Mathematical Methods for Physics and Engineering - Matematica.NET

30.8 IMPORTANT DISCRETE DISTRIBUTIONS2, is r+x−1 C x . Therefore, the total probability of obtaining x failures before therth success isf(x) =Pr(X = x) = r+x−1 C x p r q x ,which is called the negative binomial distribution (see the related discussion onp. 1137). It is straightforward to show that the MGF of this distribution is( ) p rM(t) =1 − qe t ,and that its mean and variance are given byE[X] = rq and V [X] = rqpp 2 .30.8.3 The hypergeometric distributionIn subsection 30.8.1 we saw that the probability of obtaining x successes in nindependent trials was given by the binomial distribution. Suppose that these n‘trials’ actually consist of drawing at random n balls, from a set of N such ballsof which M are red and the rest white. Let us consider the random variableX = number of red balls drawn.On the one hand, if the balls are drawn with replacement then the trials areindependent and the probability of drawing a red ball is p = M/N each time.Therefore, the probability of drawing x red balls in n trials is given by thebinomial distribution asn!Pr(X = x) =x!(n − x)! px (1 − p) n−x .On the other hand, if the balls are drawn without replacement the trials are notindependent and the probability of drawing a red ball depends on how many redballs have already been drawn. We can, however, still derive a general formulafor the probability of drawing x red balls in n trials, as follows.The number of ways of drawing x red balls from M is M C x , and the numberof ways of drawing n − x white balls from N − M is N−M C n−x . Therefore, thetotal number of ways to obtain x red balls in n trials is M C N−M x C n−x . However,the total number of ways of drawing n objects from N is simply N C n . Hence theprobability of obtaining x red balls in n trials isM C N−M x C n−xPr(X = x) =NC nM!(N − M)!=x!(M − x)! (n − x)!(N − M − n + x)!=n!(N − n)!, (30.97)N!(Np)!(Nq)! n!(N − n)!x!(Np − x)!(n − x)!(Nq − n + x)! N! , (30.98)1173