Chapter 4 Basic Probability And Probability Distributions

# Chapter 4 Basic Probability And Probability Distributions

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## Chapter 4 Basic Probability And Probability Distributions

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1. Business Statistics A First Course (3rd Edition) Chapter 4 Basic Probability And Probability Distributions

2. Chapter Topics • Basic Probability Concepts: • Sample Spaces and Events, Simple • Probability, and Joint Probability, • Conditional Probability • Bayes’ Theorem • The Probability Distribution for a Discrete Random Variable

3. Chapter Topics • Binomial and Poisson Distributions • Covariance and its Applications in Finance • The Normal Distribution • Assessing the Normality Assumption

4. Sample Spaces Collection of all Possible Outcomes e.g. All 6 faces of a die: e.g. All 52 cards of a bridge deck:

5. Events • Simple Event:Outcome from a Sample Space • with 1 Characteristic • e.g. A Red Card from a deck of cards. • Joint Event: Involves 2 Outcomes Simultaneously • e.g. An Acewhich is also aRed Cardfrom 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

7. Simple Events The Event of a Happy Face There are 5 happy faces in this collection of 18 objects

8. Joint Events The Event of a Happy Face AND Light Colored 3Happy Faces which are light in color

9.  Special Events Null Event • Null event • Club & diamond on 1 card draw • Complement of event • For event A, • All events not In A:

10. Dependent or Independent Events The Event of a Happy Face GIVEN it is Light Colored E = Happy FaceLight Color 3Items:3 Happy Faces Given they are Light Colored

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. Contingency Table 2500 Employees of Company ABC Agree Neutral Opposed | Total MALE 200 400 | 1500 900 300 600 | 1000 FEMALE 100 2500 1000 | Total 1200 300 Sample Space

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

14. 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 and collective exhaustive events is 1. 1 Certain .5 0 Impossible

15. Computing Probability • The Probability of an Event, E: • Each of the Outcome in the Sample Space • equally likely to occur. Number of Event Outcomes P(E) = Total Number of Possible Outcomes in the Sample Space X e.g. P() = 2/36 = T (There are 2 ways to get one 6 and the other 4)

16. 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 Number of Possible Outcomes in Sample Space e.g. P(Red CardandAce) =

17. Joint Probability Using Contingency Table Event Total B1 B2 Event A1 P(A1andB1) P(A1andB2) P(A1) P(A2andB2) A2 P(A2) P(A2 and B1) 1 Total P(B1) P(B2) Marginal (Simple) Probability Joint Probability

18. Computing Compound Probability The Probability of a Compound Event, A or B: e.g. P(Red Card orAce)

19. Contingency Table 2500 Employees of Company ABC Agree Neutral Opposed | Total MALE 200 400 | 1500 900 300 600 | 1000 FEMALE 100 2500 1000 | Total 1200 300 Sample Space

20. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal

21. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24

22. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24 • 2. Neutral

23. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24 • 2. Neutral 300/2500 = 0.12

24. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24 • 2. Neutral 300/2500 = 0.12 • 3. Opposed to the proposal, GIVEN that • the employee selected is a female

25. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24 • 2. Neutral 300/2500 = 0.12 • 3. Opposed to the proposal, GIVEN that • the employee selected is a female 600/1000 = 0.60

26. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24 • 2. Neutral 300/2500 = 0.12 • 3. Opposed to the proposal, GIVEN that • the employee selected is a female 600/1000 = 0.60 • 4. Either a female or opposed to the • proposal

27. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24 • 2. Neutral 300/2500 = 0.12 • 3. Opposed to the proposal, GIVEN that • the employee selected is a female 600/1000 = 0.60 • 4. Either a female or opposed to the • proposal ……….. 1000/2500 + 1000/2500 - 600/2500 = • 1400/2500 = 0.56

28. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24 • 2. Neutral 300/2500 = 0.12 • 3. Opposed to the proposal, GIVEN that • the employee selected is a female 600/1000 = 0.60 • 4. Either a female or opposed to the • proposal ……….. 1000/2500 + 1000/2500 - 600/2500 = • 1400/2500 = 0.56 • 5. Are Gender and Opinion (statistically) independent?

29. The pervious table refers to 2500 employees of ABC Company, classified by gender and by opinion on a company proposal to emphasize fringe benefits rather than wage increases in an impending contract discussion • Calculate the probability that an employee selected (at random) from this group will be: • 1. A female opposed to the proposal 600/2500 = 0.24 • 2. Neutral 300/2500 = 0.12 • 3. Opposed to the proposal, GIVEN that • the employee selected is a female 600/1000 = 0.60 • 4. Either a female or opposed to the • proposal ……….. 1000/2500 + 1000/2500 - 600/2500 = • 1400/2500 = 0.56 • 5. Are Gender and Opinion (statistically) independent? • For Opinion and Gender to be independent, the joint probability of each pair of A events (GENDER) and B events (OPINION) should equal the product of the respective unconditional probabilities….clearly this does not hold…..check, e.g., the prob. Of MALE and IN FAVOR against the prob. of MALE times the prob. of IN FAVOR …they are not equal….900/2500 does not equal 1500/2500 * 1200/2500

30. Compound ProbabilityAddition Rule P(A1orB1 ) = P(A1) +P(B1) - P(A1 andB1) Event Total B1 B2 Event A1 P(A1and B1) P(A1and B2) P(A1) P(A2 andB1) A2 P(A2 and B2) P(A2) 1 Total P(B1) P(B2) For Mutually Exclusive Events: P(A or B) = P(A) + P(B)

31. Computing Conditional Probability The Probability of Event Agiven thatEvent B has occurred: P(A B) = e.g. P(Red Cardgiven that it isanAce) =

32. Conditional Probability Using Contingency Table Conditional Event: Draw 1 Card. Note Kind & Color Color Type Total Red Black Revised Sample Space 2 2 4 Ace 24 24 48 Non-Ace 26 26 52 Total

33. Conditional Probability and Statistical Independence Conditional Probability: P(AB) = Multiplication Rule: P(A and B) = P(AB) • P(B) = P(BA) • P(A)

34. Conditional Probability and Statistical Independence (continued) Events are Independent: P(AB) = P(A) Or, P(BA) = P(B) 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.

35. Bayes’ Theorem P(Bi A) = Adding up the parts of A in all the B’s Same Event

36. A manufacturer of VCRs purchases a particular microchip, called the LS-24, from three suppliers: Hall Electronics, Spec Sales, and Crown Components. Thirty percent of the LS-24 chips are purchased from Hall, 20% from Spec, and the remaining 50% from Crown. The manufacturer has extensive past records for the three suppliers and knows that there is a 3% chance that the chips from Hall are defective, a 5% chance that chips from Spec are defective and a 4% chance that chips from Crown are defective. When LS-24 chips arrive at the manufacturer, they are placed directly into a bin and not inspected or otherwise identified as to supplier. A worker selects a chip at random. What is the probability that the chip is defective?   A worker selects a chip at random for installation into a VCR and finds it is defective. What is the probability that the chip was supplied by Spec Sales?

37. Bayes’ Theorem:Contingency Table What are the chances of repaying a loan, given a college education? Loan Status Prob. Education Repay Default .2 .05 .25 College ? ? ? No College ? ? 1 Prob. P(College and Repay)  P(RepayCollege) = P(College and Repay) + P(College and Default) .20 = = .80 .25

38. Discrete Random Variable • Random Variable: outcomes of an experiment expressed numerically • e.g. • Throw a die twice: Count the number of times 4 comes up (0, 1, or 2 times)

39. Discrete Random Variable • Discrete Random Variable: • Obtained by Counting (0, 1, 2, 3, etc.) • Usually finite by number of different values • e.g. • Toss a coin 5 times. Count the number of tails.(0, 1, 2, 3, 4, or 5 times)

40. Discrete Probability Distribution Example Event: Toss 2 Coins. Count # Tails. • Probability distribution • Valuesprobability • 0 1/4 = .25 • 1 2/4 = .50 • 2 1/4 = .25 T T T T

41. Discrete Probability Distribution • List of all possible [ xi, p(xi) ] pairs • Xi = value of random variable • P(xi) = probability associated with value • Mutually exclusive (nothing in common) • Collectively exhaustive (nothing left out) • 0 p(xi)  1 • P(xi) = 1

42. Discrete Random Variable Summary Measures • Expected value (The mean) • Weighted average of the probability distribution •  = E(X)= xi p(xi) • E.G. Toss 2 coins, count tails, compute expected value: • = 0  .25 + 1 .50 + 2  .25 = 1 Number of Tails

43. Discrete Random Variable Summary Measures • Variance • Weighted average squared deviation about mean •  = E [ (xi - )2]= (xi -  )2p(xi) • E.G. Toss 2 coins, count tails, compute variance: •  = (0 - 1)2(.25) + (1 - 1)2(.50) + (2 - 1)2(.25) = .50

44. Important Discrete Probability Distribution Models Discrete Probability Distributions Binomial Poisson

45. Binomial Distributions • ‘N’ identical trials • E.G. 15 tosses of a coin, 10 light bulbs taken from a warehouse • 2 mutually exclusive outcomes on each trial • E.G. Heads or tails in each toss of a coin, defective or not defective light bulbs

46. Binomial Distributions • 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 • 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

47. Binomial Probability Distribution Function n ! X  X n P(X)  1 ) p (  p X ! (  ) ! n X P(X) = probability that Xsuccesses given a knowledge of n and p X= number of ‘successes’ insample, (X = 0, 1, 2, ...,n) p = probability of each ‘success’ n = sample size Tails in 2 Tosses of Coin XP(X) 0 1/4 = .25 1 2/4 = .50 2 1/4 = .25

48. Binomial Distribution Characteristics n = 5p = 0.1 P(X) Mean .6  E ( X )   np .4 .2 e.g. = 5 (.1) = .5 0 X 0 1 2 3 4 5 Standard Deviation n = 5p = 0.5 P(X) )    p np ( 1 .6 .4 .2 e.g.=5(.5)(1 - .5) = 1.118 X 0 0 1 2 3 4 5

49. P ( X  x |  -  x e  x ! Poisson Distribution • Poisson process: • Discrete events in an ‘interval’ • The probability ofone successin an interval isstable • 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

50. Poisson Distribution Function  X e  P ( X )  X ! P(X) = probability ofXsuccesses given = expected (mean) number of ‘successes’ e = 2.71828 (base of natural logs) X= number of ‘successes’per unit e.g. Find the probability of4customers arriving in 3 minutes when the mean is3.6. -3.6 4 e 3.6 P(X) = =.1912 4!