De Morgan’s Law, Kolmogorov’s Axioms, Basic Probability Theorems PDF

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DeMorgan’s Law, Kolmogorov’s Axioms, Basic Probability Theorems DeMorgan's Laws a) (A ⋃ B)c= Ac ⋃ Bc b) (A ⋂ B)c= Ac ⋂ Bc Kolmogorov’s Axioms Let there be a random experiment with sample space, S. We say that P( •) is a real valued set function, if P assigns a real number P(A) to event set A. We call such a set function a probability measure, P( •) if P( •) satisfies the following axioms 1) For event E in S, P(E) ≥ 0 2) P(S)=1 3) If E1,E2, …. Is any sequence of events such that Ei ⋂ Ej = φ for i =/ j, then P(E1 ⋃ E2 ⋃ …) = P(E1) + P(E2) + ……. Jargon: If for two events E and F we have E ⋂ F = φ then we say that E and F are disjoint or mutually exclusive. Basic Theorems Theorem 1: P(Ac )= 1-P(A) Note: Sometimes it is easier to find the probability of an event A by first finding the P(Ac ) and then subtracting it from 1 We know A ⋃ Ac = S Next, A ⋂ Ac = φ so, P(A ⋃ Ac) = P(S)=1 (axiom 2) On the other hand, P(A ⋃ Ac ) = P(A) +P(Ac ) (axiom 3) Therefore, P(Ac )= 1-P(A) or P(A) = 1-P(Ac ) Theorem 2: P( φ )= 0 Theorem 3: Let A and B be two events, then P(A ⋃ Bc )= P(A)-P(A ⋂ B) Theorem 4: If ACB (A is a subset of B), then P(A) ≤ P(B) Theorem 5: Let A and B be any two events then P(A ⋃ B)= P(A)+P(B)-P(A ⋂ B) Note: If events are disjoint then P(A ⋂ B)= 0, then P(A ⋃ B)= P(A)+P(B). It follows from above that if E is any event, then 0 ≤ p(E) ≤ 1 .

Advice on doing word Problems 1. DO NOT USE tree diagrams 2. Begin by defining events in as simple form as possible. Then construct your more complicated events as unions, intersections, complements, etc of your simple events. 3. TRanslate the given information into probability statements about your simple events. 4. Use the 5 theorems (plus the axioms) to find the probabilities of your more complicated events. Example: Suppose that it is known that the probability that someone has Multiple Sclerosis (MS) is 0.001. Suppose that it is known that 10% of all people have had mononucleosis. Further, suppose that the proportion of people who have either had mono or MS is 0.1005. i) What proportion of people have MS and have had mono? ii) What is the probability that someone has had neither mono nor has MS? iii) What proportion have had mono but do not have MS? *** Proportion and probability are interchangeable and mean that same thing. Solution Les MS= event that the person has MS m= event that the person has had mono Then M S ⋃ M = event a person has either had mono or MS We know: P(MS)= 0.001, P(M)=0.10, and P( M S ⋃ M )= 0.1005 i) *** “and” in question = intersection We want P( M S ⋂ M ) . Use Theorem 5. [ P (M S ⋃ M ) = P (M S) + P (M ) − P (M S ⋂ M ) ] P (M S ⋂ M ) = P (M S) + P (M ) − P (M S ⋃ M ) P (M S ⋂ M ) = 0.001 + 0.1 − 0.1005 P (M S ⋂ M ) = 0.005

ii) We want P (M S c ⋂ M c ) = P (M S ⋂ M )c = 1- P (M S ⋃ M ) = 1-0.1005 = 0.8995

(DeMorgan’s Law) (Theorem 1)

iii) P (M S c ⋂ M ) (Use Theorem 3) = P(M)- P (M S ⋂ M ) = 0.1-0.0005 = 0.0995 Uncertainty: Once you observe an event it’s probability is 1. (ie. chance of winning lottery before attempt can be any probability. However once you won, the probability is 1) Sometimes we can compute probability by counting. In particular, the following theorem is basic Theorem: Let S be a sample dspace with a finite number, N, if equally likely outcomes. If E is any event in S, then, of ways in which E can occur P(E)= number total # of possible outcomes = N So such problems reduce to counting. We need to count in both denominator and numerator. Sometimes the counting can be difficult and so it helps to have some methods for counting. Result 1: Multiplication rule of counting. Suppose that we have K sets that contain, respectively, n1 , n 2 , ..., nk , distinct objects. Think sausages. Then number of ways in which we can select one object from each set (sausage) is n1 * n 2 * ... * nk Result 2: Terminology; when we write n! (n factorial), we mean n!= (n)(n-1)(n-2).....(2)(1) Eg. 7!=(7)(6)(5)(4)(3)(2)(1) 0!= 1...