1. 1 1 Slide STATISTICS FOR BUSINESS AND ECONOMICS Seventh Edition AndersonSweeneyWilliams Slides Prepared by John Loucks © 1999 ITP/South-Western College Publishing

2. 2 2 Slide Chapter 4 Introduction to Probability n Experiments, the Sample Space, and Counting Rules n Assigning Probabilities to Experimental Outcomes n Events and Their Probabilities n Some Basic Relationships of Probability n Conditional Probability n Bayes’ Theorem

3. 3 3 Slide An Experiment and Its Sample Space n An experiment is any process that generates well- defined outcomes. n The sample space for an experiment is the set of all experimental outcomes. n A sample point is an element of the sample space, any one particular experimental outcome.

4. 4 4 Slide Example: Bradley Investments Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that the possible outcomes of these investments three months from now are as follows. Investment Gain or Loss Investment Gain or Loss in 3 Months (in $000) in 3 Months (in $000) Markley Oil Collins Mining 10 8 5-2 5-2 0 -20 -20

5. 5 5 Slide A Counting Rule for Multiple-Step Experiments If an experiment consists of a sequence of k steps in which there are n 1 possible results for the first step, n 2 possible results for the second step, and so on, then the total number of experimental outcomes is given by ( n 1 )( n 2 )... ( n k ).

6. 6 6 Slide A Counting Rule for Multiple-Step Experiments n Example: Bradley Investments can be viewed as a two-step experiment; it involves two stocks, each with a set of experimental outcomes. Markley Oil: n 1 = 4 Collins Mining: n 2 = 2 Total Number of Experimental Outcomes: n 1 n 2 = (4)(2) = 8

7. 7 7 Slide Example: Bradley Investments n Tree Diagram Markley Oil Collins Mining Experimental Markley Oil Collins Mining Experimental (Stage 1) (Stage 2) Outcomes (Stage 1) (Stage 2) Outcomes (10, 8) Gain $18,000 (10, 8) Gain $18,000 (10, -2) Gain $8,000 (10, -2) Gain $8,000 (5, 8) Gain $13,000 (5, 8) Gain $13,000 (5, -2) Gain $3,000 (5, -2) Gain $3,000 (0, 8) Gain $8,000 (0, 8) Gain $8,000 (0, -2) Lose $2,000 (0, -2) Lose $2,000 (-20, 8) Lose $12,000 (-20, 8) Lose $12,000 (-20, -2) Lose $22,000 (-20, -2) Lose $22,000 Gain 5 Gain 8 Gain 10 Gain 8 Lose 2 Even

8. 8 8 Slide Counting Rule for Combinations Another useful counting rule enables us to count the number of experimental outcomes when n objects are to be selected from a set of N objects. n The number of combinations of N objects taken n at a time is where N ! = N ( N - 1)( N - 2)... (2)(1) n ! = n ( n - 1)( n - 2)... (2)(1) n ! = n ( n - 1)( n - 2)... (2)(1) 0! = 1 0! = 1

9. 9 9 Slide Assigning Probabilities to Experimental Outcomes n Classical Method Assigning probabilities based on the assumption of equally likely outcomes. n Relative Frequency Method Assigning probabilities based on experimentation or historical data. n Subjective Method Assigning probabilities based on the assignor’s judgment.

10. 10 Slide Classical Method If an experiment has n possible outcomes, this method would assign a probability of 1/ n to each outcome. n Example Experiment: Rolling a die Experiment: Rolling a die Sample Space: S = {1, 2, 3, 4, 5, 6} Sample Space: S = {1, 2, 3, 4, 5, 6} Probabilities: Each sample point has a 1/6 chance Probabilities: Each sample point has a 1/6 chance of occurring. of occurring.

11. 11 Slide Example: Lucas Tool Rental n Relative Frequency Method Lucas would like to assign probabilities to the number of floor polishers it rents per day. Office records show the following frequencies of daily rentals for the last 40 days. Number ofNumber Number ofNumber Polishers Rentedof Days Polishers Rentedof Days 0 4 1 6 2 18 3 10 4 2

12. 12 Slide n Relative Frequency Method The probability assignments are given by dividing the number-of-days frequencies by the total frequency (total number of days). Number of Number Number of Number Polishers Rentedof Days Probability 0 4.10 = 4/40 1 6.15 = 6/40 2 18.45 etc. 3 10.25 4 2.05 401.00 401.00 Example: Lucas Tool Rental

13. 13 Slide Subjective Method n When economic conditions and a company’s circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. n We can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occur. n The best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with the subjective estimates.

14. 14 Slide Example: Bradley Investments Applying the subjective method an analyst made the following probability assignments. Exper. Outcome Net Gain/Loss Probability Exper. Outcome Net Gain/Loss Probability (10, 8) $18,000Gain.20 (10, 8) $18,000Gain.20 (10, -2) $8,000Gain.08 (10, -2) $8,000Gain.08 (5, 8) $13,000Gain.16 (5, 8) $13,000Gain.16 (5, -2) $3,000Gain.26 (5, -2) $3,000Gain.26 (0, 8) $8,000Gain.10 (0, 8) $8,000Gain.10 (0, -2) $2,000Loss.12 (0, -2) $2,000Loss.12 (-20, 8) $12,000Loss.02 (-20, 8) $12,000Loss.02 (-20, -2) $22,000Loss.06 (-20, -2) $22,000Loss.06

15. 15 Slide Events and Their Probabilities n An event is a collection of sample points. n The probability of any event is equal to the sum of the probabilities of the sample points in the event. n Example: Event M = Markley Oil Profitable M = {(10, 8), (10, -2), (5, 8), (5, -2)} M = {(10, 8), (10, -2), (5, 8), (5, -2)} P( M ) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) P( M ) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) =.2 +.08 +.16 +.26 =.2 +.08 +.16 +.26 =.70 =.70 Event C = Collins Mining Profitable Event C = Collins Mining Profitable P( C ) =.48 (found using the same logic) P( C ) =.48 (found using the same logic)

16. 16 Slide Some Basic Relationships of Probability n Complement of an Event n Union of Two Events n Intersection of Two Events n Mutually Exclusive Events

17. 17 Slide Complement of an Event n The complement of event A is defined to be the event consisting of all sample points that are not in A. n The complement of A is denoted by A c. n The Venn diagram below illustrates the concept of a complement. Event A AcAc Sample Space S

18. 18 Slide n The union of events A and B is the event containing all sample points that are in A or B or both. The union is denoted by A  B  The union is denoted by A  B  n The union of A and B is illustrated below. Sample Space S Event AEvent B Union of Two Events

19. 19 Slide Union of Two Events n Example: Event M = Markley Oil Profitable Event C = Collins Mining Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable M  C = Markley Oil Profitable or Collins Mining Profitable or Collins Mining Profitable M  C = {(10, 8), (10, -2), (5, 8), (5, -2), (0, 8), (-20, 8)} M  C = {(10, 8), (10, -2), (5, 8), (5, -2), (0, 8), (-20, 8)} P( M  C) = P(10, 8) + P(10, -2) + P(5, 8) P( M  C) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) + P(0, 8) + P(-20, 8) =.20 +.08 +.16 +.26 +.10 +.02 =.20 +.08 +.16 +.26 +.10 +.02 =.82 =.82

20. 20 Slide Intersection of Two Events The intersection of events A and B is the set of all sample points that are in both A and B. The intersection of events A and B is the set of all sample points that are in both A and B. The intersection is denoted by A  The intersection is denoted by A  n The intersection of A and B is the area of overlap in the illustration below. Sample Space S Event BEvent A

21. 21 Slide n Example: Event M = Markley Oil Profitable Event C = Collins Mining Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable M  C = Markley Oil Profitable and Collins Mining Profitable and Collins Mining Profitable M  C = {(10, 8), (5, 8)} M  C = {(10, 8), (5, 8)} P( M  C) = P(10, 8) + P(5, 8) P( M  C) = P(10, 8) + P(5, 8) =.20 +.16 =.20 +.16 =.36 =.36 Intersection of Two Events

22. 22 Slide Addition Law n The addition law provides a way to compute the probability of event A or B or both A and B occurring. n The law is written as: P( A  B ) = P( A ) + P( B ) - P( A   P( A  B ) = P( A ) + P( B ) - P( A   Example: Markley Oil or Collins Mining Profitable Example: Markley Oil or Collins Mining Profitable We know: P( M ) =.70, P( C ) =.48, P( M  C ) =.36 We know: P( M ) =.70, P( C ) =.48, P( M  C ) =.36 Thus: P( M  C ) = P( M ) + P( C ) - P( M  C ) Thus: P( M  C ) = P( M ) + P( C ) - P( M  C ) =.70 +.48 -.36 =.70 +.48 -.36 =.82 =.82 This result is the same as that obtained earlier using This result is the same as that obtained earlier using the definition of the probability of an event. the definition of the probability of an event.

23. 23 Slide Addition Law for Mutually Exclusive Events n Two events are said to be mutually exclusive if the events have no sample points in common. That is, two events are mutually exclusive if, when one event occurs, the other cannot occur. n Addition Law for Mutually Exclusive Events: P( A  B ) = P( A ) + P( B ) P( A  B ) = P( A ) + P( B ) Sample Space S Event B Event A

24. 24 Slide Conditional Probability n The probability of an event given that another event has occurred is called a conditional probability. n The conditional probability of A given B is denoted by P( A | B ). n A conditional probability is computed as follows: n Example: Collins Mining Profitable Given Markley Oil Profitable

25. 25 Slide Multiplication Law n The multiplication law provides a way to compute the probability of an intersection of two events. n The law is written as: P( A   B ) = P( B )P( A | B ) P( A   B ) = P( B )P( A | B ) Example: Markley Oil and Collins Mining Profitable Example: Markley Oil and Collins Mining Profitable We know: P( M ) =.70, P( C | M ) =.51 We know: P( M ) =.70, P( C | M ) =.51 Thus: P( M  C ) = P( M )P( M|C ) Thus: P( M  C ) = P( M )P( M|C ) = (.70)(.51) = (.70)(.51) =.36 =.36 This result is the same as that obtained earlier using This result is the same as that obtained earlier using the definition of the probability of an event. the definition of the probability of an event.

26. 26 Slide Multiplication Law for Independent Events n Events A and B are independent if P( A | B ) = P( A ). n Multiplication Law for Independent Events: P( A   B ) = P( A )P( B ) P( A   B ) = P( A )P( B ) n The multiplication law also can be used as a test to see if two events are independent. n Example: Are M and C independent?  Does  P( M  C ) = P( M )P( C ) ? We know: P( M  C ) =.36, P( M ) =.70, P( C ) =.48 We know: P( M  C ) =.36, P( M ) =.70, P( C ) =.48 But: P( M )P( C ) = (.70)(.48) =.34 But: P( M )P( C ) = (.70)(.48) =.34.34  so  M and C are not independent..34  so  M and C are not independent.

27. 27 Slide Bayes’ Theorem n Often we begin probability analysis with initial or prior probabilities. n Then, from a sample, special report, or a product test we obtain some additional information. n Given this information, we calculate revised or posterior probabilities. n Bayes’ theorem provides the means for revising the prior probabilities. NewInformationNewInformationApplication of Bayes’ TheoremApplication TheoremPosteriorProbabilitiesPosteriorProbabilitiesPriorProbabilitiesPriorProbabilities

28. 28 Slide A proposed shopping center will provide strong competition for downtown businesses like L. S. Clothiers. If the shopping center is built, the owner of L. S. Clothiers feels it would be best to relocate. The shopping center cannot be built unless a zoning change is approved by the town council. The planning board must first make a recommendation, for or against the zoning change, to the council. Let: The shopping center cannot be built unless a zoning change is approved by the town council. The planning board must first make a recommendation, for or against the zoning change, to the council. Let: A 1 = town council approves the zoning change A 2 = town council disapproves the change n Prior Probabilities Using subjective judgment: P( A 1 ) =.7, P( A 2 ) =.3 Using subjective judgment: P( A 1 ) =.7, P( A 2 ) =.3 Example: L. S. Clothiers

29. 29 Slide Example: L. S. Clothiers n New Information The planning board has recommended against the zoning change. Let B denote the event of a negative recommendation by the planning board. The planning board has recommended against the zoning change. Let B denote the event of a negative recommendation by the planning board. Given that B has occurred, should L. S. Clothiers revise the probabilities that the town council will approve or disapprove the zoning change? Given that B has occurred, should L. S. Clothiers revise the probabilities that the town council will approve or disapprove the zoning change? Past history with the planning board and the town council indicates the following conditional probabilities: Past history with the planning board and the town council indicates the following conditional probabilities: P( B | A 1 ) =.2 P( B | A 2 ) =.9

30. 30 Slide n Probability Tree Analysis P( B c | A 1 ) =.8 P( A 1 ) =.7 P( A 2 ) =.3 P( B | A 2 ) =.9 P( B c | A 2 ) =.1 P( B | A 1 ) =.2  P( A 1  B ) =.14  P( A 2  B ) =.27  P( A 2  B c ) =.03  P( A 1  B c ) =.56 Example: L. S. Clothiers

31. 31 Slide Bayes’ Theorem n To find the posterior probability that event A i will occur given that event B has occurred we apply Bayes’ theorem. n Bayes’ theorem is applicable when the events for which we want to compute posterior probabilities are mutually exclusive and their union is the entire sample space.

32. 32 Slide Example: L. S. Clothiers n Posterior Probabilities Given the planning board’s recommendation not to approve the zoning change, we revise the prior probabilities as follows. =.34 =.34 n Conclusion The planning board’s recommendation is good news for L. S. Clothiers. The posterior probability of the town council approving the zoning change is.34 versus a prior probability of.7.

33. 33 Slide The End of Chapter 4