Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
STATISTICS Basic Concept of Probability Ir. Mahmud Sudibandriyo MSc., PhD September 24, 2008 1 Topics ¾ Concepts and Definitions ¾ Compound Event Probabilities ¾ Enumeration Technique September 24, 2008 2 Concepts and Definitions Probability Experiment: activities where an outcome, response, or measurement is obtained. Sample space (S): A set of values which covers all possibility of outcome, response, or measurement. Event : A subset of all outcome, response, or measurement in the sample space. Example: we inspect 3 pumps one by one in sequence, we note G for a pump in a good condition and F for a fail pump.. S S = {GGG,GGF,GFG,FGG,GFF,FGF,FFG,FFF} A If A is an event where we find one fail pump, then: A = {GGF, GFG, FGG} September 24, 2008 3 Probability: Probability = 1 if the event is definitely occurs Probability = 0 if the event is impossible. Classical Definition: P(A) = fA/N and P(Ā) = 1- P(A) exp: probability to get Ace in bridge cards is 4/52 probability to get non Ace is 1 – 4/52 = 48/52 Relative Frequency Definition: P (A) = lim fA/N NÆ∞ Subjective Definition: based on expert judgment exp: what is the probability of PERSIJA to win the game against PSIS ? September 24, 2008 4 Compound Event Probabilities Compound Event: a combination of 2 or more simple events Conditional probability: P (A|B) = P (A∩B)/ P(B) ; P(B) >0 exp: In PC production, 60% is installed word processor (A), 40% is installed with spreadsheet (B), 30% is installed with both of them. If one buy a PC with spreadsheet in it, what the probability this computer is also installed with word processor ? P(A) = 0.6 ; P(B) = 0.4 ; P (A∩B) = 0.3 ; P(A|B)=0.3/0.4=0.75 B A 0.3 September 24, 2008 0.3 0.1 A∩B 0.1 0.3 5 Independent and Dependent Event Independent event : P(A|B) = P(A) or P(B|A) = P(B) exp: A: “heads on the fifth toss”, B: “heads on the sixth toss” Dependent event : exp. conditional probability Mutually Exclusive P(A|B) = 0 and P(B|A) = 0 exp: A: “drawing an ace from a deck of cards” B: “drawing a King” since both ace and king cannot be drawn in a single draw, they are thus mutually exclusive A September 24, 2008 B 6 Laws of Probabilities in Compound Event Multiplication Law Independent Event P(A and B and C and…) = P(A∩B∩C∩…)= P(A).P(B).P(C)… exp: Probability of heads on both the fifth and sixth tosses P(A).P(B) = ½ . ½ = ¼ Dependent Event P(A and B) = P(A∩B) = P(A|B).P(B) = P(B|A).P(A) exp: see Ref. Book p. 51 September 24, 2008 7 Addition Law P(A or B) = P(AUB) = P(A)+P(B) – P(A∩B) P(A or B or C) = P(AUBUC) = P(A)+P(B)+P(C)-P(A∩B) -P(A∩C) – P(B∩C) + P(A∩B∩C) Exp: see Ref. book p. 52-53 Bayes Formulation A combination of conditional probability and multiplication law Exp: see Ref. Book p 55 September 24, 2008 8 Enumeration Technique Probability Tree A1 A2 P(A1) P(A2) B1 P(B1|A1) B2 P(B2|A1) B3 P(B3|A1) B1 P(B1|A2) B2 P(B2|A2) B3 P(B3|A2) P(A1∩B1) P(A1∩B2) P(A1∩B3) P(A2∩B1) P(A2∩B2) P(A2∩B3) Exp. See textbook p. 56-57 September 24, 2008 9 Combinatorial Analysis Permutation An arrangement of r out of n objects with attention given to the order of arrangement n! n Pr = P ( n, r ) = Pn , r = P = (n − r )! n r Exp: the number of permutation of the letters a,b,c taken two at a time is 3P2 = 3.2/1 = 6 Combination A selection of r out of n objects with no attention given to the n! n Pr = n Cr = r!(n − r )! r! order of arrangement Exp: 3C2 = 6/2 = 3 September 24, 2008 10