single trial. A variable $X$ takes the value $1$ (success) with probability $p$ and $0$ (failure) with probability $q=1-p$ , and the two cases are written compactly with exponents that switch one factor off.

P(X=k)=p^k q^{1-k},\qquad q=1-p,\quad k\in\{0,1\} \qquad (1)

where: $X$ — Bernoulli indicator of success ( $1$ ) or failure ( $0$ ); $p$ — success probability, $0\le p\le 1$ ; $q$ — failure probability. Reading it: at $k=1$ the mass is $p$ , at $k=0$ it is $q$ , and the mean is $p$ with variance $pq$ because $X^2=X$ for a $0/1$ variable. The binomial distribution counts successes across $n$ independent Bernoulli trials that share the same $p$ . The combinatorial factor tallies the orderings; the power tallies the probability of any one ordering.

P(X=k)=\binom{n}{k}p^k q^{n-k},\qquad k=0,1,\dots,n \qquad (2)

where: $n$ — number of independent trials; $k$ — number of successes; $\binom{n}{k}=\dfrac{n!}{k!(n-k)!}$ — the count of distinct arrangements of $k$ successes among $n$ slots. Reading it: probability of a specific run $\times$ number of runs. Because the trials share $p$ and are independent, the total mass over all $k$ is the binomial theorem evaluated at $p+q$ , which is exactly $1$ — the distribution is automatically normalized.

\sum_{k=0}^{n}\binom{n}{k}p^k q^{n-k}=(p+q)^n=1 \qquad (3)

where: $(p+q)^n$ — the binomial expansion whose terms are the masses $P(k)$ . Reading it: normalization is not an extra condition to impose; it is the binomial theorem in disguise. The mean and variance are the two numbers a physicist quotes first. They come straight from the additive Bernoulli decomposition and are re-derived below.

\mathbb{E}[X]=np,\qquad \mathrm{Var}(X)=npq,\qquad \sigma=\sqrt{npq} \qquad (4)

where: $\mathbb{E}[X]$ — expected number of successes; $\mathrm{Var}(X)$ — variance of the count; $\sigma$ — standard deviation. Reading it: the mean grows like $n$ but the spread only like $\sqrt{n}$ , so the relative fluctuation $\sigma/\mathbb{E}[X]=\sqrt{q/np}$ shrinks as $1/\sqrt{n}$ . The generating functions encode the whole distribution in one closed form. Independence turns the expectation of a product into a product of expectations, so each factorizes into an $n$ -th power.

G(s)=\mathbb{E}[s^X]=(q+ps)^n,\qquad M(t)=\mathbb{E}[e^{tX}]=(q+pe^t)^n,\qquad \varphi(t)=(q+pe^{it})^n \qquad (5)

where: $G(s)$ — probability-generating function, whose $s^k$ coefficient is $P(k)$ and with $G'(1)=\mathbb{E}[X]$ ; $M(t)$ — moment-generating function, with $M^{(m)}(0)=\mathbb{E}[X^m]$ ; $\varphi(t)$ — characteristic function, always defined. Reading it: differentiate $M$ to pull out moments; multiply two $G$ 's to add two binomials. Multiplicativity of the generating function gives the reproductive property: two independent binomials with a common $p$ combine into one bigger binomial, because their PGFs multiply to another power of $(q+ps)$ .

X_1\sim\mathrm{Bin}(n_1,p),\ X_2\sim\mathrm{Bin}(n_2,p)\ \Rightarrow\ X_1+X_2\sim\mathrm{Bin}(n_1+n_2,p) \qquad (6)

where: $X_1+X_2$ — pooled success count of $n_1+n_2$ trials. Reading it: pooling two coin-flip experiments with the same coin just makes a longer experiment — the family is closed under addition. Finally, the ratio of consecutive masses locates the peak. Where the ratio crosses unity, the distribution stops rising, fixing the most probable count.

\frac{P(k+1)}{P(k)}=\frac{n-k}{k+1}\cdot\frac{p}{q},\qquad k_{\text{mode}}=\lfloor (n+1)p\rfloor \qquad (7)

where: $P(k+1)/P(k)$ — one-step ratio of masses; $\lfloor\cdot\rfloor$ — floor function; $k_{\text{mode}}$ — most probable number of successes. Reading it: masses climb while the ratio exceeds $1$ and fall once it drops below, so the mode sits just below $(n+1)p$ , essentially at the mean $np$ .

Derivation AThe binomial law and its moments from Bernoulli bricks

The goal is to build the entire binomial — its mass function, its mean, and its variance — from one structural idea, that a binomial count is a sum of $n$ independent Bernoulli indicators, $X=\sum_i X_i$ . We first assemble $P(X=k)$ from independence and counting, then read the mean $np$ and the variance $npq$ straight off that additive decomposition, so three separate-looking results collapse into a single argument that never touches the factorials again after the first step.

Step 1. Fix one specific outcome sequence with $k$ successes and $n-k$ failures, say SS…SFF…F. By independence its probability factorizes into the product of the single-trial probabilities, $p^k q^{n-k}$ , and this value is the same for every sequence with the same tally of successes, regardless of order. Why: independence is exactly the statement that joint probabilities multiply, so only the counts of S and F matter, not their positions.

Step 2. Count how many distinct sequences carry exactly $k$ successes: it is the number of ways to choose which $k$ of the $n$ positions are the successes, namely $\binom{n}{k}$ . Why: the positions of the successes are an unordered $k$ -subset of $n$ slots, and that is the definition of the binomial coefficient.

Step 3. The distinct sequences are mutually exclusive events, so by the additivity axiom their probabilities sum, giving $P(X=k)=\binom{n}{k}p^k q^{n-k}$ — equation (2). Why: no single run of trials can be two different sequences at once, so summing over the $\binom{n}{k}$ disjoint runs is legitimate.

Step 4. Now decompose $X=\sum_{i=1}^n X_i$ with each $X_i$ the Bernoulli indicator of trial $i$ . A $0/1$ variable satisfies $X_i^2=X_i$ , so $\mathbb{E}[X_i]=p$ and $\mathrm{Var}(X_i)=\mathbb{E}[X_i^2]-p^2=p-p^2=pq$ . Why: squaring a variable that is only ever $0$ or $1$ leaves it unchanged, which collapses the second moment to the first and hands you the single-trial variance for free.

Step 5. Expectation is always additive, so $\mathbb{E}[X]=\sum_i p=np$ ; and because the trials are independent, variance is additive too, so $\mathrm{Var}(X)=\sum_i pq=npq$ . Why: linearity of expectation needs no independence, but the clean addition of variances does — cross-covariances vanish precisely because the trials do not influence one another.

Go deeper · Derivation B · 1 minThe Poisson limit of rare events

Derivation BThe Poisson limit of rare events

The goal is to show that when successes are individually rare but numerous in aggregate — many trials, tiny $p$ , with the mean $np=\lambda$ held fixed — the binomial mass collapses onto the Poisson law $\lambda^k e^{-\lambda}/k!$ . We set $p=\lambda/n$ , hold $k$ and $\lambda$ fixed, and let $n\to\infty$ , tracking the three pieces of the mass function (the coefficient, the success power, and the failure power) as they each settle to their limits.

Step 1. Write the falling factorial explicitly, $\binom{n}{k}=\dfrac{n(n-1)\cdots(n-k+1)}{k!}$ , a product of $k$ factors each close to $n$ when $n$ is large. Why: separating the $k!$ from the top lets the $k$ near- $n$ factors pair up cleanly with the $p^k=(\lambda/n)^k$ that follows.

Step 2. Combine that falling factorial with $p^k=(\lambda/n)^k$ : the $n^k$ in the denominator cancels the product's leading $n^k$ , and $\dfrac{n(n-1)\cdots(n-k+1)}{n^k}=\prod_{j=0}^{k-1}\dfrac{n-j}{n}\to 1$ , leaving $\lambda^k/k!$ . Why: each ratio $(n-j)/n\to 1$ because $j$ is fixed while $n$ diverges, so all the finite corrections wash out.

Step 3. Handle the failure factor $(1-\lambda/n)^{n-k}=(1-\lambda/n)^n(1-\lambda/n)^{-k}$ : the first piece is the classic limit $(1-\lambda/n)^n\to e^{-\lambda}$ , and the second $\to 1$ . Why: the definition of the exponential is exactly $\lim(1-x/n)^n=e^{-x}$ , and the $-k$ correction is a fixed finite power of a factor tending to one.

Step 4. Collect the surviving pieces: $P(X=k)\to \dfrac{\lambda^k}{k!}e^{-\lambda}$ , the Poisson mass with parameter $\lambda=np$ . Why: the coefficient contributed $\lambda^k/k!$ and the failure factor contributed $e^{-\lambda}$ , and nothing else survives — the rare-event count is Poisson.

Go deeper · Derivation C · 1 minThe de Moivre–Laplace Gaussian limit

Derivation CThe de Moivre–Laplace Gaussian limit

The goal is to show that for large $n$ with $p$ fixed away from $0$ and $1$ , the binomial mass near its peak is a Gaussian of mean $np$ and variance $npq$ — the earliest instance of the central limit theorem. We take the logarithm of the mass, apply Stirling's approximation to the three factorials, introduce the scaled deviation from the mean, and expand to second order, watching the first-order terms cancel exactly because we are sitting at the mode.

Step 1. Take logs and apply Stirling, $\ln m!\approx m\ln m-m+\tfrac12\ln(2\pi m)$ , to $n!$ , $k!$ , and $(n-k)!$ ; after cancellation this gives $\ln P(k)\approx -\tfrac12\ln(2\pi npq)+k\ln\dfrac{np}{k}+(n-k)\ln\dfrac{nq}{n-k}$ . Why: Stirling turns intractable factorials into smooth functions of $k$ that can be differentiated and expanded, and the $\tfrac12\ln(2\pi\cdot)$ terms assemble into the Gaussian's normalization.

Step 2. Introduce the deviation $k=np+\delta$ with $\delta=x\sqrt{npq}$ and $|\delta|\ll np$ , measuring displacement from the mean in units of the standard deviation. Why: the interesting structure lives within a few $\sqrt{npq}$ of the mean, so this is the natural zoom in which the shape stabilizes.

Step 3. Expand the two logarithmic terms to second order in $\delta$ : the first-order pieces cancel because $np$ is the stationary point (the mode), and the leading survivor is $k\ln\dfrac{np}{k}+(n-k)\ln\dfrac{nq}{n-k}\approx-\dfrac{\delta^2}{2npq}$ . Why: expanding about an extremum kills the linear term by construction, leaving the quadratic that becomes the Gaussian exponent.

Step 4. Exponentiate to recover $P(k)\approx \dfrac{1}{\sqrt{2\pi npq}}\exp\!\left[-\dfrac{(k-np)^2}{2npq}\right]$ , a normal density of mean $np$ and variance $npq$ . Why: undoing the logarithm turns the $-\delta^2/2npq$ into the bell curve, and the prefactor is precisely the normalization a Gaussian of variance $npq$ requires.

Limiting cases

The binomial is a hub: push it in different directions and it becomes the other canonical distributions.

Case	Result	Meaning
Single trial, $n=1$	$P(k)=p^k q^{1-k},\ k\in\{0,1\}$	reduces to the Bernoulli brick — mean $p$ , variance $pq$
Symmetric, $p=\tfrac12$	$P(k)=\binom{n}{k}2^{-n},\ \mathbb{E}[k]=\tfrac n2,\ \mathrm{Var}=\tfrac n4$	the fair coin and symmetric random walk; mass symmetric about $n/2$
Rare events, $np=\lambda$ fixed, $n\to\infty$	$\binom{n}{k}p^k q^{n-k}\to \dfrac{\lambda^k e^{-\lambda}}{k!}$	Poisson law — clicks, decays, arrivals with finite mean $\lambda$
Many trials, $p$ fixed, $n\to\infty$	$P(k)\approx \dfrac{1}{\sqrt{2\pi npq}}\exp\!\left[-\dfrac{(k-np)^2}{2npq}\right]$	de Moivre–Laplace Gaussian of mean $np$ , variance $npq$
Deterministic, $p\to0$ or $p\to1$	$\mathrm{Var}=npq\to 0$	all-failure or all-success; no fluctuation, the mass is a spike

Common trapThe binomial needs independence and a constant p — drop either and it fails

The two hypotheses are not decoration. If the success probability changes from trial to trial, or if the trials influence one another, $\binom{n}{k}p^k q^{n-k}$ is simply wrong. The classic trap is sampling without replacement: draw coloured balls from an urn and keep them, and each draw alters the composition, so $p$ drifts and the draws are dependent. The correct law there is the hypergeometric distribution, which the binomial only approximates when the population is far larger than the sample. A second trap is treating $P(k)$ as an area: it is a probability mass, so probabilities are sums over integer $k$ , never integrals — the Gaussian limit is a continuous approximation to a discrete object, and honest use of it near small counts needs a continuity correction (shift integer bounds by $\pm\tfrac12$ ).

Worked examples

1 · One hundred fair tosses, fully worked (confidence builder)

A fair coin is tossed $n=100$ times. Find the mean, standard deviation, the probability of exactly $50$ heads, the probability of landing between $45$ and $55$ heads, and the relative fluctuation.

Step 1. With $p=q=\tfrac12$ , the mean is $\mathbb{E}[X]=np=100\cdot\tfrac12=\mathbf{50}$ and the variance is $\mathrm{Var}(X)=npq=100\cdot\tfrac12\cdot\tfrac12=25$ , so the standard deviation is $\sigma=\sqrt{25}=\mathbf{5}$ . Heads cluster as $50\pm5$ .

Step 2. The exact peak mass is $P(50)=\binom{100}{50}2^{-100}=\mathbf{0.0796}$ . The de Moivre–Laplace formula predicts $P(50)\approx \dfrac{1}{\sqrt{2\pi npq}}=\dfrac{1}{\sqrt{2\pi\cdot25}}=\dfrac{1}{12.533}=\mathbf{0.0798}$ — agreement to better than $0.3\%$ , a vivid check on the Gaussian limit.

Step 3. For the band $45\le k\le 55$ , i.e. within one standard deviation, the Gaussian rule of thumb gives about $68\%$ . The exact binomial sum is $\sum_{k=45}^{55}\binom{100}{k}2^{-100}=\mathbf{0.7287}$ , and the continuity-corrected normal estimate $2\Phi(1.1)-1=0.7287$ matches to four decimals — the extra few percent over $68\%$ comes from including the endpoints $45$ and $55$ fully.

Step 4. The relative fluctuation is $\sigma/\mathbb{E}[X]=5/50=\mathbf{0.10}$ . Repeat with $n=10^6$ : mean $5\times10^5$ , $\sigma=500$ , relative fluctuation $\mathbf{0.001}$ . The $1/\sqrt{n}$ law in action — a million tosses pin the head-fraction to a tenth of a percent. Mean 50, σ = 5, P(50) = 0.0796, P(45–55) = 0.729, relative spread falls as 1/√n.

2 · A noisy digital channel and the Poisson limit

A packet of $n=1000$ bits crosses a channel that independently flips each bit with probability $p=0.002$ . Find the expected number of errors, the exact probability of exactly two errors, and compare it with the Poisson approximation.

Step 1. The mean error count is $\lambda=np=1000\cdot0.002=\mathbf{2}$ , with variance $npq=1000\cdot0.002\cdot0.998=1.996\approx2$ — for rare events the variance nearly equals the mean, the Poisson signature.

Step 2. The exact binomial probability of exactly two errors is $P(2)=\binom{1000}{2}(0.002)^2(0.998)^{998}$ . Here $\binom{1000}{2}=499\,500$ , $(0.002)^2=4\times10^{-6}$ , and $(0.998)^{998}=0.1356$ , giving $P(2)=499\,500\cdot4\times10^{-6}\cdot0.1356=\mathbf{0.2709}$ .

Step 3. The Poisson approximation with $\lambda=2$ gives $P(2)\approx \dfrac{\lambda^2 e^{-\lambda}}{2!}=\dfrac{4\,e^{-2}}{2}=2e^{-2}=\mathbf{0.2707}$ — a $0.1\%$ discrepancy. Likewise $P(0)$ : binomial $(0.998)^{1000}=0.1351$ versus Poisson $e^{-2}=0.1353$ .

Step 4. The lesson: with $n$ in the thousands and $p$ in the thousandths, the two-parameter binomial is unnecessary — the one-parameter Poisson with $\lambda=np$ reproduces it to three decimals, which is why counting statistics for rare independent events are quoted as Poisson. λ = 2, exact P(2) = 0.271, Poisson P(2) = 0.271 — the rare-event limit is quantitatively exact here.

Physics of the portrait

The binomial distribution is where combinatorics, the law of large numbers, and the central limit theorem were all born — and it carries the name of the mathematician who first proved that empirical frequencies converge to probabilities.

Origins · 1 minJacob Bernoulli and how this was discovered

Jacob Bernoulli

1655–1705 · the origins of the law of large numbers and the mathematics of counting chance

Jacob Bernoulli, eldest of the Basel dynasty that would fill mathematics with the family name, gave the binomial distribution its structure and its first deep theorem. In Ars Conjectandi — written in the 1690s and published posthumously by his nephew Nicolaus in 1713 — he collected the combinatorics of the numbers $\binom{n}{k}$ (which is why they carry his cousinly claim to the label "Bernoulli trials"), and then proved something no one had proved before: his theorema aureum, the golden theorem, now called the weak law of large numbers. It states that the observed fraction of successes in $n$ independent trials converges in probability to the true $p$ as $n$ grows — the first rigorous bridge from a mathematical probability to a measurable frequency. Bernoulli even estimated how large $n$ must be for a given accuracy, foreshadowing modern confidence intervals. What he could not do was describe the shape of the fluctuations around $np$ ; that fell to Abraham de Moivre, who in 1733 found the normal curve as the large- $n$ limit of the binomial, and to Pierre-Simon Laplace, who generalized it into what we now call the de Moivre–Laplace theorem, the earliest central limit theorem. A century later Siméon Denis Poisson took the complementary limit — many trials, vanishing $p$ , fixed mean — and extracted the law of rare events. The whole edifice of classical probability, and much of statistical physics after it, rests on the two-outcome trial Bernoulli first analyzed.

History

1655

— Jacob Bernoulli born in Basel

1713

— Ars Conjectandi published posthumously, stating the weak law of large numbers

1733

— Abraham de Moivre finds the normal curve as the large-n limit of the binomial

1774

— Laplace generalizes the normal approximation and founds analytic probability

1837

— Poisson derives the rare-event limit with np held fixed, the Poisson distribution

Going further

Texts. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, Ch. VI–VII (binomial, de Moivre–Laplace) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 23 (Probability and Statistics) · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 30 · Stone & Goldbart, Mathematics for Physics, probability chapter · Byron & Fuller, Mathematics of Classical and Quantum Physics, statistics sections.

Provenance. All results are standard textbook material, re-verified here. A checker should: (1) confirm normalization via $(p+q)^n=1$ from the binomial theorem; (2) recover mean $np$ and variance $npq$ both from the Bernoulli-sum decomposition and by differentiating $M(t)=(q+pe^t)^n$ at $t=0$ ; (3) verify the Poisson limit through $(1-\lambda/n)^n\to e^{-\lambda}$ ; (4) verify the de Moivre–Laplace limit via Stirling's approximation. Numerical anchors, all recomputed: $n=100,\ p=\tfrac12$ gives mean $50$ , $\sigma=5$ , $P(50)=0.0796$ , $P(45\le k\le55)=0.7287$ ; and $n=1000,\ p=0.002$ gives exact $P(2)=0.2709$ against Poisson $2e^{-2}=0.2707$ .

Prerequisites. Axioms of Probability · Random Variables and Probability Distributions · Expectation, Variance and Moments · Conditional Probability and Bayes' Theorem. Leads to. Poisson Distribution · Gaussian (Normal) Distribution · Central Limit Theorem · Moment-Generating and Characteristic Functions · Laws of Large Numbers and Convergence of Random Variables.

Ask the tutor: Ask me to walk through any binomial calculation — a coin, a detector, a poll, or a decay count — and I will show you the Bernoulli-brick decomposition and which limit (Poisson or Gaussian) applies.

Why is the binomial variance npq and not something more complicated?

Because a binomial is a sum of $n$ independent Bernoulli variables, and variance adds over independent summands. A single Bernoulli indicator has variance $pq$ : since it is only ever $0$ or $1$ , $X^2=X$ , so $\mathbb{E}[X^2]=\mathbb{E}[X]=p$ and $\mathrm{Var}=p-p^2=pq$ . Adding $n$ independent copies simply adds $n$ equal variances, giving $npq$ . There is no cross term because independence makes every covariance vanish. The same logic gives the mean $np$ by linearity — which needs no independence at all — so the two headline numbers are almost immediate once you see the sum-of-bricks structure rather than fighting the factorials directly.

How can one binomial have both a Poisson limit and a Gaussian limit?

They are two different scalings of the same object. The Poisson limit fixes the mean $np=\lambda$ while sending $n\to\infty$ and $p\to0$ : successes become individually rare, the count stays modest, and you get $\lambda^k e^{-\lambda}/k!$ . The Gaussian limit instead fixes the probability $p$ (away from $0$ and $1$ ) while $n\to\infty$ : successes are common, the mean $np$ grows large, and the standardized count flows to a bell curve of mean $np$ and variance $npq$ . The distinguishing question is whether the expected number of successes stays finite (Poisson) or grows without bound (Gaussian). Both are correct; they describe different physical regimes — a handful of rare decays versus a landslide of coin flips.

What is the difference between a Bernoulli and a binomial distribution?

A Bernoulli distribution describes one trial with two outcomes: it takes the value $1$ with probability $p$ and $0$ with probability $q=1-p$ , full stop. A binomial distribution describes the number of successes in $n$ such independent, identically weighted Bernoulli trials, taking integer values $0,1,\dots,n$ with mass $\binom{n}{k}p^k q^{n-k}$ . So Bernoulli is the special case $n=1$ , and binomial is the sum of $n$ Bernoulli variables. The relationship is the whole point of this article: everything about the binomial — its mean $np$ , variance $npq$ , generating function $(q+ps)^n$ , and limiting behaviour — is inherited additively from the Bernoulli brick it is built from.

When is the Poisson approximation to the binomial good enough to use?

As a rule of thumb, when $n$ is large and $p$ small with $np$ moderate — many texts cite $n\ge 20,\ p\le0.05$ , or the looser $n\ge100,\ np\le10$ . It replaces two parameters $(n,p)$ with a single $\lambda=np$ , and the error is controlled by $p$ : the leading correction scales like $p$ per event. In the worked channel example, $n=1000$ and $p=0.002$ agree to about $0.1\%$ , far tighter than any real measurement needs. This is why counts of independent rare events — radioactive decays, dark counts, arrivals — are reported with Poisson error bars $\sqrt\lambda$ .

Where does the binomial coefficient come from, physically?

It is pure bookkeeping of orderings. Any specific sequence of $k$ successes and $n-k$ failures has probability $p^k q^{n-k}$ by independence — the order does not change the product. But the event "exactly $k$ successes" does not care which trials succeeded, so it lumps together every sequence with the same tally. The count of such sequences is the ways to choose which $k$ of the $n$ positions are successes, $\binom{n}{k}=n!/[k!(n-k)!]$ . So $\binom{n}{k}$ is the multiplicity — the same quantity a statistical physicist calls the number of microstates for a macrostate with $k$ up-spins — and $p^k q^{n-k}$ is the weight of each microstate.

Does a phase like "68% within one sigma" hold exactly for a binomial?

Not exactly — it is the Gaussian rule, and the binomial is discrete, so it holds only asymptotically. For $n=100,\ p=\tfrac12$ the one-sigma band is $45\le k\le 55$ , and the exact binomial probability there is $0.729$ , a few points above the Gaussian $0.683$ because including the integer endpoints fully adds mass. A continuity correction — treating the band as $44.5$ to $55.5$ — brings the normal estimate to $0.729$ , matching the exact sum. So " $68\%$ " is a good mnemonic for large $n$ with $p$ not near $0$ or $1$ , but for precise work sum the actual masses or use the continuity-corrected normal, especially in the tails where the Gaussian is weakest.