Set Notation and Axioms of Probability NOT NOT X = X = X'' = X

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1 Set Notation and Axioms of Probability Memory Hints: Intersection I AND I looks like A for And Union U OR + U looks like U for Union Complement NOT X = X = X' NOT NOT X = X = X'' = X Commutative Law A AND B = B AND A A I B = B I A A B = B A A OR B = B OR A A U B = B U A A + B = B + A Distributive Law (A AND B) OR C = (A OR C) AND (B OR C) (A I B) U C = (A U C) I (B U C) (A B) + C = (A + C) (B + C) (A OR B) AND C = (A AND C) OR (B AND C) (A U B) I C = (A I C) U (B I C) (A + B) C = (A C) + (B C) A = (A AND B) OR (A AND B) B = (A AND B) OR (A AND B) A = (A I B) U (A I B) B = (A I B) U (A I B) A = (A B) + (A B) B = (A B) + (A B) DeMorgan s Law _ (A AND B) = A OR B (A OR B) = A AND B _ (A I B) = A U B (A U B) = A I B _ (A B) = A + B (A + B) = A B Axioms of Probability P(A) + P(A) = 1 P(A) = 1 - P(A) 0 < P(A) < 1 Σ P(A i ) = 1

2 Joint Probabilities Addition Rule P(A OR B) = P(A) + P(B) - P(A AND B) Mutually Exclusive If A and B are mutually exclusive then A AND B = P(A AND B) = 0 P(A OR B) = P(A) + P(B) Conditional Probability The conditional probability of an event A given an event B, denoted P(A B) is P(A B) = P(A AND B) / P(B) The conditional probability of an event B given an event A, denoted P(B A) is P(B A) = P(B AND A) / P(A) Multiplication Rule P(A AND B) = P(B AND A) P(A AND B) = P(A B) * P(B) P(B AND A) = P(B A) * P(A) P(A B) * P(B) = P(B A) * P(A) Bayes Theorem P(A B) = P(B A) * P(A) / P(B) P(A) = P(B) * [ P(A B) / P(B A) ] P(B A) = P(A B) * P(B) / P(A) P(B) = P(A) * [ P(B A) / P(A B) ] Total Probability Rule P(A) = P(A AND B) + P(A AND B) = P(A B) * P(B) + P(A B) * P(B) P(B) = P(B AND A) + P(B AND A) = P(B A) * P(A) + P(B A) * P(A) Joint Probabilities Dependent (Without Replacement) P(A AND B) = P(A) * P(B A) = P(B) * P(A B) Independent (With Replacement) P(A AND B) = P(A) * P(B) Independence Two events are independent if any one of the following equivalent statements is true. P(A B) = P(A) P(B A) = P(B) P(A AND B) = P(A) * P(B) If any one of the above statements is true, then all of them are true.

3 Probability Distribution Functions Discrete Probability Functions Continuous Probability Functions Probability Mass Function f(x) Probability Density Function f(x) f(x i ) 0 f(x) 0 Σ f(x i ) = 1 f(x) dx = 1 f(x i ) = Prob (X = x i ) Prob (a X b) = f(x) dx Prob (X = x) = 0 Cumulative Distribution Function F(x) Cumulative Distribution Function F(x) F(x) = P(X x) = Σ f(x i ) F(x) = P(X x) = f(x) dx 0 F(x) 1 0 F(x) 1 Mean & Variance Mean & Variance μ = E(x) = Σ x f(x) μ = E(x) = x f(x) dx σ 2 = V(x) = E(x - μ) 2 σ 2 = V(x) = E(x - μ) 2 σ 2 = Σ [(x - μ) 2 f(x)] = Σ x 2 f(x) - μ 2 σ 2 = (x - μ) 2 f(x) dx = x 2 f(x) dx - μ 2

4 Methods of Counting Multiplication Rule For a process or operation with a sequence of k steps or alternatives, the total number of ways to complete the process is n 1 x n 2 x n 3 x... x n k Permutations A permutation is an ordered sequence of the elements of a set. The number of permutations of n different elements is n! = n (n - 1) (n -2) (n -3)... (3) (2) (1) Example for the elements a, b, c, d: abcd, abdc, acbd, acdb, adbc, adcb, bacd, badc, bcad, bcda, bdac, bdca, cabd, cadb, cbad, cbda, cdab, cdba, dabc, dacb, dbac, dbca, dcab, dcba For n = 4 n! = 4 x 3 x 2 x 1 = 24 The number of permutations of subsets of r elements from a set of n different elements is n P = r n! (n r)! Example the four letters a, b, c, d taken two at a time, the number of permutations is 4! 4! 4x3x2x1 12 (4 2)! (2)! 2 x1 4 P = = = = 2 ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, dc The number of permutations of n = n 1 + n n r objects of which n 1 are of one type, n 2 are of a second type,..., and n r of are an r th type is n! n 1!n 2!n 3!... n r! Example: three O's and two X's (n = = 5) Permutations = 5! 120 = = 10 (3!) (2!) (6)(2) OOOXX, OOXOX, OOXXO, OXOOX, OXOXO, OXXOO, XXOOO, XOXOO, XOOXO, XOOOX Combinations A combination is an unordered subset of r elements from a set of n elements. The number of combinations, subsets of size r that can be selected from a set of n elements, is n C = r n! r!(n r)! Example the four letters a, b, c, d taken two at a time, the number of combinations is 4! 4! 2!(4 2)! (2)!(2)! 4 C = = = 2 6 ab, ac, ad, bc, bd, cd

5 Combinations & Permutations Permutations - "Rearranging" Examples: 1234 is different than 4312 ABC is different than CAB For a set of n items taken all at once, the number of permutations = n! For a set of n items taken r at a time (without repetitions) i.e., not allowed npr = n! / (n - r)! Examples: For A, B C, D taken three at a time ABC, ACB, ADE, BAC, BCA, DEA, etc, etc, etc. Combinations - "Group" where "Order" is not important; i.e., ABC is the same as BCA For a set of n items taken all at once, the number of combinations is 1. For a set of n items taken one at a time, the number of combinations is n. For a set of n items taken r at a time, the number of combinations is ncr = n! / [(r!)(n - r)!]

6 Bernoulli Processes and Binomial Distributions Conditions: 1. Only two possible outcomes per trial ("success" or "failure") 2. n independent trials 3. Probability of success on any one trial is constant = p 4. The random variable X is the number of successes in n trials, such that Probability Mass Function n n! = = C C = x x!(n x)! n x (n x) p(x x) p (1 p) where x Mean: μ = E(X) = np Variance: σ 2 = V(X) = np(1 - p)