Statistics for Managers Using Microsoft Excel (3 rd Edition)

1. Statistics for Managers Using Microsoft Excel (3rd Edition) Chapter Basic Probability and Discrete Probability Distributions

2. Chapter Topics • Basic probability concepts • Sample spaces and events, simple probability, joint probability • Conditional probability • Statistical independence, marginal probability • Bayes’s Theorem

3. Chapter Topics (continued) • The probability of a discrete random variable • Covariance and its applications in finance • Binomial distribution • Poisson distribution • Hypergeometric distribution

4. Sample Spaces • Collection of all possible outcomes • e.g.: All six faces of a die: • e.g.: All 52 cards in a deck:

5. Events • Simple event • Outcome from a sample space with one characteristic • e.g.: A red card from a deck of cards • Joint event • Involves two outcomes simultaneously • e.g.: An ace that is also red from a deck of cards

6. Visualizing Events • Contingency Tables • Tree Diagrams Ace Not Ace Total Black 2 24 26 Red 2 24 26 Total 4 48 52 Ace Red Cards Not an Ace Full Deck of Cards Ace Black Cards Not an Ace

7. Simple Events The Event of a Triangle There are 5 triangles in this collection of 18 objects

8. Joint Events The event of a triangle ANDblue in color Two triangles that are blue

9.   Special Events Null Event • Impossible event e.g.: Club & diamond on one card draw • Complement of event • For event A, all events not in A • Denoted as A’ • e.g.: A: queen of diamonds A’: all cards in a deck that are not queen of diamonds

10. Special Events (continued) • Mutually exclusive events • Two events cannot occur together • e.g.: -- A: queen of diamonds; B: queen of clubs • Events A and B are mutually exclusive • Collectively exhaustive events • One of the events must occur • The set of events covers the whole sample space • e.g.: -- A: all the aces; B: all the black cards; C: all the diamonds; D: all the hearts • Events A, B, C and D are collectively exhaustive • Events B, C and D are also collectively exhaustive

11. Contingency Table A Deck of 52 Cards Red Ace Not an Ace Total Ace Red 2 24 26 Black 2 24 26 Total 4 48 52 Sample Space

12. Tree Diagram Event Possibilities Ace Red Cards Not an Ace Full Deck of Cards Ace Black Cards Not an Ace

13. Probability • Probability is the numerical measure of the likelihood that an event will occur • Value is between 0 and 1 • Sum of the probabilities of all mutually exclusive and collective exhaustive events is 1 1 Certain .5 0 Impossible

14. Computing Probabilities • The probability of an event E: • Each of the outcomes in the sample space is equally likely to occur e.g. P() = 2/36 (There are 2 ways to get one 6 and the other 4)

15. Properties of Probability • If A is an event and A’ is its complement then P(A) = 1-P(A’) • For any two events A and B P(AUB) = P(A) + P(B)-P(AB) A B A=(AB)U(AB’) P(A)=P(AB)+P(AB’) P(AUB)=P(B)+P(AB’) = P(B)+P(A)-P(AB)

16. B A A Properties of Probability • If A subset of B then P(A)≤P(B)

17. Properties of Probability

18. Properties of Probability

19. Computing Joint Probability • The probability of a joint event, A and B:

20. Event Total B1 B2 Event A1 n(A1 and B1) n(A1 and B2) n(A1) n(A2 and B2) A2 n(A2) n(A2 and B1) N(S) Total n(B1) n(B2) Joint Probability Using Contingency Table Joint Probability Marginal (Simple) Probability

21. Event Total B1 B2 Event A1 P(A1 and B1) P(A1 and B2) P(A1) P(A2 and B2) A2 P(A2) P(A2 and B1) 1 Total P(B1) P(B2) Joint Probability Using Contingency Table Joint Probability Marginal (Simple) Probability

22. Computing Compound Probability • Probability of a compound event, A or B:

23. Event Total B1 B2 Event A1 P(A1 and B1) P(A1 and B2) P(A1) P(A2 and B2) A2 P(A2) P(A2 and B1) 1 Total P(B1) P(B2) Compound Probability (Addition Rule) P(A1 or B1 ) = P(A1) + P(B1) - P(A1 and B1) For Mutually Exclusive Events: P(A or B) = P(A) + P(B)

24. Computing Conditional Probability • The probability of event A given that event B has occurred:

25. Conditional Probability Event Total B B’ Event A P(A and B) P(A1 and B’) P(A) P(A’ and B’) A’ P(A) P(A’and B) 1 Total P(B) P(B’)

26. Conditional Probability Using Contingency Table Color Type Total Red Black 2 2 4 Ace 24 24 48 Non-Ace 26 26 52 Total Revised Sample Space

27. Example • A family has two children. What is the conditional probability that both are boys given that at least one of them is a boy ? Assume that the sample space S is given by S={(b,b),(b,g),(g,b),(g,g)}, and all outcomes are equally likely. [(b,g) means for instance that the older child is boy and the younger child is a girl.]

28. Letting E denote the event that both children are boys, and F the event that at least one of them is a boy, then the desired probability is given by Solution

29. Example • Bety can either take a course in mathematics or in statistics. If She takes the statistic course, then she will receive an A grade with probability ½ , while if she takes the math course then she will receive an A grade with prob. 1/3 . Bety decides to base her decision on the flip of fair coin. What is the prob that Bety will get an A in math ?

30. Solution • If we let F be the event that Bety takes math and E denote the event that she receives an A in whatever course she takes, then the prob is P(EF) = P(E|F)P(F) = 1/3.1/2 = 1/6. • P(F) =1/2 , because Bety decides to base her decision on the flip of fair coin.

31. Example • Suppose that each of three men at the party throws his hat into the center of the room. The hats are first mixed up and then each man randomly selects a hat. What is the probability that none of the three men selects his own hat ?

32. Solution • Let us denote by Ei ,i=1,2,3, the event that the ith man selects his own hat. The probability that none selects his own hat is • Now we compute

34. Conditional Probability and Statistical Independence • Conditional probability: • Multiplication rule:

35. Conditional Probability and Statistical Independence (continued) • Events A and B are independent if • Events A and B are independent when the probability of one event, A, is not affected by another event, B

36. C1 C2 Example • A series system of two components, C1 and C2. The probability C1 fail is 0.1 and C2 fail is 0.2 and both of them are independent. • The probability that the system fails is • P(C1 fail U C2 fail) =P(C1 fail) + P(C2 fail) - P(C1,C2 fail) = P(C1 fail) + P(C2 fail) - P(C1 fail)xP(C2 fail)

37. C1 C2 Example • A paralel system of two components, C1 and C2. The probability C1 fail is 0.1 and C2 fail is 0.2 and both of them are independent. • The probability that the system fails is • P(C1 fail and C2 fail) =P(C1 fail).P(C2 fail) =0.1x0.2 = 0.02

38. Total Probability • Let E and F be events. We may express E as E = EF U EF’ , since both of them are abviously mutually exclusive, we have that P(E) = P(E|F)P(F) + P(E|F’)P(F’) • If F can be separated by F1 , F2 , …, F k and each of them mutually exclusive then P(E) = P(E|F1)P(F1) + …+ P(E|Fk)P(Fk )

39. Bayes’s Theorem Adding up the parts of A in all the B’s Same Event

40. Bayes’s Theorem Using Contingency Table Fifty percent of borrowers repaid their loans. Out of those who repaid, 40% had a college degree. Ten percent of those who defaulted had a college degree. What is the probability that a randomly selected borrower who has a college degree will repay the loan? R = Repaid ; C = College

41. Bayes’s Theorem Using Contingency Table (continued) Repay Repay Total College .2 .05 .25 .3 .45 .75 College Total .5 .5 1.0

42. Example • In answering a question on a multiple choice test, a student either knowns the answer of he guesses . Let p be the prob that she knows the answer. There are m multiple-choice alternatives. What is the conditional that a student knew the answer to a question given that she answered it correctly ?

43. Let C and K denote respectively the event that the student answers the question correctly and the event that she actually knows the answer. Now Solution

44. Example • A laboratory blood test is 95 percent effective in detecting a certain disease when it is, in fact present. However, the test also yields a “ false positive” result for 1 percent of the healthy persons tested. If 0.5 percent of the population actually has the disease, what is the prob a person has the disease given that his test result is positive ?

45. Solution • Let D be the event that the tested person has the disease, and E the event that his test result is positive.

46. Random Variable • Random Variable • Outcomes of an experiment expressed numerically • e.g.: Toss a die twice; count the number of times the number 4 appears (0, 1 or 2 times)

47. Discrete Random Variable • Discrete random variable • Obtained by counting (1, 2, 3, etc.) • Usually a finite number of different values • e.g.: Toss a coin five times; count the number of tails (0, 1, 2, 3, 4, or 5 times)

48. Probability Distribution ValuesProbability 0 1/4 = .25 1 2/4 = .50 2 1/4 = .25 Discrete Probability Distribution Example Event: Toss two coins Count the number of tails T T T T

49. Example • Suppose we toss a coin having a prob p of coming up heads, until the first head appears. Letting N denote the number of flips required, then assuming that the outcome of successive flips are independent, N is a random variable taking on one of the values 1,2,3,…with respective probabilities

50. Solution • P(N=1) = P(H) = p; P(N=2) = P({T,H}) = (1-p)p ; : P(N=n) = P({T,…,T,H})= (1-p)n-1 p, n>=1 • As a check, note that