Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITYwhere in the last line p = M/N and q =1− p. This is called the hypergeometricdistribution.By performing the relevant summations directly, it may be shown that thehypergeometric distribution has meanE[X] =n M N = npand varianceV [X] =nM(N − M)(N − n)N 2 (N − 1)= N − nN − 1 npq.◮In the UK National Lottery each participant chooses six different numbers between 1and 49. In each weekly draw six numbered winning balls are subsequently drawn. Find theprobabilities that a participant chooses 0, 1, 2, 3, 4, 5, 6 winning numbers correctly.The probabilities are given by a hypergeometric distribution with N (the total number ofballs) = 49, M (the number of winning balls drawn) = 6, and n (the number of numberschosen by each participant) = 6. Thus, substituting in (30.97), we find6 C 43 0 C 6Pr(0) = = 16 49C 6 2.29 , Pr(1) = C 43 1 C 5= 149C 6 2.42 ,6 C 43 2 C 4Pr(2) = = 16 49C 6 7.55 , Pr(3) = C 43 3 C 3= 149C 6 56.6 ,6 C 43 4 C 2Pr(4) = = 16 49C 6 1032 , Pr(5) = C 43 5 C 1= 149C 6 54 200 ,It can easily be seen thatPr(6) =6 C 43 6 C 0 1=49C 6 13.98 × 10 . 6as expected. ◭6∑Pr(i) =0.44 + 0.41 + 0.13 + 0.02 + O(10 −3 )=1,i=0Note that if the number of trials (balls drawn) is small compared with N, Mand N − M then not replacing the balls is of little consequence, and we mayapproximate the hypergeometric distribution by the binomial distribution (withp = M/N); this is much easier to evaluate.30.8.4 The Poisson distributionWe have seen that the binomial distribution describes the number of successfuloutcomes in a certain number of trials n. The Poisson distribution also describesthe probability of obtaining a given number of successes but for situationsin which the number of ‘trials’ cannot be enumerated; rather it describes thesituation in which discrete events occur in a continuum. Typical examples of1174