IS 310 Business Statistics CSU Long Beach

1. IS 310 Business Statistics CSU Long Beach

2. Chapter 4 Introduction to Probability Few events are certain to happen. Probability offers a numeric measure of the likelihood of an event occurring

3. Probability as a Numerical Measureof the Likelihood of Occurrence Increasing Likelihood of Occurrence 0 .5 1 Probability: The event is very unlikely to occur. The occurrence of the event is just as likely as it is unlikely. The event is almost certain to occur.

4. Probability Concepts • To understand probability, one needs to understand the following concepts: • Experiment • Experimental outcome • Sample space • Sample point

5. An Experiment and Its Sample Space An experimentis any process that generates well-defined outcomes. The sample space for an experiment is the set of all experimental outcomes. An experimental outcome is also called a sample point.

6. Experiment and Outcomes • Let’s look at the following experiments and their outcomes. • ExperimentOutcome • Tossing a coin Head, tail • Selecting a part Defective, non-defective • for inspection • Playing ball game Win, lose, tie • Conducting a sales Sale, no sale • call

7. Counting Outcomes • In order to calculate probability, we must be able to count the total number of outcomes in an experiment. In simple experiments, it is easy to know the number of outcomes. In complex experiments, we need some rules for counting the number of outcomes. • There are three rules: • Rule for multiple-step experiments • Combination • Permutation

8. A Counting Rule for Multiple-Step Experiments • If an experiment consists of a sequence of k steps • in which there are n1 possible results for the first step, • n2 possible results for the second step, and so on, • then the total number of experimental outcomes is • given by (n1)(n2) . . . (nk). • A helpful graphical representation of a multiple-step experiment is a tree diagram.

9. Counting Rule for Multi-step Experiments • Let’s take the experiment of tossing two coins. How many total outcomes are possible in this experiment? • The first coin has two outcomes (H, T) • The second coin has also two outcomes (H, T) • Total outcomes is: (2) (2) = 4 • Often, a tree diagram helps. Look at Figure 4.2 (10-Page 145; 11-Page 152).

10. Counting Rule for Multi-step Experiments • Let’s take another example. • Kentucky Power & Light Company is starting an expansion project. The project consists of two stages: design and construction. • Design work can be completed in 2, 3 or 4 months (three outcomes) • Construction work can be completed in 6, 7 or 8 months (three outcomes) • Total number of outcomes in this experiment is: • (3) (3) = 9 • Look at Table 4.1 and Figure 4.3 (10-Page 146; 11-Page 153).

11. Counting Rule for Combinations A second 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. Number of Combinations of N Objects Taken n at a Time where: N! = N(N- 1)(N- 2) . . . (2)(1) n! = n(n- 1)(n- 2) . . . (2)(1) 0! = 1

12. Counting Rule for Permutations A third 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, where the order of selection is important. Number of Permutations of N Objects Taken n at a Time where: N! = N(N- 1)(N- 2) . . . (2)(1) n! = n(n- 1)(n- 2) . . . (2)(1) 0! = 1

13. Example Problems A home builder offers four different floor plans, three different exterior designs and two types of roof. How many ways can a home buyer choose his/her home? 4 x 3 x 2 = 24

14. Examples on Combinations and Permutations Combinations: How many ways can three Republicans be chosen as members of a committee from a group of seven Republicans? 7 7! C = ----------- = 35 3 3! (7 – 3)! Permutations: Automobile license plates in a state consists of seven letters of the alphabets; a letter may not be repeated. How many license plates are possible? 26 26! P = --------- = 26 x 25 x 24 x 23 x 22 x 21 x 20 7 (26-7)!

15. Examples on Combinations and Permutations Automobile license plates in a state consist of five letters and two digits. A letter or a digit may be repeated. How many license plates are possible? 26 26! The first position may be filled in P = -------------- = 26 1 (26 – 1)! 26 x 26 x 26 x 26 x 26 x 10 x 10 Automobile license plates in CA consist of any combination of seven letters and/or digits. A letter or a digit may be repeated seven times. How many license plates are possible? 36 x 36 x 36 x 36 x 36 x 36 x 36

16. Examples on Combinations and Permutations Ten individuals are interested in serving on a board of directors. There are three positions available. How many ways can the selection be made? 10 10! C = -------------- = 120 3 3! (10 – 3)! In a class of 40 students, there are 25 women and 15 men. A committee is to be selected with 5 women and three men. How many ways can this selection be made? 25 15 25! 15! C x C = -------------- x ------------- 5 3 5! (25 – 5)! 3! (15 – 3)!

17. Examples on Combinations and Permutations Five different scholarships are to be awarded to five students. How many ways can these scholarships be awarded? 5 5! P = -------- = 5! = 5 x 4 x 3 x 2 = 120 5 (5 – 5)!

18. Assigning Probabilities Classical Method Assigning probabilities based on the assumption of equally likely outcomes Relative Frequency Method Assigning probabilities based on experimentation or historical data Subjective Method Assigning probabilities based on judgment

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

20. Relative Frequency Method • Assigns probabilities based on relative frequencies. • Let’s take the example (10-Page 149; 11-Page 156) • Number of patients waiting for service in the X-ray department of a hospital: • Number WaitingNumber of Days • 0 2 • 1 5 • 2 6 • 3 4 • 4 3 • Total 20

21. Relative Frequency Method • The probability of zero patient waiting for service is: • 2/20 = 0.1 • The probability of one patient waiting for service is: • 5/20 = 0. 25 • The probability of two patients waiting for service is: • 6/20 = 0.30 • And so on.

22. Subjective Method • When economic conditions and a company’s • circumstances change rapidly it might be • inappropriate to assign probabilities based solely on • historical data. • 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. • The best probability estimates often are obtained by • combining the estimates from the classical or relative • frequency approach with the subjective estimate.

23. Sample Problems • Problem # 11 (10-Page 152; 11-Page 159) • 858/(858+228) = 858/1,086 = 0.79 • Yes. Problem # 12 (10-Page 153; 11-Page 160) • 55!/(5!) x (55-5)! • 1/[55!/(5!)x(55-50!] • 1/[55!/(5!)x(55-5)!] x 1/[42!/(1!)x(42-1)!]

24. Events and Their Probabilities An eventis a collection of sample points. The probability of any event is equal to the sum of the probabilities of the sample points in the event. If we can identify all the sample points of an experiment and assign a probability to each, we can compute the probability of an event.

25. Events and Their Probabilities • Let’s go back to the example of Kentucky Power & Light Company. • Suppose, we want to find out the probability of completing the expansion project in 10 months or less. • According to Table 4.3, the following sample points (2,6), (2,7), (2,8), (3,6), (3,7) and (4,6) provide a completion time of 10 months or less. • Thus, the event C that the project can be completed in 10 months or less is: • P(C) = P(2,6) + P(2,7) + P(2,8) + P(3,6) + P(3,7) + P(4,6)

26. Sample Problem • Problem #20 (10-Page 157; 11-Page 164) • Given: • StateNumber of Companies • New York 54 • California 52 • Texas 48 • Illinois 33 • Ohio 30 • Find: P(N) = Find: P(T) = • Find P(B) =

27. Some Basic Relationships of Probability There are some basic probability relationships that can be used to compute the probability of an event without knowledge of all the sample point probabilities. Complement of an Event Union of Two Events Intersection of Two Events Mutually Exclusive Events

28. Complement of an Event The complement of event A is defined to be the event consisting of all sample points that are not in A. The complement of A is denoted by Ac. Sample Space S Event A Ac Venn Diagram

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

30. 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 denoted by A Sample Space S Event A Event B Intersection of A and B

31. Addition Law The addition law provides a way to compute the probability of event A, or B, or both A and B occurring. The law is written as: P(AB) = P(A) + P(B) -P(AB

32. Mutually Exclusive Events Two events are said to be mutually exclusive if the events have no sample points in common. Two events are mutually exclusive if, when one event occurs, the other cannot occur. Sample Space S Event A Event B

33. Mutually Exclusive Events If events A and B are mutually exclusive, P(AB = 0. The addition law for mutually exclusive events is: P(AB) = P(A) + P(B) There is no need to include “-P(AB”

34. Sample Problem • Problem #24 (10-Page 162; 11-Page 169) • a. Let S = Experience surpassed expectation • N = No response • M= Met expectation • F = Fell short of expectation • P(S) = 1 – P(N) – P(M) – P(F) = 1- 0.04 – 0.65 – 0.26 = 0.05 • b. P(M or S) = P(M) + P(S) = 0.65 + 0.05 = 0.70

35. Sample Problem • Problem # 28 (10-Page 163; 11-Page 170) • a. P(B or P) = P(B) + P(P) – P(B and P) • = 0.458 + 0.54 – 0.3 = 0.698 • b. P(neither B nor P) = 1 – P(B or P) • = 1 – 0.698 • = 0.302

36. Conditional Probability The probability of an event given that another event has occurred is called a conditional probability. The conditional probability of A given B is denoted by P(A|B). A conditional probability is computed as follows :

37. Sample Problem • Problem #33 (10-Page 169-170; 11-Page 176-177) • Reason • Quality Cost/Conv Other Total • FT 0.22 0.20 0.04 0.46 • PT 0.21 0.31 0.02 0.54 • _________________________________________ • 0.43 0.51 0.06 1.00 • b. School cost/convenience followed by quality • c. P(Q given FT) = 0.22/0.46 = 0.47 • d. P(Q given PT) = 0.21/0.54 = 0.39 • e. P(A) = 0.46 P(B) = 0.43 P(A and B) = 0.22 • Since P(A and B) not equal to P(A) * P(B), the events are • NOT independent

38. Multiplication Law The multiplication law provides a way to compute the probability of the intersection of two events. The law is written as: P(AB) = P(B)P(A|B)

39. Independent Events If the probability of event A is not changed by the existence of event B, we would say that events A and B are independent. Two events A and B are independent if: P(A|B) = P(A) P(B|A) = P(B) or

40. Multiplication Lawfor Independent Events The multiplication law also can be used as a test to see if two events are independent. The law is written as: P(AB) = P(A)P(B)

41. Miscellaneous Problems Problem # 50 (10-Page 180; 11-Page 187) Rating Frequency Poor 4 Below Average 8 Average 11 Above Average 14 Excellent 13 • What is the probability that a randomly selected viewer will rate the new show as average or better? (11 + 14 + 13) / 50 = 0.76 • What is the probability that a viewer will rate the show as below average or worse? (8 + 4) / 50 = 0.24

42. Miscellaneous Problems Problem # 57 (10-Page 183; 11-Page 189) Given: six percent employees suffered lost-time accidents last year. Fifteen percent who had lost-time accidents last year will experience the same this year. • What percent employees will experience lost-time accidents in both years? (0.06) (0.15) = 0.009 b. (0.06) + (0.15) = 0.21

43. Miscellaneous Problems A box contains 20 units of a product of which four are defective and 16 are good. Four units are selected at random. Calculate the following probabilities: • All four units are defective P(4D) = P(D) . P(D). P(D). P(D) = (4/20) (3/19) (2/18) (1/17) = 1/4845 • Two units are defective and two units are good Two defective units must come Two good units must come from from the four defective units: the sixteen good units: 4 16 C = 6 ways C = 120 ways 2 2

44. Miscellaneous Problems Continued Four units can be selected from the original 20 units in 20 C = 4845 ways 4 (6) (120) P(2D and 2G) = ----------- = 144/969 4845

45. Miscellaneous Problems Continued • At least three units are defective P(at least 3D) means P(3D or 4D) = P(3D) + P(4D) 4 16 4 C . C C 3 1 4 P(3D) = ----------------- P(4D) = ------------------ 20 20 C C 4 4

46. Miscellaneous Problems In how many ways can a television director schedule six different commercials during the six time slots allocated to commercials during the telecast of the first quarter of a football game? 6 P = 6! = 720 6 Seven wooden dowels of varying lengths are arranged in a row. What is the probability that the dowels will be arranged in order of size? 2 P(arranged in order of size) = ----------------- 7!

47. End of Chapter 4