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Probability Notes Math 309
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Sample spaces, events, axioms Math 309 Chapter 1
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Some Definitions Experiment - means of making an observation Sample Space (S) - set of all outcomes of an experiment listed in a mutually exclusive and exhaustive manner Event - subset of a sample space Simple Event - an event which can only happen in one way; )
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Since events are sets, we need to understand the basic set operations Intersection (A B) - everything in A and B Union (A B) - everything in A or B or both Complement (A C ) - everything not in A Difference A – B = A B C – everything in A that is not in B
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You should be able to sketch Venn diagrams to describe the intersections, unions, & complements of sets. Note that these set operations obey the commutative, associative, and distributive laws
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DeMorgan’s Laws (A B) C = (A C B C ) (A B) C = (A C B C ) Convince yourself that these are reasonable with Venn diagrams!
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Another definition - A and B are mutually exclusive iff A B =
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Axioms of Probability (these are FACT, no proof needed!) Let A represent an event, S the sample space, P(S) = 1 For pairwise mutually exclusive events, the probability of their union is the sum of their respective probabilities, i.e. P(A 1 A 2 ... A n ...) = P(A 1 )+P(A 2 )+... +P(A n ) +...
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Theorems (You should be able to prove these using the axioms and definitions.) Thm 1.1 - The probability of the empty set is zero. Thm 1.2 – Let {A 1, A 2,...,A n } be a mutually exclusive set of events. Then P(A 1 A 2 ... A n ) = P(A 1 ) + P(A 2 ) +... + P(A n )
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From Axiom 3, it can be shown that: (Prop. 1*) Let {A 1, A 2,...,A n } be a mutually exclusive set of events. Then P(A 1 A 2 ... A n ) = P(A 1 ) + P(A 2 ) +... + P(A n )
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Let A and B be mutually exclusive, our last theorem with n = 2 gives: P(A B) = P(A) +P(B) Letting B = A C along with axioms 1 & 2 gives: 0 <= P(A)<= 1. (Can you show this?)
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Let A and B be mutually exclusive, our last theorem with n = 2 gives: P(A B) = P(A) +P(B)
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More Theorems Let A and B be any two events. Thm 1.4 - P(A C ) = 1 - P(A) Thm 1.5- If A B, then P(B-A) = P(B A C ) = P(B) – P(A) – Corollary: If A is a subset of B, then P(A) <= P(B). Thm 1.6 P(A B) = P(A) + P(B) - P(A B) Thm 1.7 P(A) = P(A B) + P(A B C )
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More Theorems (Propositions) Let A and B be any two events. Prop. 4.1 - P(A C ) = 1 - P(A) Prop. 4.2 - If A is a subset of B, then P(A) <= P(B) Prop. 4.3 - P(A B) = P(A) + P(B) - P(A B) Prop. 2* - P(A) = P(A B) + P(A B C )
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Unions get complicated if events are not mutually exclusive! P(A B C) = P(A) + P(B) + P(C) - P(A B) - P(A C) - P(B C) + P(A B C) A C B
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Sample Spaces with Equally Likely Outcomes In an experiment where all sample points are equally likely, one can find the probability of an event by counting two sets.
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Combinatorial Methods Math 309 Chapter 1
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Combinatorics Basic Principle of Counting –(a.k.a. Multiplication Principle) Permutations –Permutations with indistinguishable objects Combinations
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Basic Counting Principle If experiment 1 has m outcomes and experiment 2 has n outcomes, then there are mn outcomes for both experiments. The principle can be generalized for r experiments. The number of outcomes of r experiments is the product of the number of outcomes of each experiment.
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We define experiment as a means of making an observation (e.g. flip a coin, choose a color). Each experiment could be making a choice from a different set.
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Permutations # of arrangements of one set, order matters application of the basic counting principle where we return to the same set for the next selection P(n,r) = n!/(n-r)!
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Permutations with Indistinguishable Objects Order the objects as if they were distinguishable Then “divide out” those arrangements that look identical.
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Combinations the number of selections, order doesn’t matter – C(n,r) = n!/[(n-r)!r!] the number of arrangements can be counted by selecting the objects and then ordering them –i.e. P(n,r) = C(n,r)*r!
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Observations about Combinations C(n, r) = C(n, n-r) C(n, n) = C(n, 0) = 1 C(n, 1) = n = C(n, n-1) C(n, 2) = n(n-1)/2
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Combining Counting Techniques If we are careful with language, –when we say “AND”, we multiply –“AND” multiplication intersection –when we say “OR”, we add –“OR” addition union
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Conditional Probability and Independence Math 309 Chapter 3
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Conditional Probability P(A|B) P(A|B) is read, “the probability of A given B” B is known to occur. P(A|B) = P(A B) / P(B), if P(B) > 0 i.e. the conditional probability is the probability that both occur divided by what is given occurs
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The multiplication rule and intersection multiply P(A B) = P(A)*P(B|A) = P(B)*P(A|B) (Note that this is an algebraic manipulation of the formula for conditional probability.) Intersections get more complicated when there are more events, e.g. P(A B C D) = P(A)* P(B|A)*P(C|A B)*P(D|A B C)
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Independent Events A and B are independent if any of the following are true: –P(A B) = P(A)*P(B) –P(A|B) = P(A) –P(B|A) = P(B) You need to check probabilities to determine if events are independent. If A, B, C, & D are pairwise independent, –P (A B C D) = P(A)*P(B)*P(C)*P(D)
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