BTRY 408 FINAL NOTES - WAYS TO CHOOSE SAMPLES For a group...

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WAYS TO CHOOSE SAMPLES: For a group of size n, there are: n! ways to arrange them all without replacement n r ways to choose r of them with replacement For indistinguishable subgroups n 1 , n 2 , n 3 …. n r , there are n!/(n 1 !n 2 !n 3 !...n r !) ways to arrange them r n = n choose r = n!/((n-r)!r!) = - + - - r n r n 1 1 1 Num of unordered samples of size r when drawing with replacement from set of size n is - + r r n 1 Num of ways to place m objects between n other objects with no more than one of the m objects between every two of the n objects is + m n 1 Num of ways to distribute n indistinguishable items into r distinguishable categories is - - + 1 1 r r n INDEPENDENCE: Two events are independent if P(AB) = P(A) * P(B) For more than two events, every subset of the set of events must be independent as per test above. P(A|B) = P(AB)/P(B) P(A BEFORE B): Ignore all non A and B events, to get: P(A|A U B) / (P(A|A U B) + P(B|A U B)) P(A U B) = P(A) + P(B) – P(AB) E(X) = xp(x) E(g(X)) = sum over i of (g(x i )p(x i )) Var(X) = E(X 2 ) – E(X) 2 Var(X) = E((X-mu) 2 ) E(aX+b) = aE(X) + b Var(aX+b) = a 2 Var(X) Geometric Series : r a ar ar ar ar a i i - = = + + + + = 1 ... 0 3 2 for -1 < r < 1 Exponential Series : λ λ λ λ λ e i i i = = + + + + = 0 3 2 ! ... ! 3 ! 2 1 log(1+x) = ... 5 4 3 2 5 4 3 2 - + - + - x x x x x -log(1-a) = ... 5 4 3 2 5 4 3 2 1 + + + + + = = a a a a a x a x x Binomial Series : k i a x k a x = = + 0 ) 1 ( , -1<x<1 Binomial Theorem: k n k n k n y x k n y x - = = + 0 ) (
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BINOMIAL DISTRIBUTION : i n i p p i n i p - - = ) 1 ( ) ( np X E = ) ( , ) 1 ( ) ( p np X Var - = Sum of two Binomial variables is binomial with parameters (n 1 +n 2 , p) BERNOULLI DISTRIBUTION: equals binomial with parameters (1,p) POISSON DISTRIBUTION : !
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