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probsum-1

# probsum-1 - summary of probability rules basic terms to...

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Unformatted text preview: summary of probability rules basic terms to know: outcome event = a set of outcomes sample space = the set of all possible outcomes for k equally likely outcomes, assign probability l/k to each for an event A consisting of m of these outcomes, assign probability. P(A) = m/k conditional probability P(AIB) restricts the sample space to be the set B, a subset of the original sample space thus, we are no longer considering all the outcomes in event A but only those outcomes in A that are also in B, or A n B for equally likely outcomes, assign P(AIB) = number of outcomes in A n B / number of outcomes inB O s P(A) s 1 P(A) = 1 if event A is logically certain P(A) = 0 if event A is logically impossible events related to A and B: not A Ac, the complement of the set A A or B ‘ I equivalent to A u B, the union of sets A and B A or B I unless speciﬁcally stated, this is A or B or both A and B both A and B equivalent to A m B, the intersection of sets A and B saying A but B, for contrast, is the same as A and B P(A U B) = P(A) + P(B) — P(A n B) called the Addition Rule important special case of the addition rule: the sample space = A u Ac and these are non-overlapping, ll so 1 = P(A u AC ) P(A) + P(A“) from which we can show that P(Ac ) = 1 — P(A) probability of A complement P(AlB) = P(A n B) / P(B) general formula for conditional probability if P(B) : 0, then we deﬁne P( A | B ) = O P(A m B) = P(A) P(BlA) ' called the Multiplication Rule also = P(B) P(AIB) from which we derive P(B)P(A|B) = P(A)P(B|A) and so P(AIB) = P(A)P(B|A) / P(B) and P(B|A) = P(B)P(A|B) / P(A) these formulas, or alternate versions of them, are called Bayes’Rule if P( A I B ) and P(A | Bc ) turn out to be equal to P(A), events A and B are said to be independent events and the Multiplication Rule simpliﬁes to P(A n B) = P(A) P(B) P(A) = P(A n B) + P(A 0 BC) 3 called Law of Total Probability more generally, if the sample space = Bl u B2 u u B j with the Bi non-overlapping thenP(A) = Z 1,-=I,P(A o B.—) for any sets A and C, C = (A n C) u (Ac 0 C ) so we can show P(A | C) + P(AC | C) = 1 and P(Ac | C) = 1 — P(A I C) there is no rule relating P(A | Bc ) and P(A | B) ...
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probsum-1 - summary of probability rules basic terms to...

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