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Business Statistics, Can. ed. By Black, Chakrapani & Castillo Discrete Distributions Chapter 4 Probability Prepared by Dr. Clarence S. Bayne JMSB, Concordia University Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Learning Objectives • Understanding what is probability • Comprehend the different ways of assigning probability. • Understand and apply marginal, union, joint, and conditional probabilities. • Select the appropriate laws of probability to use in solving problems. • Solve problems using the laws of probability including the laws of addition, multiplication and conditional probability • Revise probabilities using Bayes’ rule. Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Methods of Assigning Probabilities There are three methods for assigning probabilities: • The classical method (mathematical rules and laws) • Relative frequency of occurrence (based on historical data: empirical ) • Subjective probability (based on personal intuition or reasoning) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Classical Probability • Number of outcomes leading to the event divided by the total number of outcomes possible • Each outcome is equally likely P (E ) Where : N total n • Determined a priori ‐‐ before performing the experiment • Applicable to games of chance • Objective ‐‐ everyone correctly using the method assigns an identical probability Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. e n e N number number of outcomes of outcomes in E Relative Frequency Probability • Based on historical data • Computed after performing the experiment n P (E ) Where N : N total • Number of times an event occurred divided by the number of trials • Objective ‐‐ everyone correctly using the method assigns an identical probability Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. n e e number number producing of trials of outcomes E Subjective Probability • Subjective probability comes from a person’s intuition or reasoning • However different individuals may (correctly) assign different numeric probabilities to the same event • Expresses an individual’s degree of belief • Useful for unique (single‐trial) experiments – New product introduction – Initial public offering of common stock – Site selection decisions – Sporting events Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Fundamental Concepts and Laws of Probability Theory • • • • • • • • • Experiment Event Elementary Events Sample Space Unions and Intersections Mutually Exclusive Events Independent Events Collectively Exhaustive Events Complementary Events Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Experiment • Experiment: a process that produces outcomes – More than one possible outcome – Only one outcome per trial • Trial: one repetition of the process • Elementary Event: cannot be decomposed or broken down into other events • Event: an outcome of an experiment – May be an elementary event, or – May be an aggregate of elementary events – Usually represented by an uppercase letter, e.g., A, E1 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. An Example Experiment Experiment: randomly select, without replacement, two families from the residents of Tiny Town Elementary Event: the sample includes families A and C Event: each family in the sample has children in the household Event: the sample families own a total of four automobiles Tiny Town Population Family Children in Household Number of Automobiles A B C D Yes Yes No Yes 3 2 1 2 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Sample Space • The set of all elementary events for an experiment • Methods for describing a sample space – roster or listing – tree diagram – set builder notation – Venn diagram Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Sample Space: Roster Example • Experiment: randomly select, without replacement, two families from the residents of Tiny Town • Each ordered pair in the sample space is an elementary event, for example ‐‐ (D,C) Family A B C D Children in Household Number of Automobiles Yes Yes No Yes 3 2 1 2 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Listing of Sample Space (A,B), (A,C), (A,D), (B,A), (B,C), (B,D), (C,A), (C,B), (C,D), (D,A), (D,B), (D,C) Sample Space: Tree Diagram for Random Sample of Two Families A B C D Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. B C D A C D A B D A B C Sample Space: Set Notation for Random Sample of Two Families • S = {(x,y) | x is the family selected on the first draw, and y is the family selected on the second draw} • Concise description of large sample spaces Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Sample Space • Useful for discussion of general principles and concepts Listing of Sample Space (A,B), (A,C), (A,D), (B,A), (B,C), (B,D), (C,A), (C,B), (C,D), (D,A), (D,B), (D,C) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Venn Diagram Union of Sets • The union of two sets contains an instance of each element of the two sets. X 1,4 ,7 ,9 Y 2 ,3,4 ,5,6 X Y X Y 1,2 ,3,4 ,5,6,7 ,9 C IBM , DEC , Apple F Apple, Grape, Lime C F IBM , DEC , Apple, Grape, Lime Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. XY Intersection of Sets • The intersection of two sets contains only those element common to the two sets. X Y 1 , 4 , 7 , 9 2 , 3 , 4 , 5 , 6 X Y X Y 4 C IBM , DEC , Apple F Apple , Grape , Lime C F Apple Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. XY Mutually Exclusive Events • Events with no common outcomes • Occurrence of one event precludes the occurrence of the other event C IBM , DEC , Apple P( X Y) 0 X Y X 1,7 ,9 F Grape , Lime Y 2 ,3,4 ,5,6 CF X Y Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Independent Events • Occurrence of one event does not affect the occurrence or nonoccurrence of the other event • The conditional probability of X given Y is equal to the marginal probability of X. • The conditional probability of Y given X is equal to the marginal probability of Y. P( X| Y ) P( X) and P(Y| X) P(Y ) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Collectively Exhaustive Events • Contains all elementary events for an experiment E1 E2 E3 Sample Space with three collectively exhaustive events Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Complementary Events • All elementary events not in the event ‘A’ are in its complementary event. Sample Space A A P(SampleSpace) 1 P( A) 1 P( A) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Counting the Possibilities • mn Rule • Sampling from a Population with Replacement • Combinations: Sampling from a Population without Replacement Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. mn Rule • If an operation can be done in m ways and a second operation can be done in n ways, then there are mn ways for the two operations to occur in order. • A cafeteria offers 5 salads, 4 meats, 8 vegetables, 3 breads, 4 desserts, and 3 drinks. A meal consists of one serving of each of the items. How many meals are available? • (Ans: 548343 = 5,760 meals.) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Sampling from a Population with Replacement • A tray contains 1,000 individual tax returns. If 3 returns are randomly selected with replacement from the tray, how many possible samples are there? • (N)n = (1,000)3 = 1,000,000,000 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Combinations: Sampling from a Population without Replacement • This counting method uses combinations • Selecting n items from a population of N without replacement Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Combinations: Sampling from a Population without Replacement • For example, suppose a small law firm has 16 employees and three are to be selected randomly to represent the company at the annual meeting of the Bar Association. • How many different combinations of lawyers could be sent to the meeting? • Answer: NCn = 16C3 = 16!/(3!13!) = 560. Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Four Types of Probability • Marginal Probability • Union Probability • Joint Probability • Conditional Probability Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Four Types of Probability Marginal Union Joint Conditional P( X ) P( X Y ) P( X Y ) P( X| Y ) The probability of X occurring X The probability of X or Y occurring The probability of X and Y occurring X Y The probability of X occurring given that Y has occurred X Y Y Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. General Law of Addition P( X Y) P( X) P(Y) P( X Y) X Y Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. General Law of Addition -- Example P( N S) P( N ) P(S) P( N S) .70 P ( N ) .7 0 S N .56 .67 P ( S ) .6 7 P ( N S ) .5 6 P ( N S ) .7 0 .6 7 .5 6 0 .8 1 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Office Design Problem Probability Matrix Noise Reduction Yes No Total Increase Storage Space Yes No .14 .56 .19 .11 .33 .67 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Total .70 .30 1.00 Office Design Problem Probability Matrix Noise Reduction Yes No Total Increase Storage Space Yes No .14 .56 .19 .11 .33 .67 Total .70 .30 1.00 P( N S ) P( N ) P(S ) P( N S ) .70 .67 .56 .81 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Venn Diagram of the X or Y but not Both Case X Y Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. The Neither/Nor Region X Y P( X Y ) 1 P( X Y ) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. The Neither/Nor Region N S P( N S ) 1 P( N S ) 1.81 .19 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Special Law of Addition If X and Y are mutually exclusive, P( X Y ) P( X ) P(Y ) Y X Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Special Law of Multiplication for Independent Events • General Law P( X Y ) P( X) P(Y| X) P(Y ) P( X| Y ) • Special Law If events X and Y are independent, P( X ) P( X | Y ), and P(Y ) P(Y | X ). Consequently, P( X Y ) P( X ) P( Y ) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Law of Conditional Probability • The conditional probability of X given Y is the joint probability of X and Y divided by the marginal probability of Y. P( X Y ) P(Y| X) P( X) P( X| Y ) P(Y) P(Y ) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Law of Conditional Probability N S .56 .70 P ( N ) .7 0 P ( N S ) .5 6 P(N S) P (S| N ) P(N ) .5 6 .7 0 .8 0 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Independent Events • If X and Y are independent events, the occurrence of Y does not affect the probability of X occurring. • If X and Y are independent events, the occurrence of X does not affect the probability of Y occurring. If X and Y are independent events, P( X | Y ) P( X ), and P(Y | X ) P(Y ). Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Independent Events Demonstration Problem 4.10 P( A G) 14 0.33 42 P(G) P( A | G) 0.33 P( A) 0.28 P( A | G) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. P( A) 0.28 Independent Events Demonstration Problem 4.11 D E A 8 12 20 B 20 30 50 C 6 9 15 34 51 85 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. 8 P( A| D) .2353 34 20 P( A) .2353 85 P( A| D) P( A) 0.2353 Revision of Probabilities: Bayes’ Rule • An extension to the conditional law of probabilities • Enables revision of original probabilities with new information P ( X i| Y ) P ( Y | Xi ) P ( Xi ) P ( Y | X 1) P ( X 1) P ( Y | X 2 ) P ( X 2 ) P ( Y | X n ) P ( X n ) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Revision of Probabilities with Bayes' Rule: Ribbon Problem P(Pr airie) 0.65 P( Badlands) 0.35 P(d | Pr airie) 0.08 P(d | Badlands) 0.12 P(d | Prairie) P(Pr airie) P(d | Pr airie) P(Pr airie) P(d | Badlands) P( Badlands) (0.08)(0.65) 0.553 (0.08)(0.65) (0.12)(0.35) P(d | Badlands) P( Badlands) P( Badlands | d ) P(d | Pr airie) P(Pr airie) P(d | Badlands) P( Badlands) (0.12)(0.35) 0.447 (0.08)(0.65) (0.12)(0.35) P(Pr airie | d ) Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Revision of Probabilities with Bayes’ Rule: Ribbon Problem Prior Probability Event Prairie P ( Ei ) 0.65 Conditional Probability Joint Probability Revised Probability P(d| Ei ) P(Ei d) P( Ei| d ) 0.08 0.052 0.052 0.094 =0.553 Badlands 0.35 0.12 0.042 0.042 0.094 0.094 =0.447 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Probability for a Sequence of Independent Trials • 25 percent of a bank’s customers are commercial (C) and 75 percent are retail (R). • Experiment: Record the category (C or R) for each of the next three customers arriving at the bank. • Sequences with 1 commercial and 2 retail customers. – C1 R2 R3 – R1 C2 R3 – R1 R2 C3 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Probability for a Sequence of Independent Trials • Probability of specific sequences containing 1 commercial and 2 retail customers, assuming the events C and R are independent 1 3 3 9 P(C1 R2 R3) P(C) P( R) P( R) 4 4 4 64 3 1 3 9 P( R1 C2 R3) P( R) P(C) P( R) 4 4 4 64 3 3 1 9 P( R1 R2 C3) P( R) P( R) P(C) 4 4 4 64 Business Statistics, Can. Ed. © 2010 John Wiley & Sons Canada, Ltd. Check whether independent X and Y in the following table are Y | | 0 1 p(x) ----|-----------------------1 | 1/8 1/8 1/4 | | X 0 | 2/8 2/8 2/4 | | 1 | 1/8 1/8 1/4 | P(y) 2/4 2/4 1 Anaswer: P(Xi and Yi) = P(Xi) P(Yi) for all i , hence they are indepwndent. textbook example page 141 E1=Prairie E2=Badlands D=side effects P(E1)=0.65 P(E2)=0.35 P(D/E1)=0.08 P(D/E2)=0.12 P(E1/D)= P(D/E1)P(E1)/(P(D/E1)P(E1)+P(D/E2)P(E2)) = (0.08 * 0.65) /(0.08*0.65+0.12*0.35) = 0.553 P(E2/D)= P(D/E2)P(E2)/(P(D/E1)P(E1)+P(D/E2)P(E2)) = (0.12 * 0.35) /(0.08*0.65+0.12*0.35) = 0.447 Ch4 probability examples Consider an ordinary deck of playing cards. What is the probability of drawing, in a random drawing of a single card, a face card (jack, queen, or king) or a diamond? F=face card D=diamond P(F) = 12/52 P(D) = 13/52 P(F and D)=3/52 P(F or D)=P(F) + P(D) - P(F and D) =12/52 + 13/52 - 3/52 = 11/26 Suppose two cards are dealt from a well-shuffled deck. They are placed side-by-side face down on the table, the first card to the left of the second. a) What is the probability that at least one of the cards is an ace? P(A1 or A2) = P(A1 and A2)+P(notA1 and A2)+(A1 and notA2) = (4/52)*(3/51)+(48/52)*(4/51)+(4/52)*(48/51) = 33/221 b) If the first card is turned over and seen to be an ace, what is now the probability that the second card is an ace? P(A2/A1) = P(A1 and A2)/P(A1) = (4/52)*(3/51)/(4/52) = 3/51 = 1/17 Suppose Mr. Jones chooses at random one of the integers 1,2, or 3. Then he throws as many dice as indicated by the chosen number. What is the probability that he will score a total of 4 points? P(4)=P(1d and 4)+P(2d and 4) +P(3d and 4) 1d => 4 2d => (2,2)(1,3) (3,1) 3d => (1,1,2) (1,2,1) (2,1,1) P(4) = (1/3)*(1/6+3/36+3/216) = 19/216 JR Lfw P(l<l~)-- ~ f( v4 /~ ).:::~ f (~ )-::: '· r- ! (P; J ~ o , 5 f(rpl Ir( J~ P(~ ~~~{) f (f. ___ f(K?!~r) l P (~ J f~/#JJI(~)-r F(K: (!Jzlf(f'._) .. ---- c%)( ~ \ ,-------· 3 \ •. 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