To be sure, the right hand side of Eq. (3.24) is a kind of mathematical trick and most readers will not have seen in advance that this is the way to proceed. That is fine, part of learning how to use the tools is to apprentice with a skilled craft person and watch what he or she does and thus learn how to do it oneself. Note that some of the terms on the right hand side of Eq. (3.24) comprise the probability that K = k — 1. When we combine those terms and examine what remains, we see that

Equation (3.25) is an iterative relationship between the probability that K = k — 1 and the probability that K = k. From Eq. (3.23), we know explicitly the probability that K = 0. Starting with this probability, we can compute all of the other probabilities using Eq. (3.25). We will use this method in the numerical examples discussed below.

Although Eq. (3.24) seems to be based on a bit of a trick, here's an insight that is not: when we examine the outcome of Ntrials, something must happen. That is 0 Pr{K = k} = 1. We can use this observation to find the mean and variance of the random variable K. The expected value of K is

EfK} = £ kPrfK = k} = £ k i / (1 — p)N—k k=0 k=0 \ k /

There is nothing tricky about what we have done thus far, but another trick now comes into play. We know how to evaluate the binomial sum from k = 0, but not from k = 1. So, we will manipulate terms accordingly by first writing the binomial coefficient explicitly and then factoring out Np from the expression on the right hand side of Eq. (3.26)

= NpY (N — 1)! pk—1(1 — )N—k (k — 1)!(N — k)!P (1 P)

and we now set j = k — 1. When k = 1, j = 0 and when k=N, j = N — 1. The last expression in Eq. (3.27) becomes a recognizable summation:

0 0

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