By Jfry KSmit

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Given that A1 = p1 for all i, and show that n n−1 1 = np1 − p2 2 Deduce that p1 ≥ 1/n and p2 ≤ 2/n. Ai ∩ Aj = p2 for all i j i = j , 30 Discrete outcomes Solution As Ai1 ∩ · · · ∩ Aik = 0 for k > 2, the exclusion–inclusion formula gives 1 = np1 − n n − 1 p2 /2. Rearranging: p1 = 1 n−1 1 + p2 ≥ n 2 n as p2 ≥ 0. 27 State the exclusion–inclusion formula for n i=1 Ai . A large and cheerful crowd of n junior wizards leave their staff in the Porter’s Lodge on the way to a long night in the Mended Drum.

Finally n all turn into bullfrogs = r=0 for n large enough. n Pr n pr ≈ r=0 e−1 pr = e−1 ep = ep−1 r! 3 The exclusion–inclusion formula 31 n pr → ep−1 one needs an assertion guaranteeing that the term-wise convergence (in our case Pr n pr → e−1 pr /r! ∀ r) implies the convergence of the sum of the series. For example, the following theorem will do: Remark To formally prove the convergence If an m → an ∀ n as m → S m = n an m → S = n r=0 Pr and an m ≤ bn , where n bn < , then the sum n an . In fact, Pr n pr = P0 n − r pr /r!

In view of independence, Bnc ∩ An+1 = Bnc An+1 Bn ∩ Acn+1 = and Bn Acn+1 which implies Bn+1 = with An+1 + Bnc Acn+1 = 1 − Bn Bn p + Bn 1 − p B0 = 1. That is, n+1 = 1−p n +p 1− n = 1 − 2p n +p Substituting n = a 1 − 2p n + b gives that b = 1/2, and the condition 0 = 1 that a = 1/2. Then n = 1 − 2p n + 1 /2. A shorter way to derive the recursion is by conditioning on Bn and Bnc : n+1 = Bn+1 = Bn+1 ∩ Bn + = Acn+1 Bn Bn + = 1−p n +p 1− Bn+1 ∩ Bnc An+1 Bnc Bnc n Writing recursive equations like the one in the statement of the current problem is a convenient instrument used in a great many situations.

### Applied Statistics and the SAS Programming Language by Jfry KSmit

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