1. Statistics for Managers Using Microsoft Excel (3 rd 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 The probability of a discrete random variable Covariance and its applications in finance Binomial distribution Poisson distribution Hypergeometric distribution (continued)

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 Red 2 24 26 Black 2 24 26 Total 4 48 52 Ace Not Ace Total Full Deck of Cards Red Cards Black Cards Not an Ace Ace Not an Ace

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

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

9. Special Events 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   Null Event

10. Special Events 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 (continued)

11. Contingency Table A Deck of 52 Cards Ace Not an Ace Total Red Black Total 224 2 26 44852 Sample Space Red Ace

12. Full Deck of Cards Tree Diagram Event Possibilities Red Cards Black Cards Ace Not an Ace Ace 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 Certain Impossible.5 1 0

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

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) AB A=(AB)U(AB’) P(A)=P(AB)+P(AB’) P(AUB)=P(B)+P(AB’) = P(B)+P(A)-P(AB)

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

17. Properties of Probability

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

20. Joint Probability Using Contingency Table Joint Probability Marginal (Simple) Probability n(A 1 and B 2 )n(A 1 ) Total Event n(A 2 and B 1 ) n(A 1 and B 1 ) Event Total N(S) A1A1 A2A2 B1B1 B2B2 n(B 1 ) n(B 2 ) n(A 2 and B 2 ) n(A 2 )

21. Joint Probability Using Contingency Table Joint Probability Marginal (Simple) Probability P(A 1 and B 2 )P(A 1 ) Total Event P(A 2 and B 1 ) P(A 1 and B 1 ) Event Total 1 A1A1 A2A2 B1B1 B2B2 P(B 1 ) P(B 2 ) P(A 2 and B 2 ) P(A 2 )

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

23. Compound Probability (Addition Rule) P(A 1 or B 1 ) = P(A 1 ) + P(B 1 ) - P(A 1 and B 1 ) P(A 1 ) P(B 2 ) P(A 1 and B 1 )P(A 1 and B 2 ) Total Event P(A 2 and B 1 ) Event Total 1 A1A1 A2A2 B1B1 B2B2 P(B 1 ) P(A 2 and B 2 ) P(A 2 ) 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. P(A) P(B’) P(A and B) Conditional Probability P(A 1 and B’) Total Event P(A’ and B) Event Total 1 A A’ B B’ P(B) P(A’ and B’) P(A)

26. Conditional Probability Using Contingency Table Black Color Type Red Total Ace 224 Non-Ace 24 48 Total 26 52 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. Solution 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

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 E i,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 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 (continued)

36. Example A series system of two components, C 1 and C 2. The probability C 1 fail is 0.1 and C 2 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) C1C2

37. Example A paralel system of two components, C 1 and C 2. The probability C 1 fail is 0.1 and C 2 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 C1 C2

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 F 1, F 2, …, F k and each of them mutually exclusive then P(E) = P(E|F 1 )P(F 1 ) + …+ P(E|F k )P(F k )

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 College 1.0.5.2.3.05.45.25.75 Total

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. Solution 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

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 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 Values Probability 01/4 =.25 12/4 =.50 21/4 =.25 Discrete Probability Distribution Example T T TT Event: Toss two coinsCount the number of tails

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

51. Discrete Probability Distribution List of all possible [X j, p(X j ) ] pairs X j = value of random variable P(X j ) = probability associated with value Mutually exclusive (nothing in common) Collectively exhaustive (nothing left out)

52. Summary Measures Expected value (the mean) Weighted average of the probability distribution e.g.: Toss 2 coins, count the number of tails, compute expected value

53. Summary Measures Variance Weight average squared deviation about the mean e.g. Toss two coins, count number of tails, compute variance (continued)

54. Covariance and its Application

55. Correlation The correlation coefficient of X and Y is

56. Computing the Mean for Investment Returns Return per $1,000 for two types of investments P(X i Y i ) Economic condition Dow Jones fund X Growth Stock Y.2 Recession-$100 -$200.5 Stable Economy+ 100 + 50.3 Expanding Economy + 250 + 350 Investment

57. Computing the Variance for Investment Returns P(X i Y i ) Economic condition Dow Jones fund X Growth Stock Y.2 Recession-$100 -$200.5 Stable Economy+ 100 + 50.3 Expanding Economy + 250 + 350 Investment

58. Computing the Covariance for Investment Returns P(X i Y i ) Economic condition Dow Jones fund X Growth Stock Y.2 Recession-$100 -$200.5 Stable Economy+ 100 + 50.3 Expanding Economy + 250 + 350 Investment The Covariance of 23,000 indicates that the two investments are positively related and will vary together in the same direction.

59. Correlation The correlation coefficient of X and Y is If the value of X increase, then the value of Y increase too.

60. Cumulative Distribution Function The cumulative distribution function of a random variable X is defined for any real x by

61. Example Consider the distribution of lifetimes, X (in months), of a particular type of component. We will assume that the CDF has the form The median lifetime is

62. It is desired to find the time t such that 10% of the component fail before t. This is the 10 th percentile : Thus if the components are guaranteed for one month, slightly more than 10% will need to be replaced

63. Important Discrete Probability Distributions Discrete Probability Distributions BinomialHypergeometricPoisson

64. Binomial Probability Distribution ‘n’ identical trials e.g.: 15 tosses of a coin; ten light bulbs taken from a warehouse Two mutually exclusive outcomes on each trials e.g.: Head or tail in each toss of a coin; defective or not defective light bulb Trials are independent The outcome of one trial does not affect the outcome of the other

65. Binomial Probability Distribution Constant probability for each trial e.g.: Probability of getting a tail is the same each time we toss the coin Two sampling methods Infinite population without replacement Finite population with replacement (continued)

66. Binomial Probability Distribution Function Tails in 2 Tosses of Coin X P(X) 0 1/4 =.25 1 2/4 =.50 2 1/4 =.25

67. Proof of the Probability Note that, by the binomial theorem, the probabilities sum to one, that is

68. Binomial Distribution Characteristics Mean E.g. Variance and Standard Deviation E.g. n = 5 p = 0.1 0.2.4.6 012345 X P(X)

69. Expectation

70. Variance

71. Binomial Distribution in PHStat PHStat | probability & prob. Distributions | binomial Example in excel spreadsheet

72. Example S uppose that an airplane engine will fall, when in flight, with prob 1-p independently from engine to engine; suppose that the airplane will make a succesful flight if at least 50 percent of its engines remain operative. For what values of p is a four-engine plane preferable to a two-engine plane ?

73. Solution The probaility that a four-engine plane makes a successful flight is Whereas the corresponding probability for a two-engine plane is

74. Solution Hence the four-engine is safer if Hence, the four-engine plane is safer when the engine success probability is at least as large as 2/3, whereas the two-engine plane is safer when this probability falls below 2/3

75. Poisson Distribution Poisson Process: Discrete events in an “interval” The probability of One Success in an interval is stable The probability of More than One Success in this interval is 0 The probability of success is independent from interval to interval e.g.: number of customers arriving in 15 minutes e.g.: number of defects per case of light bulbs PXx x x (| !  e -

76. Poisson Probability Distribution Function e.g.: Find the probability of 4 customers arriving in 3 minutes when the mean is 3.6.

77. Poisson Distribution in PHStat PHStat | probability & prob. Distributions | Poisson Example in excel spreadsheet

78. Poisson Distribution Characteristics Mean Standard Deviation and Variance  = 0.5  = 6 0.2.4.6 012345 X P(X) 0.2.4.6 0246810 X P(X)

79. Approximate a binomial to poisson An important property of the poisson random variable is that it may be used to approximate a binomial random variabel when the binomial parameter n is large and p is small. To see this, suppose that X is a binomial r.v. with parameters (n,p), and let µ = np. Then

80. Proof

81. Expectation

82. Variance

83. Hypergeometric Distribution “n” trials in a sample taken from a finite population of size N Sample taken without replacement Trials are dependent Concerned with finding the probability of “X” successes in the sample where there are “A” successes in the population

84. Hypergeometric Distribution Function E.g. 3 Light bulbs were selected from 10. Of the 10 there were 4 defective. What is the probability that 2 of the 3 selected are defective?

85. Hypergeometric Distribution Characteristics Mean Variance and Standard Deviation Finite Population Correction Factor

86. Hypergeometric Distribution in PHStat PHStat | probability & prob. Distributions | Hypergeometric … Example in excel spreadsheet

87. Expectation

88. Variance

89. Jointly Distributed Random Variables The joint probability mass function of X and Y is p(x,y)=P(X=x,Y=y) The probability mass function of X The probability mass function of Y

90. Expectation

91. Example As another example of the usefulness of equation above, let us use it to obtain the expectation of a binomial r.v.

92. Example At a party N men throw their hats into the center of a room. The hats are mixed up and each man randomly selects one. Find the expected number of men that select their own hats

93. Solution Letting X denote the number of men that select their own hats, we can best compute E(X) by noting that X = X 1 +…+X N ; where X i is indicator function if the ith man select his own hat. So P(X i = 1) = 1/N. And so E(X i ) = 1/N. Hence We obtain that E(X) = 1. So, no matter how many people are at the party, on the average exactly one of the men will select his own hat.

94. Independent R.V X and Y are independent if

95. Covariance and Variance of Sums of Random Variables The covariance of any two random variables X and Y, denoted by Cov(X,Y), is defined by Cov(X,Y)= E[(X-E[X])(Y-E[Y])] = E(XY)-E(X)E(Y) If X and Y are independent Cov(X,Y) = 0

96. Properties of Covariance Cov(x,X) = Var(X) Cov(cX,Y) = c Cov(X,Y) Cov(X,Y+Z)= E[X(Y+Z)]-E[X]E[Y+Z] = E[XY]-E[X]E[Y] + E[XZ]-E[X]E[Z] = (Cov(X,Y) + Cov(X,Z) The last property easily generalizes to give

97. Variance of Sum Variabel

98. Proposition Suppose that X 1,…,X n are independent and identically distributed with expected value µ and variance σ 2. Then

99. Example Sums of independent Poisson Random Variables : Let A and Y be independent Poisson random variables wirh respective means λ 1 and λ 2. Calculate the distribution of X + Y. Solution : Since the event {X+Y = n} may be written as the union of the disjoint events {X=k,Y=n-k}, 0≤k≤n, we have