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By Klotz J.H.

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Then if we denote the number of outcomes in the set A by ||A|| we have P (A) = i: ei ∈A 1 1 = N N 1= i: ei ∈A ||A|| . ||S|| For such probabilities, it is necessary to count the number of outcomes ||A|| and ||S|| = N. We give some standard counting rules. 1 Product rule. If one operation can be done in n1 ways and the second operation can be done in n2 ways independently of the first operation then the pair of operations can be done in n1 × n2 ways. If we have K operations, each of which can be done independently in nk ways for k = 1, 2, .

N i=1 n (Xi − C)2 = i=1 ¯ +X ¯ − C)2 (Xi − X n n = i=1 ¯ 2 + 2(X ¯ − C) (Xi − X) n i=1 (Xi and using i=1 ¯ + n(X ¯ − C)2 (Xi − X) ¯ = 0, − X) n n = i=1 ¯ 2 + n(X ¯ − C)2 ≥ (Xi − X) i=1 ¯ 2. (Xi − X) ¯ minimizes. 3 If C ≥ 1 then the proportion of observations outside the ˜ − CD, X ˜ + CD) does not exceed 1/C. interval (X Proof. ˜ − CD or Xi ≥ X ˜ + CD}. Then the proportion outside Let A = {i : Xi ≤ X the interval is i∈A 1 n = 1 n 1≤ i: 1 n ˜ |Xi −X|/(CD)≥1 n i=1 ˜ |Xi − X|/(CD) = 1 . 4 (Chebyshev’s Proposition for sample data).

Using S = A ∪ Ac , where A ∩ Ac = φ, and finite additivity 1 = P (S) = P (A ∪ Ac ) = P (A) + P (Ac ) . Subtracting P (A) from both sides gives the result. 4 If A ⊂ B, then A = A ∩ B and P (A) ≤ P (B). 3. PROBABILITY Proof. e ∈ A =⇒ e ∈ A and e ∈ B =⇒ e ∈ A ∩ B . Converseley e ∈ A ∩ B =⇒ e ∈ A and e ∈ B =⇒ e ∈ A so that A = A ∩ B. Next using the distributive law for events P (B) = P (B ∩ S) = P (B ∩ (A ∪ Ac )) = P ((B ∩ A) ∪ (B ∩ Ac )) and using A = A ∩ B and finite additivity for disjoint A and B ∩ Ac = P (A ∪ (B ∩ Ac )) = P (A) + P (B ∩ Ac ) ≥ P (A) .

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