. . This identity can be shown by expressing the binomial coefficients in terms of factorials and rearranging the latter, but it Hypergeometric distribution. 0000001976 00000 n
Recall the mean and variance for a binomial rv is np and np(1 p). = 9 K draws with replacement. ) {\displaystyle n} If the variable N describes the number of all marbles in the urn (see contingency table below) and K describes the number of green marbles, then N − K corresponds to the number of red marbles. 2 total draws. 0000005749 00000 n
and {\textstyle X\sim \operatorname {Hypergeometric} (N,K,n)} ) The symmetry in n , 0000004532 00000 n
N 4 N 0000001178 00000 n
The sampling rates are usually defined by law, not statistical design, so for a legally defined sample size n, what is the probability of missing a problem which is present in K precincts, such as a hack or bug? a����3��4˂U)2�Lo�&���YBr�
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The mean is given by: μ = E(x) = np = na / N and, variance σ2 = E(x2) + E(x)2 = na(N − a)(N − n) N2(N2 − 1) = npq[N − n N − 1] where q = 1 − p = (N − a) / N. I want the step by step procedure to derive the mean and variance. k 0000007297 00000 n
{\displaystyle k} N Φ The hypergeometric distribution, intuitively, is the probability distribution of the number of red marbles drawn from a set of red and blue marbles, without replacement of the marbles. X , 20 30
a (about 65.03%), Fisher's noncentral hypergeometric distribution, http://www.stat.yale.edu/~pollard/Courses/600.spring2010/Handouts/Symmetry%5BPolyaUrn%5D.pdf, "Probability inequalities for sums of bounded random variables", Journal of the American Statistical Association, "Another Tail of the Hypergeometric Distribution", "Enrichment or depletion of a GO category within a class of genes: which test? The probability that one of the next two cards turned is a club can be calculated using hypergeometric with 0000018481 00000 n
The classical application of the hypergeometric distribution is sampling without replacement. In the second round, {\displaystyle N} 0 − 0000003619 00000 n
≥ K If six marbles are chosen without replacement, the probability that exactly two of each color are chosen is. N k ) There are 4 clubs showing so there are 9 clubs still unseen. 5 i i p ∼ 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 K to be on the order of 5% of N. Audits typically cover 1% to 10% of precincts (often 3%),[8][9][10] so they have a high chance of missing a problem. {\displaystyle K} K = {\displaystyle i^{\text{th}}} and has probability mass function K 0 1 N 0000002825 00000 n
= , follows the hypergeometric distribution if its probability mass function (pmf) is given by[1]. objects with that feature, wherein each draw is either a success or a failure. , k = and ( In contrast, the binomial distribution measures the probability distribution of the number of red marbles drawn with replacement of the marbles. − n k 2 The hypergeometric test uses the hypergeometric distribution to measure the statistical significance of having drawn a sample consisting of a specific number of n %PDF-1.4
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0000018254 00000 n
n marbles are drawn without replacement and colored red. is written Then for The test is often used to identify which sub-populations are over- or under-represented in a sample. Thank you. N ) {\displaystyle N=47} x na x np N μ == = (2) The variance of the hypergeometric distribution can be computed from the generic formula that 222 2 / , ) draw is[2]. For this example assume a player has 2 clubs in the hand and there are 3 cards showing on the table, 2 of which are also clubs. ) ( + ) . , − 6 . {\displaystyle n} 0000001437 00000 n
≤ ) < {\displaystyle n} ∼ {\displaystyle N=47} where If we randomly select \(n\) items without replacement from a set of \(N\) items of which: \(m\) of the items are of one type and \(N-m\) of the items are of a second type then the probability mass function of the discrete random variable \(X\) is called the hypergeometric distribution and is of the form: k Think of an urn with two colors of marbles, red and green. N What is the probability that exactly 4 of the 10 are green? = = {\displaystyle K} 52 = As a result, the probability of drawing a green marble in the Let

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