Binomial distribution ($\mathrm{Bin}(n,p)$)

The binomial distribution counts successes in a fixed number $n$ of independent Bernoulli trials with common success probability $p$. Equivalently,Here $n$ is a nonnegative integer and $0\le p\le 1$. The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n!/[k!(n-k)!]$ counts placements of the $k$ successes; each particular arrangement has probability $p^k(1-p{)}^{n-k}$. The binomial theorem makes the masses sum to $[p+(1-p){]}^n=1$. At $p=0$ all mass is at zero, and at $p=1$ all mass is at $n$; these endpoint laws can be read directly without interpreting $0^0$.

The linearity of expectation and addition of variances for independent random variables giveThe fixed trial count, common success probability, and joint independence are substantive model assumptions. Sampling uniformly without replacement from a finite population instead leads to a hypergeometric distribution. Independent trials with different success probabilities generally give a Poisson binomial distribution. Common marginal success probabilities alone are insufficient: if all ten indicators equal the same Bernoulli random variable with $p=0.3$, their sum is zero with probability $0.7$ or ten with probability $0.3$. Its mean is still three, but its variance is $100(0.21)=21$. The standard construction is given in Penn State STAT 504, binomial distribution.

For an exact constructed example, let each of ten independent trials choose uniformly from ten labelled tokens, with replacement, and mark three tokens as successes. Then $p=3/10$ and $K\sim \mathrm{Bin}(10,0.3)$. Exactly three successes have probabilityThe mean is $10(0.3)=3$, the variance is $10(0.3)(0.7)=2.1$, and the standard deviation is $\sqrt{2.1}\approx 1.4491$. Zero successes have probability $0.7^{10}=0.0282475249$, so at least one success has probability $1-0.7^{10}=0.9717524751$. A single count calculation replaces enumeration of all $10^{10}$ equally likely token sequences.

Each bar's height is one probability mass. Three is the most likely count, but its mass is only about $26.7\%$: a most likely outcome need not occur most of the time.

A random variable with a binomial distribution.

For a binomial distribution with parameters $n$ and $p$, the binomial theorem gives the displayed moment-generating function, finite for every real $t$. Products of these moment-generating functions show that the sum of independent binomial random variables with the same success probability again has a binomial distribution, with trial counts added.

If $N$ independent trials give a specified digit with probability $p$, its observed fraction is $f=K/N$ with binomial count $K$. Since $\mathbb{E}[f]=p$ and $\mathrm{Var}(K)=Np(1-p)$, the mean squared discrepancy is the displayed inverse-sample-size law. Dividing by $p^2$ converts it to a relative squared discrepancy without changing the log-log slope. Nested prefixes are correlated; a single deterministic expansion's plotted discrepancies do not prove that its digits are independent or normally distributed in the number-theoretic sense.

For independent random variables $X,Y$ with the same binomial distribution $\mathrm{Bin}(n,p)$, partition the tie event into the mutually exclusive events $X=Y=j$. Independence makes each probability the square of its binomial probability, proving the formula. At $p=1/2$, Vandermonde's identity gives $\mathrm{Pr}(X=Y)=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. Symmetry under exchange of the two variables then gives $\mathrm{Pr}(X>Y)=\mathrm{Pr}(Y>X)=(1-\mathrm{Pr}(X=Y))/2$.

The binomial distribution with parameters $n$ and $p$ assigns the displayed probability to $k$ successes, for integers $0\le k\le n$. Each specified sequence of independent trials has probability $p^k(1-p{)}^{n-k}$, and the binomial coefficient counts the sequences with exactly $k$ successes. Outside this range the probability is zero.

Let $X$ and $Y$ be independent random variables with binomial distributions $\mathrm{Bin}(n,p)$ and $\mathrm{Bin}(m,q)$. Conditional on $X+Y=k$ having positive probability, the possible values of $X$ satisfy $\max (0,k-m)\le j\le \min (n,k)$. Their probabilities arewhere the denominator sums over that feasible integer range. The formula follows by multiplying independent binomial probabilities and dividing by the probability of the disjoint union of all feasible count pairs.

For $p,q\in (0,1)$, common factors cancel and the weights are proportional to $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{m}{k-j}\right){\theta }^j$, where $\theta =p(1-q)/[q(1-p)]$. If $p=q$, this reduces to the hypergeometric distribution, with probabilities $\left(\genfrac{}{}{0ex}{}{n}{j}\right)\left(\genfrac{}{}{0ex}{}{m}{k-j}\right)/\left(\genfrac{}{}{0ex}{}{n+m}{k}\right)$. At parameter endpoints, use the original probability weights and check that the conditioning event has positive probability before dividing.

For $D\sim \mathrm{Bin}(n,p)$ with $n>0$, the sample binomial proportion is $D/n$. Its expectation is p and its variance is $p(1-p)/n$. The central limit theorem gives a normal approximation when both expected outcome counts are sufficiently large. These moments determine a standard error and model-based control limits.

With $m$ groups of $q$ trials and common marginal failure probability $p$, each count $X_j$ has expectation $qp$, so the displayed estimate is unbiased without assuming independence within groups. Independent individual trials give variance $p(1-p)/(mq)$. If only groups are independent, estimate the variance instead by $s_X^2/(m{q}^2)$, retaining within-group dependence. Equal marginal probabilities alone do not justify a binomial confidence interval.

If $X\sim \mathrm{Bin}(n,p)$ and $X=0$ is observed, the maximum-likelihood estimate of $p$ is zero. A one-sided $1-\alpha$ upper confidence bound solves $P_{p_U}(X=0)=(1-p_U{)}^n=\alpha$, so $p_U=1-{\alpha }^{1/n}$. For large $n$, $p_U\simeq -\log (\alpha )/n$, giving approximately $3/n$ at 95% confidence. This describes uncertainty after no observed events, rather than proving that the underlying probability is zero. It assumes independent observations with a common event probability; clustered data require a different uncertainty calculation.

The zero-truncated binomial distribution has probabilities $\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x(1-p{)}^{n-x}/[1-(1-p{)}^n]$ for $1\le x\le n$. Its natural parameter is $\varphi =\log [p/(1-p)]$ and its cumulant is $\log [(1+e^{\varphi }{)}^n-1]$. Mean and variance are $np/s$ and $np(1-p)/s-n^2{p}^2(1-p{)}^n/s^2$, where $s=1-(1-p{)}^n$. The conditional normalizer depends on $p$, so its likelihood is not proportional to the untruncated likelihood even after a positive observation. At $X=1$ and $n>1$, its likelihood supremum is at $p\downarrow 0$ rather than at $1/n$.

For $W\sim \mathrm{Bin}(n,p)$, expand the fourth power of the sum of centered independent Bernoulli variables. Terms with an index occurring once vanish; single-index fourth moments contribute $nq(1-3q)$ and paired second moments contribute $3n(n-1)q^2$. Their sum is the displayed formula. Scaling by $n^4$ yields a uniform $O(n^{-2})$ fourth moment for $W/n-p$.

For independent Bernoulli observations with success probability $p$, the sample proportion is the success count divided by the sample size. It is an unbiased estimator with variance $p(1-p)/n$. It supplies the arm-specific risks in a risk ratio and the mean difference in a two-proportion comparison.

For two independent samples of Bernoulli random variables, the difference of their sample proportions estimates the difference of population probabilities. Since the samples are independent,Substitution of the sample proportions gives a standard error for a large-sample Wald confidence interval. This interval is a descriptive comparison of the specified groups; a causal effect needs further assumptions. Sparse counts and probabilities near zero or one can make the normal approximation poor. Testing equality can instead use a null-pooled estimate of the common probability for the test's standard error.

If $x$ successes occur in $n$ independent Bernoulli trials, the binomial likelihood for the success probability $p$ is proportional to $p^x(1-p{)}^{n-x}$.