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17 Cards in this Set
- Front
- Back
Proof: If A⊂B and B⊂C, imply A⊂C |
Suppose x ∈ A. Since A⊂B, x ∈ B. Since B⊂C, x ∈ C ∴ A⊂C |
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DeMorgan's Law 1: (A∩B)' = ? |
A'UB'
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DeMorgan's Law 2: (AUB)' = ? |
A'∩B' |
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Multiplicative Law: The probability of the INTERSECTION of two events A and B. P(A∩B) = ? |
P(A) * P(B|A) and P(B) * P(A|B) |
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Multiplicative Law for 3 Events P(A∩B∩C) = ? |
P[(A∩B) U C] = P(A∩B) * P(C|A∩B) = P(A) * P(B|A) * P(C|A∩B) |
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Additive Law : The probability of the UNION of two events A and B. P(AUB) = ? |
P(A) + P(B) - P(A∩B) |
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If P(A)>0 and P(B)>0, then mutually exclusive implies... |
dependence. |
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If P(A∩B)=0, then ... |
... A and B are mutually exclusive. |
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P(A|B) ⇔ (formula) |
⇔ P(A∩B) / P(B) |
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P(B|A) ⇔ (formula) |
⇔ P(B∩A) / P(A) |
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Mutually Exclusive ⇒ |
Dependence (But dependence ≠ M.E.) |
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Independence ⇒ |
Not Mutually Exclusive (But Not M.E. ≠ Independence) |
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Law of Total Probability P(A) = formula |
P(A) = ∑ P(A|Bi) * P(Bi) for i=1 to k. (Conditions: Let B1, B2, ... , Bk be a partition of the sample space Si. B1∪B2∪...∪Bk = Si and Bi∩Bj = ∅ for i≠j) |
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Baye's Rule P(Bj|A) = formula |
P(Bj|A) = [ P(Bj)*P(A|Bj)] / [∑(P(Bi)*P(A|Bi)] where Bj is constant and i=1 to n |
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P(A|AUB) = (equivalent formula) |
P(A) / P(AUB) |
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P(A|A∩B) = (equivalent formula) |
P(A∩B) / P(A∩B) = 1 This is because we already know that event A has occurred. |
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P(A∩B|AUB) = (equivalent formula) |
P(A∩B) / P(AUB) |