Binomial distribution: Difference between revisions

   | notation  = <math>B(n,p)</math>

In [[probability theory]] and [[statistics]], the '''binomial distribution''' with parameters {{mvar|n}} and {{mvar|p}} is the [[discrete probability distribution]] of the number of successes in a sequence of {{mvar|n}} [[statistical independence|independent]] [[experiment (probability theory)|experiment]]s, each asking a [[yes–no question]], and each with its own [[Boolean-valued function|Boolean]]-valued [[outcome (probability)|outcome]]: ''success'' (with probability {{mvar|p}}) or ''failure'' (with probability {{math|''q'' {{=}} 1 − ''p''}}). A single success/failure experiment is also called a [[Bernoulli trial]] or Bernoulli experiment, and a sequence of outcomes is called a [[Bernoulli process]]; for a single trial, i.e., {{math|''n'' {{=}} 1}}, the binomial distribution is a [[Bernoulli distribution]]. The binomial distribution is the basis for the [[binomial test]] of [[statistical significance]].<ref>{{Cite book|last=Westland|first=J. Christopher|title=Audit Analytics: Data Science for the Accounting Profession|publisher=Springer|year=2020|isbn=978-3-030-49091-1|location=Chicago, IL, USA|pages=53}}</ref>

The binomial distribution is frequently used to model the number of successes in a sample of size {{mvar|n}} drawn [[with replacement]] from a population of size {{mvar|N}}. If the sampling is carried out without replacement, the draws are not independent and so the resulting distribution is a [[hypergeometric distribution]], not a binomial one.  However, for {{mvar|N}} much larger than {{mvar|n}}, the binomial distribution remains a good approximation, and is widely used.

: <math>f(k,n,p) = \Pr(X = k) = \binom{n}{k}p^k(1-p)^{n-k}</math>

: <math>\binom{n}{k} =\frac{n!}{k!(n-k)!}</math>

is the [[binomial coefficient]]. The formula can be understood as follows: {{math|''p''{{sup|''k''}} ''q''{{sup|''n''−''k''}}}} is the probability of obtaining the sequence of {{mvar|n}} independent Bernoulli trials in which {{mvar|k}} trials are "successes" and the remaining {{math|''n'' − ''k''}} trials result in "failure". Since the trials are independent with probabilities remaining constant between them, any sequence of {{mvar|n}} trials with {{mvar|k}} successes (and {{math|''n'' − ''k''}} failures) has the same probability of being achieved (regardless of positions of successes within the sequence). There are <math display="inline">\binom{n}{k}</math> such sequences, since the binomial coefficient <math display="inline">\binom{n}{k}</math> counts the number of ways to choose the positions of the {{mvar|k}} successes among the {{mvar|n}} trials. The binomial distribution is concerned with the probability of obtaining ''any'' of these sequences, meaning the probability of obtaining one of them ({{math|''p''{{sup|''k''}} ''q''{{sup|''n''−''k''}}}}) must be added <math display="inline">\binom{n}{k}</math> times, hence <math display="inline">\Pr(X = k) = \binom{n}{k} p^k (1-p)^{n-k}</math>.

In creating reference tables for binomial distribution probability, usually, the table is filled in up to {{math|''n''/2}} values. This is because for {{math|''k'' > ''n''/2}}, the probability can be calculated by its complement as

: <math>f(k,n,p)=f(n-k,n,1-p). </math>

: <math> \frac{f(k+1,n,p)}{f(k,n,p)}=\frac{(n-k)p}{(k+1)(1-p)} </math>

and comparing it to 1. There is always an integer {{mvar|M}} that satisfies<ref>{{cite book |last=Feller |first=W. |title=An Introduction to Probability Theory and Its Applications |url=https://archive.org/details/introductiontopr01wfel |url-access=limited |year=1968 |publisher=Wiley |location=New York |edition=Third |page=[https://archive.org/details/introductiontopr01wfel/page/n167 151] (theorem in section VI.3) }}</ref>

: <math>(n+1)p-1 \leq M < (n+1)p.</math>

{{math|''f''(''k'', ''n'', ''p'')}} is monotone increasing for {{math|''k'' < ''M''}} and monotone decreasing for {{math|''k'' > ''M''}}, with the exception of the case where {{math|(''n'' + 1)''p''}} is an integer. In this case, there are two values for which {{mvar|f}} is maximal: {{math|(''n'' + 1) ''p''}} and {{math|(''n'' + 1) ''p'' − 1}}. {{mvar|M}} is the ''most probable'' outcome (that is, the most likely, although this can still be unlikely overall) of the Bernoulli trials and is called the [[Mode (statistics)|mode]].

 

: <math>f(4,6,0.3) = \binom{6}{4}0.3^4 (1-0.3)^{6-4}= 0.059535.</math>

: <math>F(k;n,p) = \Pr(X \le k) = \sum_{i=0}^{\lfloor k \rfloor} {n\choose i}p^i(1-p)^{n-i},</math>

where <math>\lfloor k\rfloor</math> is the "floor" under {{math|''k''}}, i.e. the [[floor and ceiling functions|greatest integer]] less than or equal to {{math|''k''}}.

: <math>\begin{align}

: <math>F(k;n,p) = F_{\text{beta-distribution}}\left(x=1-p;\alpha=n-k,\beta=k+1\right)</math>

: <math>F(k;n,p) = F_{F\text{-distribution}}\left(x=\frac{1-p}{p}\frac{k+1}{n-k};d_1=2(n-k),d_2=2(k+1)\right).</math>

If {{math|''X'' ~ ''B''(''n'', ''p'')}}, that is, {{math|''X''}} is a binomially distributed random variable, {{mvar|n}} being the total number of experiments and ''p'' the probability of each experiment yielding a successful result, then the [[expected value]] of {{math|''X''}} is:<ref>See [https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution Proof Wiki]</ref>

: <math> \operatorname{E}[X] = np.</math>

This follows from the linearity of the expected value along with the fact that {{mvar|X}} is the sum of {{mvar|n}} identical Bernoulli random variables, each with expected value {{mvar|p}}.  In other words, if <math>X_1, \ldots, X_n</math> are identical (and independent) Bernoulli random variables with parameter {{mvar|p}}, then {{math|1=''X'' = ''X''₁ + ... + ''X''_''n''}} and

: <math>\operatorname{E}[X] = \operatorname{E}[X_1 + \cdots + X_n] = \operatorname{E}[X_1] + \cdots + \operatorname{E}[X_n] = p + \cdots + p = np.</math>

: <math> \operatorname{Var}(X) = npq = np(1 - p).</math>

: <math>\begin{align}

\mu_2 &= np(1-p),\\

\mu_3 &= np(1-p)(1-2p),\\

\mu_4 &= np(1-p)(1+(3n-6)p(1-p)),\\

\mu_5 &= np(1-p)(1-2p)(1+(10n-12)p(1-p)),\\

\mu_6 &= np(1-p)(1-30p(1-p)(1-4p(1-p))+5np(1-p)(5-26p(1-p))+15n^2 p^2 (1-p)^2).

: <math>\begin{align}

and in general

<ref name="Andreas2008">

: <math>

where <math>\textstyle \left\{{c\atop k}\right\}</math> are the [[Stirling numbers of the second kind]], and <math>n^{\underline{k}} = n(n-1)\cdots(n-k+1)</math> is the <math>k</math>th [[Falling and rising factorials|falling power]] of <math>n</math>.

|page=109306 |arxiv=2103.17027 }}</ref> follows by bounding the Binomial moments via the [[Poisson distribution#Higher moments|higher Poisson moments]]:  

: <math>

\left(\frac{c}{\ln(c/(np)+1)}\right)^c \le (np)^c \exp\left(\frac{c^2}{2np}\right).

This shows that if <math>c=O(\sqrt{np})</math>, then <math>\operatorname {E}[X^c]</math> is at most a constant factor away from <math>\operatorname {E}[X]^c</math>

Usually the [[mode (statistics)|mode]] of a binomial {{math|''B''(''n'', ''p'')}} distribution is equal to <math>\lfloor (n+1)p\rfloor</math>, where <math>\lfloor\cdot\rfloor</math> is the [[floor function]]. However, when {{math|(''n'' + 1)''p''}} is an integer and {{math|''p''}} is neither 0 nor 1, then the distribution has two modes: {{math|(''n'' + 1)''p''}} and {{math|(''n'' + 1)''p'' − 1}}. When {{math|''p''}} is equal to 0 or 1, the mode will be 0 and {{math|''n''}} correspondingly. These cases can be summarized as follows:

: <math>\text{mode} =

: <math>f(k)=\binom nk p^k q^{n-k}.</math>

:<math>\frac{f(k+1)}{f(k)} = \frac{(n-k)p}{(k+1)(1-p)}</math>.

: <math>\begin{align}

* Any median {{math|''m''}} must lie within the interval <math>\lfloor np \rfloor\leq m \leq \lceil np \rceil</math>.<ref name="KaasBuhrman">{{cite journal|first1=R.|last1=Kaas|first2=J.M.|last2=Buhrman|title=Mean, Median and Mode in Binomial Distributions|journal=Statistica Neerlandica|year=1980|volume=34|issue=1|pages=13–18|doi=10.1111/j.1467-9574.1980.tb00681.x}}</ref>

* A median {{math|''m''}} cannot lie too far away from the mean:<math>|m-np|\leq \min\{{\ln2}, \max\{p,1-p\}\}</math> .<ref name="Hamza">

* The median is unique and equal to {{math|1=''m'' = [[Rounding|round]](''np'')}} when {{math|1={{abs|''m'' − ''np''}} ≤ min{{brace|''p'', 1 − ''p''}}}} (except for the case when {{math|1=''p'' = 1/2}} and {{math|''n''}} is odd).<ref name="KaasBuhrman"/>

* When {{math|''p''}} is a rational number (with the exception of {{math|1=''p'' = 1/2}}\ and {{math|''n''}} odd) the median is unique.<ref name="Nowakowski">

* When <math>p= \frac{1}{2} </math> and {{math|''n''}} is odd, any number {{math|''m''}} in the interval <math> \frac{1}{2} \bigl(n-1\bigr)\leq m \leq  \frac{1}{2} \bigl(n+1\bigr)</math>  is a median of the binomial distribution. If <math>p= \frac{1}{2}  </math> and {{math|''n''}} is even, then  <math>m= \frac{n}{2} </math> is the unique median.

For {{math|''k'' ≤ ''np''}}, upper bounds can be derived for the lower tail of the cumulative distribution function <math>F(k;n,p) = \Pr(X \le k)</math>, the probability that there are at most {{math|''k''}} successes. Since <math>\Pr(X \ge k) = F(n-k;n,1-p) </math>, these bounds can also be seen as bounds for the upper tail of the cumulative distribution function for {{math|''k'' ≥ ''np''}}.

: <math> F(k;n,p) \leq \exp\left(-2 n\left(p-\frac{k}{n}\right)^2\right), \!</math>

which is however not very tight. In particular, for {{math|1=''p'' = 1}}, we have that {{math|1=''F''(''k''; ''n'', ''p'') = 0}} (for fixed {{math|''k''}}, {{math|''n''}} with {{math|''k'' < ''n''}}), but Hoeffding's bound evaluates to a positive constant.

: <math> F(k;n,p) \leq \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right)  </math>

where {{math|''D''(''a'' ∥ ''p'')}} is the [[Kullback–Leibler divergence|relative entropy (or Kullback-Leibler divergence)]] between an {{math|''a''}}-coin and a {{math|''p''}}-coin (i.e. between the {{math|Bernoulli(''a'')}} and {{math|Bernoulli(''p'')}} distribution):

: <math> D(a\parallel p)=(a)\ln\frac{a}{p}+(1-a)\ln\frac{1-a}{1-p}. \!</math>

One can also obtain ''lower'' bounds on the tail {{math|''F''(''k''; ''n'', ''p'')}}, known as anti-concentration bounds. By approximating the binomial coefficient with [[Stirling's approximation|Stirling's formula]] it can be shown that<ref>{{cite book |author1=Robert B. Ash |title=Information Theory |url=https://archive.org/details/informationtheor00ashr |url-access=limited |date=1990 |publisher=Dover Publications |page=[https://archive.org/details/informationtheor00ashr/page/n81 115]|isbn=9780486665214 }}</ref>

: <math> F(k;n,p) \geq \frac{1}{\sqrt{8n\tfrac{k}{n}(1-\tfrac{k}{n})}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right),</math>

: <math> F(k;n,p) \geq \frac1{\sqrt{2n}} \exp\left(-nD\left(\frac{k}{n}\parallel p\right)\right).</math>

For {{math|1=''p'' = 1/2}} and {{math|''k'' ≥ 3''n''/8}} for even {{math|''n''}}, it is possible to make the denominator constant:<ref>{{cite web |last1=Matoušek |first1=J. |last2=Vondrak |first2=J. |title=The Probabilistic Method |work=lecture notes |url=https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf |archive-url=https://ghostarchive.org/archive/20221009/https://www.cs.cmu.edu/afs/cs.cmu.edu/academic/class/15859-f09/www/handouts/matousek-vondrak-prob-ln.pdf |archive-date=2022-10-09 |url-status=live }}</ref>

: <math> F(k;n,\tfrac{1}{2}) \geq \frac{1}{15} \exp\left(- 16n \left(\frac{1}{2} -\frac{k}{n}\right)^2\right). \!</math>

When {{math|''n''}} is known, the parameter {{math|''p''}} can be estimated using the proportion of successes:

: <math> \widehat{p} = \frac{x}{n}.</math>

This estimator is found using [[maximum likelihood estimator]] and also the [[method of moments (statistics)|method of moments]]. This estimator is [[Bias of an estimator|unbiased]] and uniformly with [[Minimum-variance unbiased estimator|minimum variance]], proven using [[Lehmann–Scheffé theorem]], since it is based on a [[minimal sufficient]] and [[Completeness (statistics)|complete]] statistic (i.e.: {{math|''x''}}). It is also [[Consistent estimator|consistent]] both in probability and in [[Mean squared error|MSE]]. This statistic is [[Asymptotic distribution|asymptotically]] [[normal distribution|normal]] thanks to the [[central limit theorem]], because it is the same as taking the [[arithmetic mean|mean]] over Bernoulli samples.  It has a variance of <math> var(\widehat{p}) = \frac{p(1-p)}{n}</math>, a property which is used in various ways, such as in [[Binomial_proportion_confidence_interval#Wald_interval|Wald's confidence intervals]].

A closed form [[Bayes estimator]] for {{math|''p''}} also exists when using the [[Beta distribution]] as a [[Conjugate prior|conjugate]] [[prior distribution]]. When using a general <math>\operatorname{Beta}(\alpha, \beta)</math> as a prior, the [[Bayes estimator#Posterior mean|posterior mean]] estimator is:

: <math> \widehat{p}_b = \frac{x+\alpha}{n+\alpha+\beta}.</math>

For the special case of using the [[standard uniform distribution]] as a [[non-informative prior]], <math>\operatorname{Beta}(\alpha=1, \beta=1) = U(0,1)</math>, the posterior mean estimator becomes:

:<math> \widehat{p}_b = \frac{x+1}{n+2}.</math>

When relying on [[Jeffreys prior]], the prior is <math>\operatorname{Beta}(\alpha=\frac{1}{2}, \beta=\frac{1}{2})</math>,<ref>Marko Lalovic (https://stats.stackexchange.com/users/105848/marko-lalovic), Jeffreys prior for binomial likelihood, URL (version: 2019-03-04): https://stats.stackexchange.com/q/275608</ref> which leads to the estimator:

: <math> \widehat{p}_{Jeffreys} = \frac{x+\frac{1}{2}}{n+1}.</math>

When estimating {{math|''p''}} with very rare events and a small {{math|''n''}} (e.g.: if {{math|1=''x'' = 0}}), then using the standard estimator leads to <math> \widehat{p} = 0,</math> which sometimes is unrealistic and undesirable. In such cases there are various alternative estimators.<ref>{{cite journal |last=Razzaghi |first=Mehdi |title=On the estimation of binomial success probability with zero occurrence in sample |journal=Journal of Modern Applied Statistical Methods |volume=1 |issue=2 |year=2002 |pages=326–332 |doi=10.22237/jmasm/1036110000 |doi-access=free }}</ref> One way is to use the Bayes estimator <math> \widehat{p}_b</math>, leading to:

: <math> \widehat{p}_b = \frac{1}{n+2}.</math>

: <math> \widehat{p}_{\text{rule of 3}} = \frac{3}{n}.</math>

Even for quite large values of ''n'', the actual distribution of the mean is significantly nonnormal.<ref name=Brown2001>{{Citation |first1=Lawrence D. |last1=Brown |first2=T. Tony |last2=Cai |first3=Anirban |last3=DasGupta |year=2001 |title = Interval Estimation for a Binomial Proportion |url=http://www-stat.wharton.upenn.edu/~tcai/paper/html/Binomial-StatSci.html |journal=Statistical Science |volume=16 |issue=2 |pages=101–133 |access-date = 2015-01-05 |doi=10.1214/ss/1009213286|citeseerx=10.1.1.323.7752 }}</ref> Because of this problem several methods to estimate confidence intervals have been proposed.

* ''n''₁ is the number of successes out of ''n'', the total number of trials

* <math>z</math> is the <math>1 - \tfrac{1}{2}\alpha</math> [[quantile]] of a [[standard normal distribution]] (i.e., [[probit]]) corresponding to the target error rate <math>\alpha</math>. For example, for a 95% [[confidence level]] the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so <math>1 - \tfrac{1}{2}\alpha</math>&nbsp;=&nbsp;0.975 and <math>z</math>&nbsp;=&nbsp;1.96.

: <math> \widehat{p\,} \pm z \sqrt{ \frac{ \widehat{p\,} ( 1 -\widehat{p\,} )}{ n } } .</math>

A [[continuity correction]] of {{math|0.5/''n''}} may be added.{{clarify|date=July 2012}}

: <math> \tilde{p} \pm z \sqrt{ \frac{ \tilde{p} ( 1 - \tilde{p} )}{ n + z^2 } }</math>

Here the estimate of {{math|''p''}} is modified to

: <math> \tilde{p}= \frac{ n_1 + \frac{1}{2} z^2}{ n + z^2 } </math>

This method works well for {{math|''n'' > 10}} and {{math|''n''₁ ≠ 0, ''n''}}.<ref>{{cite web|last1=Gulotta|first1=Joseph|title=Agresti-Coull Interval Method|url=https://pellucid.atlassian.net/wiki/spaces/PEL/pages/25722894/Agresti-Coull+Interval+Method#:~:text=The%20Agresti%2DCoull%20Interval%20Method,%2C%20or%20per%20100%2C000%2C%20etc|website=pellucid.atlassian.net|access-date=18 May 2021}}</ref> See here for <math>n\leq 10</math>.<ref>{{cite web|title=Confidence intervals|url=https://www.itl.nist.gov/div898/handbook/prc/section2/prc241.htm|website=itl.nist.gov|access-date=18 May 2021}}</ref> For {{math|1=''n''₁ = 0, ''n''}} use the Wilson (score) method below.

: <math>\sin^2 \left(\arcsin \left(\sqrt{\widehat{p\,}}\right) \pm \frac{z}{2\sqrt{n}} \right).</math>

The notation in the formula below differs from the previous formulas in two respects:<ref name="Wilson1927">{{Citation |last = Wilson |first=Edwin B. |date = June 1927 |title = Probable inference, the law of succession, and statistical inference |url = http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |journal = Journal of the American Statistical Association |volume=22 |issue=158 |pages=209–212 |access-date= 2015-01-05 |doi = 10.2307/2276774 |url-status=dead |archive-url = https://web.archive.org/web/20150113082307/http://psych.stanford.edu/~jlm/pdfs/Wison27SingleProportion.pdf |archive-date = 2015-01-13 |jstor = 2276774 }}</ref>

* Firstly, {{math|''z''_''x''}} has a slightly different interpretation in the formula below: it has its ordinary meaning of 'the {{math|''x''}}th quantile of the standard normal distribution', rather than being a shorthand for 'the {{math|(1 − ''x'')}}th quantile'.

* Secondly, this formula does not use a plus-minus to define the two bounds. Instead, one may use <math>z = z_{\alpha / 2}</math> to get the lower bound, or use <math>z = z_{1 - \alpha/2}</math> to get the upper bound. For example: for a 95% confidence level the error <math>\alpha</math>&nbsp;=&nbsp;0.05, so one gets the lower bound by using <math>z = z_{\alpha/2} = z_{0.025} = - 1.96</math>, and one gets the upper bound by using <math>z = z_{1 - \alpha/2} = z_{0.975} = 1.96</math>.

: <math>\frac{

     \widehat{p\,} + \frac{z^2}{2n} + z

         \frac{\widehat{p\,}(1 - \widehat{p\,})}{n} +

If {{math|''X'' ~ B(''n'', ''p'')}} and {{math|''Y'' ~ B(''m'', ''p'')}} are independent binomial variables with the same probability {{math|''p''}}, then {{math|''X'' + ''Y''}} is again a binomial variable; its distribution is {{math|1=''Z'' = ''X'' + ''Y'' ~ B(''n'' + ''m'', ''p'')}}:<ref>{{cite book |last1=Dekking |first1=F.M. |last2=Kraaikamp |first2=C. |last3=Lopohaa |first3=H.P. |last4=Meester |first4=L.E. |title=A Modern Introduction of Probability and Statistics |date=2005 |publisher=Springer-Verlag London |isbn=978-1-84628-168-6 |edition=1 |url=https://www.springer.com/gp/book/9781852338961}}</ref>

: <math>\begin{align}

A Binomial distributed random variable {{math|''X'' ~ B(''n'', ''p'')}} can be considered as the sum of {{math|''n''}} Bernoulli distributed random variables. So the sum of two Binomial distributed random variables {{math|''X'' ~ B(''n'', ''p'')}} and {{math|''Y'' ~ B(''m'', ''p'')}} is equivalent to the sum of {{math|''n'' + ''m''}} Bernoulli distributed random variables, which means {{math|1=''Z'' = ''X'' + ''Y'' ~ B(''n'' + ''m'', ''p'')}}. This can also be proven directly using the addition rule.

However, if {{math|''X''}} and {{math|''Y''}} do not have the same probability {{math|''p''}}, then the variance of the sum will be [[Binomial sum variance inequality|smaller than the variance of a binomial variable]] distributed as {{math|B(''n'' + ''m'', {{overline|''p''}})}}.

The binomial distribution is a special case of the [[Poisson binomial distribution]], which is the distribution of a sum of {{math|''n''}} independent non-identical [[Bernoulli trials]] {{math|B(''p''_''i'')}}.<ref>

This result was first derived by Katz and coauthors in 1978.<ref name=Katz1978>{{cite journal |last1=Katz |first1=D. |display-authors=1 |first2=J. |last2=Baptista |first3=S. P. |last3=Azen |first4=M. C. |last4=Pike |year=1978 |title=Obtaining confidence intervals for the risk ratio in cohort studies |journal=Biometrics |volume=34 |issue=3 |pages=469–474 |doi=10.2307/2530610 |jstor=2530610 }}</ref>

Let {{nowrap|''X'' ~ B(''n'', ''p''₁)}} and {{nowrap|''Y'' ~ B(''m'', ''p''₂)}} be independent. Let {{nowrap|1=''T'' = (''X''/''n'') / (''Y''/''m'')}}.

Then log(''T'') is approximately normally distributed with mean log(''p''₁/''p''₂) and variance {{nowrap|((1/''p''₁) − 1)/''n'' + ((1/''p''₂) − 1)/''m''}}.

If ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') and ''Y''&nbsp;|&nbsp;''X''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') (the conditional distribution of ''Y'', given&nbsp;''X''), then ''Y'' is a simple binomial random variable with distribution ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').

For example, imagine throwing ''n'' balls to a basket ''U_X'' and taking the balls that hit and throwing them to another basket ''U_Y''. If ''p'' is the probability to hit ''U_X'' then ''X''&nbsp;~&nbsp;B(''n'',&nbsp;''p'') is the number of balls that hit ''U_X''. If ''q'' is the probability to hit ''U_Y'' then the number of balls that hit ''U_Y'' is ''Y''&nbsp;~&nbsp;B(''X'',&nbsp;''q'') and therefore ''Y''&nbsp;~&nbsp;B(''n'',&nbsp;''pq'').

Since <math> X \sim B(n, p) </math> and <math> Y \sim B(X, q) </math>, by the [[law of total probability]],

: <math>\begin{align}

: <math> \Pr[Y = m] = \sum_{k=m}^{n} \binom{n}{m} \binom{n-m}{k-m} p^k q^m (1-p)^{n-k} (1-q)^{k-m} </math>

: <math>\begin{align}

: <math> \Pr[Y = m] = \binom{n}{m} (pq)^m \left( \sum_{i=0}^{n-m} \binom{n-m}{i} (p - pq)^i (1-p)^{n-m - i} \right) </math>

: <math>\begin{align}

and thus <math> Y \sim B(n, pq) </math> as desired.

The [[Bernoulli distribution]] is a special case of the binomial distribution, where {{math|1=''n'' = 1}}. Symbolically, {{math|''X'' ~ B(1, ''p'')}} has the same meaning as {{math|''X'' ~ Bernoulli(''p'')}}. Conversely, any binomial distribution, {{math|B(''n'', ''p'')}}, is the distribution of the sum of {{math|''n''}} independent [[Bernoulli trials]], {{math|Bernoulli(''p'')}}, each with the same probability {{math|''p''}}.<ref>{{cite web|last1=Taboga|first1=Marco|title=Lectures on Probability Theory and Mathematical Statistics|url=https://www.statlect.com/probability-distributions/binomial-distribution#hid3|website=statlect.com|access-date=18 December 2017}}</ref>

If {{math|''n''}} is large enough, then the skew of the distribution is not too great. In this case a reasonable approximation to {{math|B(''n'', ''p'')}} is given by the [[normal distribution]]

: <math> \mathcal{N}(np,\,np(1-p)),</math>

The basic approximation generally improves as {{math|''n''}} increases (at least 20) and is better when {{math|''p''}} is not near to 0 or 1.<ref name="bhh">{{cite book|title=Statistics for experimenters|url=https://archive.org/details/statisticsforexp00geor|url-access=registration|author=Box, Hunter and Hunter|publisher=Wiley|year=1978|page=[https://archive.org/details/statisticsforexp00geor/page/130 130]|isbn=9780471093152}}</ref> Various [[Rule of thumb|rules of thumb]] may be used to decide whether {{math|''n''}} is large enough, and {{math|''p''}} is far enough from the extremes of zero or one:

* One rule<ref name="bhh"/> is that for {{math|''n'' > 5}} the normal approximation is adequate if the absolute value of the skewness is strictly less than 0.3; that is, if

*: <math>\frac{|1-2p|}{\sqrt{np(1-p)}}=\frac1{\sqrt{n}}\left|\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}\,\right|<0.3.</math>

* A stronger rule states that the normal approximation is appropriate only if everything within 3 standard deviations of its mean is within the range of possible values; that is, only if

*: <math>\mu\pm3\sigma=np\pm3\sqrt{np(1-p)}\in(0,n).</math>

: This 3-standard-deviation rule is equivalent to the following conditions, which also imply the first rule above.

:: <math>n>9 \left(\frac{1-p}{p} \right)\quad\text{and}\quad n>9\left(\frac{p}{1-p}\right).</math>

: <math>np-3\sqrt{np(1-p)}>0\quad\text{and}\quad np+3\sqrt{np(1-p)}<n.</math>

: <math>np>3\sqrt{np(1-p)}\quad\text{and}\quad n(1-p)>3\sqrt{np(1-p)}.</math>

: <math>n>9 \left(\frac{1-p}p\right) \quad\text{and}\quad n>9 \left(\frac{p}{1-p}\right).</math>

: <math>\frac{\sqrt{n}}3>\sqrt{\frac{1-p}p}>0 \quad \text{and} \quad \frac{\sqrt{n}}3 > \sqrt{\frac{p}{1-p}}>0.</math>

: <math>\frac{\sqrt{n}}3>\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}>-\frac{\sqrt{n}}3;</math>

: <math>\left|\sqrt{\frac{1-p}p}-\sqrt{\frac{p}{1-p}}\,\right|<\frac{\sqrt{n}}3.</math>

* Another commonly used rule is that both values {{math|''np''}} and {{math|''n''(1 − ''p'')}} must be greater than<ref>{{Cite book |last=Chen |first=Zac |title=H2 Mathematics Handbook |publisher=Educational Publishing House |year=2011 |isbn=9789814288484 |edition=1 |location=Singapore |pages=350}}</ref><ref>{{Cite web |date=2023-05-29 |title=6.4: Normal Approximation to the Binomial Distribution - Statistics LibreTexts |url=https://stats.libretexts.org/Courses/Las_Positas_College/Math_40:_Statistics_and_Probability/06:_Continuous_Random_Variables_and_the_Normal_Distribution/6.04:_Normal_Approximation_to_the_Binomial_Distribution |access-date=2023-10-07 |archive-date=2023-05-29 |archive-url=https://web.archive.org/web/20230529211919/https://stats.libretexts.org/Courses/Las_Positas_College/Math_40:_Statistics_and_Probability/06:_Continuous_Random_Variables_and_the_Normal_Distribution/6.04:_Normal_Approximation_to_the_Binomial_Distribution |url-status=bot: unknown }}</ref> or equal to&nbsp;5. However, the specific number varies from source to source, and depends on how good an approximation one wants. In particular, if one uses&nbsp;9 instead of&nbsp;5, the rule implies the results stated in the previous paragraphs.

: <math>np\geq9>9(1-p)\quad\text{and}\quad n(1-p)\geq9>9p.</math>

: <math>n>9 \left(\frac{1-p}p\right) \quad\text{and}\quad n>9 \left(\frac{p}{1-p}\right).</math>

The following is an example of applying a [[continuity correction]]. Suppose one wishes to calculate {{math|Pr(''X'' ≤ 8)}} for a binomial random variable {{math|''X''}}. If {{math|''Y''}} has a distribution given by the normal approximation, then {{math|Pr(''X'' ≤ 8)}} is approximated by {{math|Pr(''Y'' ≤ 8.5)}}. The addition of 0.5 is the continuity correction; the uncorrected normal approximation gives considerably less accurate results.

This approximation, known as [[de Moivre–Laplace theorem]], is a huge time-saver when undertaking calculations by hand (exact calculations with large {{math|''n''}} are very onerous); historically, it was the first use of the normal distribution, introduced in [[Abraham de Moivre]]'s book ''[[The Doctrine of Chances]]'' in 1738. Nowadays, it can be seen as a consequence of the [[central limit theorem]] since {{math|B(''n'', ''p'')}} is a sum of {{math|''n''}} independent, identically distributed [[Bernoulli distribution|Bernoulli variables]] with parameter&nbsp;{{math|''p''}}. This fact is the basis of a [[hypothesis test]], a "proportion z-test", for the value of {{math|''p''}} using {{math|''x''/''n''}}, the sample proportion and estimator of {{math|''p''}}, in a [[common test statistics|common test statistic]].<ref>[[NIST]]/[[SEMATECH]], [http://www.itl.nist.gov/div898/handbook/prc/section2/prc24.htm "7.2.4. Does the proportion of defectives meet requirements?"] ''e-Handbook of Statistical Methods.''</ref>

For example, suppose one randomly samples {{math|''n''}} people out of a large population and ask them whether they agree with a certain statement. The proportion of people who agree will of course depend on the sample. If groups of ''n'' people were sampled repeatedly and truly randomly, the proportions would follow an approximate normal distribution with mean equal to the true proportion ''p'' of agreement in the population and with standard deviation

: <math>\sigma = \sqrt{\frac{p(1-p)}{n}}</math>

The binomial distribution converges towards the [[Poisson distribution]] as the number of trials goes to infinity while the product {{math|''np''}} converges to a finite limit. Therefore, the Poisson distribution with parameter {{math|1=''λ'' = ''np''}} can be used as an approximation to {{math|B(''n'', ''p'')}} of the binomial distribution if {{math|''n''}} is sufficiently large and {{math|''p''}} is sufficiently small.  According to rules of thumb, this approximation is good if {{math|''n'' ≥ 20}} and {{math|''p'' ≤ 0.05}}<ref>{{cite news |date=2023-03-28 |title=12.4 – Approximating the Binomial Distribution {{!}} STAT 414 |newspaper=Pennstate: Statistics Online Courses |url=https://online.stat.psu.edu/stat414/lesson/12/12.4 |access-date=2023-10-08 |archive-date=2023-03-28 |archive-url=https://web.archive.org/web/20230328081322/https://online.stat.psu.edu/stat414/lesson/12/12.4 |url-status=bot: unknown }}</ref> such that {{math|''np'' ≤ 1}}, or if {{math|''n'' > 50}} and {{math|''p'' < 0.1}} such that {{math|''np'' < 5}},<ref>{{Cite book |last=Chen |first=Zac |title=H2 mathematics handbook |publisher=Educational publishing house |year=2011 |isbn=9789814288484 |edition=1 |location=Singapore |pages=348}}</ref> or if {{math|''n'' ≥ 100}} and {{math|''np'' ≤ 10}}.<ref name="nist">[[NIST]]/[[SEMATECH]], [http://www.itl.nist.gov/div898/handbook/pmc/section3/pmc331.htm "6.3.3.1. Counts Control Charts"], ''e-Handbook of Statistical Methods.''</ref><ref>{{Cite web |date=2023-03-13 |title=The Connection Between the Poisson and Binomial Distributions |url=https://mathcenter.oxford.emory.edu/site/math117/connectingPoissonAndBinomial/ |access-date=2023-10-08 |archive-date=2023-03-13 |archive-url=https://web.archive.org/web/20230313085931/https://mathcenter.oxford.emory.edu/site/math117/connectingPoissonAndBinomial/ |url-status=bot: unknown }}</ref>

* ''[[Poisson limit theorem]]'': As {{math|''n''}} approaches {{math|∞}} and {{math|''p''}} approaches 0 with the product {{math|''np''}} held fixed, the {{math|Binomial(''n'', ''p'')}} distribution approaches the [[Poisson distribution]] with [[expected value]] {{math|1=''λ'' = ''np''}}.<ref name="nist"/>

* ''[[de Moivre–Laplace theorem]]'': As {{math|''n''}} approaches {{math|∞}} while {{math|''p''}} remains fixed, the distribution of

*: <math>\frac{X-np}{\sqrt{np(1-p)}}</math>

: approaches the [[normal distribution]] with expected value&nbsp;0 and [[variance]]&nbsp;1. This result is sometimes loosely stated by saying that the distribution of {{math|''X''}} is [[Asymptotic normality|asymptotically normal]] with expected value&nbsp;0 and [[variance]]&nbsp;1. This result is a specific case of the [[central limit theorem]].

Mathematically, when {{math|1=''α'' = ''k'' + 1}} and {{math|1=''β'' = ''n'' &minus; ''k'' + 1}}, the beta distribution and the binomial distribution are related by{{clarification needed|date=March 2023|

: <math>\operatorname{Beta}(p;\alpha;\beta) = (n+1)B(k;n;p)</math>

[[Beta distribution]]s also provide a family of [[prior distribution|prior probability distribution]]s for binomial distributions in [[Bayesian inference]]:<ref name=MacKay>{{cite book| last=MacKay| first=David| title = Information Theory, Inference and Learning Algorithms|year=2003| publisher=Cambridge University Press; First Edition |isbn=978-0521642989}}</ref>  

: <math>P(p;\alpha,\beta) = \frac{p^{\alpha-1}(1-p)^{\beta-1}}{\operatorname{Beta}(\alpha,\beta)}.</math>

Methods for [[random number generation]] where the [[marginal distribution]] is a binomial distribution are well-established.<ref>Devroye, Luc (1986) ''Non-Uniform Random Variate Generation'', New York: Springer-Verlag. (See especially [http://luc.devroye.org/chapter_ten.pdf Chapter X, Discrete Univariate Distributions])</ref><ref>

This distribution was derived by [[Jacob Bernoulli]]. He considered the case where {{math|1=''p'' = ''r''/(''r'' + ''s'')}} where {{math|''p''}} is the probability of success and {{math|''r''}} and {{math|''s''}} are positive integers. [[Blaise Pascal]] had earlier considered the case where {{math|1=''p'' = 1/2}}, tabulating the corresponding binomial coefficients in what is now recognized as [[Pascal's triangle]].<ref name=":1">{{cite book|last=Katz|first=Victor|title=A History of Mathematics: An Introduction|publisher=Addison-Wesley|year=2009|isbn=978-0-321-38700-4|pages=491|chapter=14.3: Elementary Probability}}</ref>

* Binomial measure, an example of a [[Multifractal system|multifractal]] [[measure (mathematics)|measure]].<ref>Mandelbrot, B. B., Fisher, A. J., & Calvet, L. E. (1997). A multifractal model of asset returns. ''3.2 The Binomial Measure is the Simplest Example of a Multifractal''</ref>