Statistics for Managers Using Microsoft Excel/SPSS Chapter 4 Basic Probability And Discrete Probability Distributions 1999 Prentice-Hall, Inc. Chap. 4-1
Chapter Topics Basic Probability Concepts: Sample Spaces and Events, Simple Probability, and Joint Probability, Conditional Probability Bayes Theorem The Probability of a Discrete Random Variable Binomial, Poisson, and Hypergeometric Distributions Covariance and its Applications in Finance 1999 Prentice-Hall, Inc. Chap. 4-2
Sample Spaces Collection of all Possible Outcomes e.g. All 6 faces of a die: e.g. All 52 cards of a bridge deck: 1999 Prentice-Hall, Inc. Chap. 4-3
Events Simple Event: Outcome from a Sample Space with 1 Characteristic e.g. A Red Card from a deck of cards. Joint or Compound Event: Involves 2 Outcomes Simultaneously e.g. An Ace which is also a Red Card from a deck of cards. An Ace given that it is a Red Card. 1999 Prentice-Hall, Inc. Chap. 4-4
Visualizing Events Contingency Tables Ace Not Ace Total Black 2 24 26 Red 2 24 26 Total 4 48 52 Tree Diagrams 1999 Prentice-Hall, Inc. Chap. 4-5
Simple Events The Event of a Happy Face There are 5 happy faces in this collection of 18 objects 1999 Prentice-Hall, Inc. Chap. 4-6
Joint Events The Event of a Happy Face AND Light Colored 3 Happy Faces which are light in color 1999 Prentice-Hall, Inc. Chap. 4-7
Compound Events The Event of Happy Face OR Light Colored 12 Items, 5 happy faces and 7 other light objects 1999 Prentice-Hall, Inc. Chap. 4-8
Special Events Null Event Club & Diamond on 1 Card Draw Null Event Complement of Event For Event A, All Events Not In A: A 1999 Prentice-Hall, Inc. Chap. 4-9
Dependent or Independent Events The Event of a Happy Face GIVEN it is Light Colored E = Happy Face Light Color 3 Items, 3 Happy Faces Given they are Light Colored 1999 Prentice-Hall, Inc. Chap. 4-10
Contingency Table Red Ace Red Black Total A Deck of 52 Cards Ace Not an Ace 2 24 2 24 Total 1999 Prentice-Hall, Inc. Chap. 4-11 26 26 4 48 52 Sample Space
Tree Diagram Event Possibilities Full Deck of Cards Red Cards Black Cards Ace Not an Ace Ace Not an Ace 1999 Prentice-Hall, Inc. Chap. 4-12
Probability Probability is the numerical measure of the likelihood that the event will occur. Value is between 0 and 1. Sum of the probabilities of all mutually exclusive events is 1. 1.5 0 Certain Impossible 1999 Prentice-Hall, Inc. Chap. 4-13
Computing Probability The Probability of an Event, E: P(E) Number of Event Outcomes Total Outcomes in Sample Space X T e.g. P( ) = 2/36 (There are 2 ways to get one 6 and the other 4) Each of the Outcome in the Sample Space equally likely to occur. 1999 Prentice-Hall, Inc. Chap. 4-14
Computing Joint Probability The Probability of a Joint Event, A and B: P(A and B) = Number of Event Outcomes from both A and B Total Outcomes in Sample Space e.g. P(Red Card and Ace) = 52 2 Re d Aces Total Number of Cards 1 26 1999 Prentice-Hall, Inc. Chap. 4-15
Joint Probability Using Contingency Table Event A 1 Event B 1 B 2 P(A 1 and B 1 ) P(A 1 and B 2 ) Total P(A 1 ) A 2 P(A 2 and B 1 ) P(A 2 and B 2 ) P(A 2 ) Total P(B 1 ) P(B 2 ) 1 Joint Probability Marginal (Simple) Probability 1999 Prentice-Hall, Inc. Chap. 4-16
Computing Compound Probability The Probability of a Compound Event, A or B: P( A or B ) Number of Event Outcomes from either Total Outcomes in Sample Space A or B e.g. P(Red Card or Ace) 4 Aces 26 Red Cards 2 Red Aces 52 Total NumberofCards 28 52 7 13 1999 Prentice-Hall, Inc. Chap. 4-17
Compound Probability Addition Rule P(A 1 or B 1 ) = P(A 1 ) +P(B 1 ) - P(A 1 and B 1 ) Event A 1 Event B 1 B 2 Total P(A 1 and B 1 ) P(A 1 and B 2 ) P(A 1 ) A 2 P(A 2 and B 1 ) P(A 2 and B 2 ) P(A 2 ) Total P(B 1 ) P(B 2 ) 1 For Mutually Exclusive Events: P(A or B) = P(A) + P(B) 1999 Prentice-Hall, Inc. Chap. 4-18
Computing Conditional Probability The Probability of the Event: Event A given that Event B has occurred P(A B) = P( A and P( B ) B ) e.g. P(Red Card given that it is an Ace) = 2 Red Aces 4 Aces 1 2 1999 Prentice-Hall, Inc. Chap. 4-19
Conditional Probability Using Contingency Table Conditional Event: Draw 1 Card. Note Kind & Color Color Type Red Black Total Ace 2 2 4 Non-Ace 24 24 48 Revised Sample Space Total 26 26 52 P(Ace Red) = P(Ace AND P(Red) Red) 2 26 / / 52 52 2 26 1999 Prentice-Hall, Inc. Chap. 4-20
Conditional Probability and Statistical Independence Conditional Probability: P(A B) = P( A and P( B ) B ) Multiplication Rule: Events are Independent: P(A and B) = P(A B) P(B) P(A B) = P(A) Or, P(A and B) = P(A) P(B) Events A and B are Independent when the probability of one event, A is not affected by another event, B. 1999 Prentice-Hall, Inc. Chap. 4-21
Bayes Theorem P(B i A) = P( AB 1 ) P(B P( AB ) P(B ) 1 i ) i P( AB k ) P(B k ) Same Event P(B anda) i P( A) Adding up the parts of A in all the B s 1999 Prentice-Hall, Inc. Chap. 4-22
Bayes Theorem: Contingency Table What are the chances of repaying a loan, given a college education? Loan Status Education Repay Default Prob. College.2.05.25??? No College Prob.?? 1 P(Collegeand Re pay) P(Collegeand Re pay) P(Collegeand Default) P(Repay. 08 College) = 1999 Prentice-Hall, Inc. Chap. 4-23
Discrete Random Variable Random Variable: represents outcomes of an experiment. e.g. Throw a die twice: Count the number of times 4 comes up (0, 1, or 2 times) Discrete Random Variable: Obtained by Counting (0, 1, 2, 3, etc.) Usually finite by number of values e.g. Toss a coin 5 times. Count the number of tails. (0, 1, 2, 3, 4, or 5 times) 1999 Prentice-Hall, Inc. Chap. 4-24
Discrete Probability Distribution Example Event: Toss 2 Coins. Count # Tails. T T T T Probability Distribution Values Probability 0 1/4 =.25 1 2/4 =.50 2 1/4 =.25 1999 Prentice-Hall, Inc. Chap. 4-25
Discrete Probability Distribution List of All Possible [ X i, P(X i ) ] Pairs X i = Value of Random Variable (Outcome) P(X i ) = Probability Associated with Value Mutually Exclusive (No Overlap) Collectively Exhaustive (Nothing Left Out) 0 P(X i ) 1 S P(X i ) = 1 1999 Prentice-Hall, Inc. Chap. 4-26
Discrete Random Variable Expected Value Summary Measures The Mean of the Probability Distribution Weighted Average m = E(X) = X i P(X i ) e.g. Toss 2 coins, Count tails, Compute Expected Value: m = 0.25 + 1.50 + 2.25 = 1 Variance Number of Tails Weighted Average Squared Deviation about Mean s 2 = E [ (X i - m ) 2 ]=S (X i - m ) 2 P(X i ) e.g. Toss 2 coins, Count tails, Compute Variance: s 2 = (0-1) 2 (.25) + (1-1) 2 (.50) + (2-1) 2 (.25) =.50 1999 Prentice-Hall, Inc. Chap. 4-27
Important Discrete Probability Distribution Models Discrete Probability Distributions Binomial Hypergeometric Poisson 1999 Prentice-Hall, Inc. Chap. 4-28
Binomial Probability Distributions n Identical Trials, e.g. 15 tosses of a coin, 10 light bulbs taken from a warehouse 2 Mutually Exclusive Outcomes, e.g. heads or tails in each toss of a coin, defective or not defective light bulbs Constant Probability for each Trial, e.g. probability of getting a tail is the same each time we toss the coin and each light bulb has the same probability of being defective 1999 Prentice-Hall, Inc. Chap. 4-29
Binomial Probability Distributions 2 Sampling Methods: Infinite Population Without Replacement Finite Population With Replacement Trials are Independent: The Outcome of One Trial Does Not Affect the Outcome of Another 1999 Prentice-Hall, Inc. Chap. 4-30
Binomial Probability Distribution Function P(X) n! X! n X p ( 1 ( )! p ) X n X P(X) = probability that X successes given a knowledge of n and p X = number of successes in sample, (X = 0, 1, 2,..., n) p = probability of success n = sample size Tails in 2 Toss of Coin X P(X) 0 1/4 =.25 1 2/4 =.50 2 1/4 =.25 1999 Prentice-Hall, Inc. Chap. 4-31
Binomial Distribution Characteristics Mean m Standard Deviation s np E ( X ) np e.g. m = 5 (.1) =.5 ( p ) e.g. s = 5(.5)(1 -.5) = 1.118 P(X).6.4.2 0 P(X).6.4.2 0 n = 5 p = 0.1 0 1 2 3 4 5 n = 5 p = 0.5 0 1 2 3 4 5 X X 1999 Prentice-Hall, Inc. Chap. 4-32
Poisson Distribution Poisson Process: Discrete Events in an Interval The Probability of One Success in Interval is Stable The Probability of More than One Success in this Interval is 0 Probability of Success is Independent from Interval to Interval e.g. # Customers Arriving in 15 min. # Defects Per Case of Light Bulbs. P ( X x l l e - 1999 Prentice-Hall, Inc. Chap. 4-33 x! l x
Poisson Probability Distribution Function P ( X ) e X P(X ) = probability of X successes given l l = expected (mean) number of successes e = 2.71828 (base of natural logs) X = number of successes per unit l X l! e.g. Find the probability of 4 customers arriving in 3 minutes when the mean is 3.6. P(X) = e -3.6 3.6 4! 4 =.1912 1999 Prentice-Hall, Inc. Chap. 4-34
Poisson Distribution Characteristics Mean m E ( X ) l Standard Deviation s N i 1 l X P ( X ) i i P(X).6.4.2 0 P(X).6.4.2 0 l = 0.5 0 1 2 3 4 5 l = 6 0 2 4 6 8 10 X X 1999 Prentice-Hall, Inc. Chap. 4-35
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 1999 Prentice-Hall, Inc. Chap. 4-36
Hypergeometric Distribution ( )( ) P ( X ) X n - X ( N ) P(X) = probability that X successes given n, N, and A n = sample size N = population size A = number of successes in population X = number of successes in sample (X = 0, 1, 2,..., n) A N - A 1999 Prentice-Hall, Inc. Chap. 4-37 n 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? 4 6 P(2) = ( 2 )( 1 ) ( ) 10 3 =.30
Hypergeometric Characteristics Mean m E ( X ) n A N Standard Deviation Finite Population Correction s na ( N N 2 A ) N n N 1 1999 Prentice-Hall, Inc. Chap. 4-38
Covariance s XY N X E( X ) Y E(Y ) P( X Y ) i 1 i i i i X = discrete random variable X X i = ith outcome of X P(X i Y i ) = probability of occurrence of the ith outcome of Y Y = discrete random variable Y Y i = ith outcome of Y i = 1, 2,, N 1999 Prentice-Hall, Inc. Chap. 4-39
Computing the Mean for Investment Returns Return per $1,000 for two types of investments Investment 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 E(X) = m X = (-100)(.2) + (100)(.5) + (250)(.3) = $105 E(Y) = m Y = (-200)(.2) + (50)(.5) + (350)(.3) = $90 1999 Prentice-Hall, Inc. Chap. 4-40
Computing the Variance for Investment Returns Investment 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 2 s X Var(X) = = (.2)(-100-105 )2 + (.5)(100-105) 2 + (.3)(250-105) 2 2 s Y = 14,725, s X = 121.35 Var(Y) = = (.2)(-200-90 )2 + (.5)(50-90) 2 + (.3)(350-90 )2 = 37,900, s Y = 194.68 1999 Prentice-Hall, Inc. Chap. 4-41
Computing the Covariance for Investment Returns Investment 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 s XY = (.2)(-100-105)(-200-90) + (.5)(100-105)(50-90) + (.3)(250-105)(350-90) = 23,300 The Covariance of 23,000 indicates that the two investments are strongly related and will vary together in the same direction. 1999 Prentice-Hall, Inc. Chap. 4-42
Chapter Summary Discussed Basic Probability Concepts: Sample Spaces and Events, Simple Probability, and Joint Probability Defined Conditional Probability Discussed Bayes Theorem Addressed the Probability of a Discrete Random Variable Discussed Binomial, Poisson, and Hypergeometric Distributions Addressed Covariance and its Applications in Finance 1999 Prentice-Hall, Inc. Chap. 4-43