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4.2 The Poisson Distribution 223 Proof We begin by rewriting the binomial probability in terms of Î»: lim nââ n k p (1 â p) k nâk n Î» k Î» nâk 1â = lim nââ k n n 1 n! Î» âk Î» n k Î» = lim 1â 1â nââ k!(n â k)! nk n n n k 1 n! Î» Î» = 1â lim k! nââ (n â k)! (n â Î»)k n But since [1 â (Î»/n)]n â eâÎ» as n â â, we need show only that n! â1 (n â k)!(n â Î»)k to prove the theorem. However, note that n! n(n â 1) Â· Â· Â· (n â k + 1) = (n â k)!(n â Î»)k (n â Î»)(n â Î») Â· Â· Â· (n â Î») a quantity that, indeed, tends to 1 as n â â (since Î» remains constant). Example 4.2.1 Theorem 4.2.1 is an asymptotic result. Left unanswered is the question of the relevance of the Poisson limit for ï¬nite n and p. That is, how large does n have to be and how small does p have to be before eânp (np)k /k! becomes a good approximation to the binomial probability, p X (k)? Since âgood approximationâ is undeï¬ned, there is no way to answer that question in any completely speciï¬c way. Tables 4.2.1 and 4.2.2, though, offer a partial solution by comparing the closeness of the approximation for two particular sets of values for n and p. In both cases Î» = np is equal to 1, but in the former, n is set equal to 5âin the latter, to 100. We see in Table 4.2.1 (n = 5) that for some k the agreement between the binomial probability and Poissonâs limit is not very good. If n is as large as 100, though (Table 4.2.2), the agreement is remarkably good for all k. Table 4.2.1 Binomial Probabilities and Poisson k 0 1 2 3 4 5 6+ Limits; n = 5 and p = 15 (Î» = 1) 5 eâ1 (1)k (0.2)k (0.8)5âk k k! 0.328 0.410 0.205 0.051 0.006 0.000 0 1.000 0.368 0.368 0.184 0.061 0.015 0.003 0.001 1.000