Editing Hypergeometric distribution (section)

===Application to auditing elections===
[[File:Election Samples.png|thumb|Samples used for election audits and resulting chance of missing a problem]]

[[Election audits]] typically test a sample of machine-counted precincts to see if recounts by hand or machine match the original counts. Mismatches result in either a report or a larger recount. The sampling rates are usually defined by law, not statistical design, so for a legally defined sample size {{mvar|n}}, what is the probability of missing a problem which is present in {{mvar|K}} precincts, such as a hack or bug? This is the probability that {{math|''k'' {{=}} 0 .}} Bugs are often obscure, and a hacker can minimize detection by affecting only a few precincts, which will still affect close elections, so a plausible scenario is for {{mvar|K}} to be on the order of 5% of {{mvar|N}}. Audits typically cover 1% to 10% of precincts (often 3%),<ref name=newyorkaudit>{{cite SSRN |first1=Amanda |last1=Glazer |first2=Jacob |last2=Spertus |date=2020-02-10 |df=dmy-all |title=Start spreading the news: New York's post-election audit has major flaws |ssrn=3536011}}</ref><ref name=vvstates>{{cite web |title=State audit laws |date=2017-02-10 |df=dmy-all |website=Verified Voting |lang=en-US |url=https://www.verifiedvoting.org/state-audit-laws/ |access-date=2018-04-02 |archive-date=2020-01-04 |archive-url=https://web.archive.org/web/20200104201852/https://www.verifiedvoting.org/state-audit-laws/ }}</ref><ref name=ncsl>{{cite web |title=Post-election audits |publisher=National Conference of State Legislatures |website=ncsl.org |lang=en-US |url=http://www.ncsl.org/research/elections-and-campaigns/post-election-audits635926066.aspx#state |access-date=2018-04-02 |df=dmy-all}}</ref> so they have a high chance of missing a problem. For example, if a problem is present in 5 of 100 precincts, a 3% sample has 86% probability that {{nobr| {{math| ''k'' {{=}} 0 }} }} so the problem would not be noticed, and only 14% probability of the problem appearing in the sample (positive {{mvar| k }}):
: <math>
\begin{align}
\operatorname{\boldsymbol\mathcal P}\{\ X = 0\ \} & = \frac{\ \left[\ \binom{\text{Hack}}{0} \binom{ N\ -\ \text{Hack}}{ n\ -\ 0 }\ \right]\ }{\left[\ \binom{N}{n}\ \right]} =  \frac{\ \left[\ \binom{N\ -\ \text{Hack}}{n}\ \right]}{\ \left[\ \binom{N}{n}\ \right]\ } = \frac{\ \left[\ \frac{\ (N\ -\ \text{Hack})!\ }{n!(N\ -\ \text{Hack}-n)!}\ \right]\ }{\left[\ \frac{N!}{n!(N\ -\ n)!}\ \right]} = \frac{\ \left[\ \frac{(N-\text{Hack})!}{(N\ -\ \text{Hack}\ -\ n)!}\ \right]\ }{\left[\ \frac{N!}{(N\ -\ n)!}\ \right]} \\[8pt]
& =  \frac{\ \left[\ \binom{100-5}{3}\ \right]\ }{\ \left[\ \binom{100}{3}\ \right]\ } = \frac{\ \left[\ \frac{(100-5)!}{(100-5-3)!}\ \right]\ }{\left[\ \frac{100!}{(100-3)!}\ \right]} = \frac{\ \left[\ \frac{95!}{92!}\ \right]\ }{\ \left[\ \frac{100!}{97!}\ \right]\ } = \frac{\ 95\times94\times93\ }{100\times99\times98} = 86\%
\end{align}
</math>

The sample would need 45 precincts in order to have probability under 5% that ''k''&nbsp;=&nbsp;0 in the sample, and thus have probability over 95% of finding the problem:
: <math>\operatorname{\boldsymbol\mathcal P}\{\ X = 0\ \} =  \frac{\ \left[\ \binom{100-5}{45}\ \right]\ }{\left[\  \binom{100}{45}\ \right]} = \frac{\ \left[\ \frac{95!}{50!}\ \right]\ }{\left[\ \frac{100!}{55!}\ \right]} = \frac{\ 95\times 94\times \cdots \times 51\ }{\ 100\times 99\times \cdots \times 56\ } = \frac{\ 55\times 54\times 53\times 52\times 51\ }{\ 100\times 99\times 98\times 97\times 96\ } = 4.6\% ~.</math>