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Probability Theory and Distribution

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  1. 1. Probabilities (What are your chances?) Presented by: Sheryl B. Satorre ( --- TM 705 – Operations Research
  2. 2. Terminology:• Experiment • A systematic investigation where the answer is 1/26/2013 unknown• Trial Prepared by: SB Satorre • One specific instance of an experiment• Outcome • Result of a single trial 2
  3. 3. Terminology:• Event • A selected outcome 1/26/2013• Sample Space Prepared by: SB Satorre • Set of all possible outcomes of an experiement 3
  4. 4. Games of Chance Games of chance commonly involve the toss of a coin, the roll of a die, or the use of a pack of cards. Examples:• Rolling a six-sided die will generate 1/26/2013 S {1, 2,3, 4,5, 6} Prepared by: SB Satorre• If two dice are rolled, the following will be the Sample Space 4
  5. 5. Probability Concepts• Probability is the study of random events.• The probability, or chance, that an event will happen can be 1/26/2013 described by a number between 0 and 1: Prepared by: SB Satorre • A probability of 0, or 0%, means the event has no chance of happening. • A probability of 1/2 , or 50%, means the event is just as likely to happen as not to happen. • A probability of 1, or 100%, means the event is certain to happen. 5
  6. 6. You can represent the probability of an event by 1/26/2013 marking it on a number line like this one Prepared by: SB SatorreImpossible 50 – 50 Chance Certain 0 = 0% ½ , .5, 50% 1 = 100% 6
  7. 7. General Formula Let S be a Sample Space E be an Event 1/26/2013 P(E) be the Probability of event E Prepared by: SB SatorreP(E) = the number of ways event E will occur (/E/) Total number of possible outcomes (/S/) 7
  8. 8. Any other probability is between 0 and 1.In symbol: 0 <= P(A) <= 1 for any event Aand that 1/26/2013 P( f ) + P( s ) = 1 for any eventSo that, Prepared by: SB Satorre P ( f ) = 1 – P( s ) f means failure P ( s ) = 1 – P( f ) s means success 8
  9. 9. Probability ValuesA fair die will have each of the six outcomes equally likely. 1/26/2013 1 P(1) P(2) P(3) P(4) P(5) P(6) 6 Prepared by: SB SatorreAn example of a biased die would be one of which 9
  10. 10. If two dice are thrown and each of the 36 outcomes is equally likely ( as will be the case 1/26/2013 two fair dice that are shaken properly), the Prepared by: SB Satorre probability value of each outcome will necessarily be 1/36. 10
  11. 11. Basic Rules of Probability1. If A is any event, then 0 <= P(A) <= 1.2. The probability of a null space P(Ø) = 0 while 1/26/2013 the probability of the sample space P(S) = 1. Prepared by: SB Satorre3. If an experiment can result in any one of N different equally likely outcomes and if exactly n of these outcomes correspond to event A, then the probability of event A is P(A) = n/N. 11
  12. 12. EVENTS AND COMPLEMENTS Prepared by: SB Satorre 1/26/2013 12
  13. 13. EventsAn event A is a subset of the sample space S. Itcollects outcomes of particular interest. 1/26/2013The probability of an event is obtained by summing the probabilities of the outcomes Prepared by: SB Satorre contained within the event A.An event is said to occur if one of the outcomescontained within the event occurs. 13
  14. 14. ComplementThe event A’ , the complement of event A, is the event consisting of everything in the sample 1/26/2013 space S that is not contained within the event A. In all cases Prepared by: SB Satorre P( A) P( A ) 1• Events that consist of an individual outcome are sometimes referred to as elementary events or 14 simple events
  15. 15. Example: ComplementSuppose that twodice arethrown, what isthe probability 1/26/2013that at least oneof the two dice is Prepared by: SB Satorre6? 11P( B) 36 11 25P( B ) 1 36 36 15
  16. 16. Properties of Events• Mutually Exclusive Events• Non-Mutually Exclusive Events 1/26/2013• Independent Events Prepared by: SB Satorre• Dependent Events 16
  17. 17. Mutually Exclusive EventsLet A and B be Events. 1/26/2013Events A and B are said to be mutually exclusive if Prepared by: SB Satorreand only if they cannot happen at the same time.Examples:• Movement: Turning left and turning right• Tossing a coin: Heads and Tails• Drawing a card: King and Queen 17
  18. 18. Non-Mutually Exclusive EventsLet A and B be Events. 1/26/2013Events A and B are said to be non-mutuallyexclusive if and only if they can happen at the Prepared by: SB Satorresame time.Examples:• Movement: Turning left and turning right• Drawing a Card: King and Heart 18
  19. 19. Independent EventsLet A and B be events. 1/26/2013If event A cannot influence the outcome of eventB, then two events are said to be independent. Prepared by: SB SatorreExample: Flipping a coin 5 times• The chance that HEAD will occur does not affect the occurrence of the TAIL from one trial to another. 19
  20. 20. Dependent EventsLet A and B be events. 1/26/2013If event A can influence the occurrence of eventB, then these two events are said to be Prepared by: SB Satorredependent.Example: Drawing 5 balls from the box• The chance of the next ball to be drawn is dependent on the outcome of the previous draw. 20
  21. 21. THE ADDITIVE RULE Prepared by: SB Satorre 1/26/2013 21
  22. 22. Mutually Exclusive EventsLet A and B be Mutually Exclusive events. 1/26/2013 P (either A or B) = P (A) + P (B) Prepared by: SB Satorre 22
  23. 23. Example: Mutually Exclusive EventsLet P(A) and P(B) be the probabilities of getting the sum of 6, and thesum of 8, respectively. (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) 1/26/2013 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) P(A) = 5/36 P(B) = 5/36 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) Prepared by: SB Satorre (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) P(A or B) = P(A) + P(B) P(A or B) = (5/36) + (5/36) 23 P(A or B) = 5/18
  24. 24. Non-Mutually Exclusive EventsLet A and B be Non-Mutually Exclusive events. 1/26/2013P (either A or B) = P (A) + P (B) – P(A and B) Prepared by: SB Satorre 24
  25. 25. Example: Non-Mutually ExclusiveEventsIf a card is drawn from an ordinary well-shuffled deck ofcards, what is the probability that it is either a spade or a 1/26/2013face card? Let A be an event of drawing a SPADE Prepared by: SB Satorre Let B be an event of drawing a FACE card Let P(A or B) be the probability of getting a Spade or a Face card P(A or B) = P(A) + P(B) – P(A and B) P(A or B) = (13/52) + (12/52) – (3/52) 25 P(A or B) = 11/26
  26. 26. MULTIPLICATION OF EVENTS Prepared by: SB Satorre 1/26/2013 26
  27. 27. Independent EventsIf two events A and B are independent, that is, P(B/A) =P(B) or P(A/B) = P(A) then the probability that both A and B 1/26/2013occurs is Prepared by: SB Satorre P(A and B) = P(A) x P(B) 27
  28. 28. Example: Independent EventsIf a coin is tossed two times in succession, what is theprobability that a head appears in the first toss and a tail in 1/26/2013the second? P(HT) = P(H) * P(T) Prepared by: SB Satorre st nd 1 toss 2 toss P(HT) = (1/2) * (1/2) P(HT) = 1/4 28
  29. 29. Dependent EventsLet A and B be dependent events.If events A and B can both occur then the probability that 1/26/2013both A and B occur is Prepared by: SB Satorre P(A and B) = P(A) x P(B/A) 29
  30. 30. Example: Dependent EventsA box contains one dozen Casio calculators, 3 of which aredefective and the rest are good. What is the probabilitythat if 3 calculators are drawn at random without 1/26/2013replacement, they will be defective? Let D = 3 total defective Casio calculators Prepared by: SB Satorre Let D’ = 9 total non-defective Casio calculators Let P(3D) be the chance that the 3 calculators drawn are all defective P(3D) = P(1D) * P(2D/1D) * P(3D/2D) P(3D) = (3/12) * (2/12) * (1/12) P(3D) = 1/288 30
  31. 31. CONDITIONAL PROBABILITY Prepared by: SB Satorre 1/26/2013 31
  32. 32. Basic ConceptsConditional probability refers to the probability ofan event B occurring given the condition that 1/26/2013some other event A has occurred. Prepared by: SB SatorreThis is denoted by the symbol P(B/A) which isusually read as “the probability that B occurs giventhat A occurs” or simply, “the probability ofB, given A”. 32
  33. 33. Conditional Probability The conditional probability of event A conditional on event B is P( A B) P( A | B) P( B) 1/26/2013 for P(B)>0. It measures the probability that event A occurs when it is known that event B occurs. Prepared by: SB Satorre A B P( A B) 0 P( A | B) 0 P( B) P( B) B A A B B P( A B) P( B) 33 P( A | B) 1 P( B) P( B)
  34. 34. Example: Conditional ProbabilityThe probability that the lady of the house is homewhen the Avon representative calls is 0.60. Giventhat the lady of the house is home, the probability 1/26/2013that she makes a purchase is 0.40. Find theprobability that the lady of the house is home and Prepared by: SB Satorremakes a purchase when the Avon representativecalls. Given: P(H) = 0.60 ; the chance that the lady of the house is home when Avon rep. calls P(P|H) = 0.40 ; the chance that the lady of the house makes a purchase given that she is home 34
  35. 35. Given:P(H) = 0.60 ; the chance that the lady of the house is homewhen Avon rep. callsP(P|H) = 0.40 ; the chance that the lady of the house makes apurchase given that she is home 1/26/2013Required: Prepared by: SB SatorreP(H and P) = the chance that the lady of the house is home andmakes purchase P(P|H) = P(H and P) P(H) 0.40 = P(H and P) 0.60 35 P(H and P) = 0.24
  36. 36. Probability Tree• a.k.a Tree Diagram 1/26/2013• Is a decision-tool that helps visualize all Prepared by: SB Satorre possible choices with their corresponding chances of outcomes in order to avoid mathematical errors 36
  37. 37. Probability Tree: Car WarrantiesA company sells a certain type of car, which it assembles in one of four possible locations. Plant 1/26/2013 I supplies 20%; plant II, 24%; plant III, 25%; and plant IV, 31%. Prepared by: SB SatorreA customer buying a car does not know where the car has been assembled, and so the probabilities of a purchased car being from each of the four plants can be thought of as being 0.20, 0.24, 0.25, and 0.31. 37
  38. 38. Probability Tree: Car Warranties(cont.)Each new car sold carries a 1-year bumper-to-bumper warranty. 1/26/2013 P( claim | plant I ) = 0.05, P( claim | plant II ) = 0.11 P( claim | plant III ) = 0.03, P( claim | plant IV ) = 0.08 Prepared by: SB SatorreFor example, a car assembled in plant I has a probability of 0.05 of receiving a claim on its warranty. Notice that claims are clearly not independent of assembly location because these four conditional probabilities are unequalProblem: If a car was purchased from this company, whatis the chance that it’s warranty will be claimed? 38
  39. 39. P( claim ) = P( plant I, claim ) + P( plant II, claim ) + P( plant III, claim) + P( plant IV, claim ) P(claim) = 0.0687 1/26/2013 Prepared by: SB Satorre 39
  40. 40. Probability Tree: FreewayIn a certain freeway, your chance of getting off atExit I is 0.60. 1/26/2013If you get off at Exit I, your chance of getting lost is0.30. Prepared by: SB SatorreIf you miss Exit I, and have to get off at Exit II, yourchance of getting lost is 0.70.What is the probability of not getting lost? 40
  41. 41. PROBABILITY DISTRIBUTIONS Prepared by: SB Satorre 1/26/2013 41
  42. 42. Basic ConceptsProbability distributions are theoretical models that describe thebehavior of the population associated with statisticalexperiments. The concept of probability distributions can bestbe understood by means of the following experiment. 1/26/2013Let us consider the probability distributions of the number of Prepared by: SB Satorreheads which shows in three tosses of a balanced coin. No. of Heads Probability 0 1/8 1 3/8 2 3/8 42 3 1/8
  43. 43. Probability distribution may be defined as a listing ofprobabilities associated with all possible outcomes thatcould result if the experiment were done.A probability distribution involves random variable. 1/26/2013A random variable is a variable whose numerical values are Prepared by: SB Satorredetermined by chance.The representation of the values of a random variable andthe probabilities associated with those values is called aprobability distribution 43
  44. 44. Random Variable• Discrete Random Variable • This is a random variable that takes on a finite number of possible values or an unending sequence with as many values as there are 1/26/2013 whole numbers. Examples include the number of heads or tails that occur when a coin is tossed a specific number of times and the number of patients coming in to a particular clinic in a given Prepared by: SB Satorre day.• Continuous Random Variable • This is a random variable that takes on an infinite number of possible values equal to the number of points on a line segment. Examples include measured quantities such as weight, height, length and distance. 44
  45. 45. Types of Probability Distribution1. Binomial2. Poisson 1/26/20133. Hypergeometric Prepared by: SB Satorre4. Trinomial 45
  46. 46. Binomial Probability DistributionThe binomial distribution is a distribution that describesdiscrete data that results from an experiment called theBernoulli Process. 1/26/2013The Bernoulli Process is a sampling process that conforms to the Prepared by: SB Satorrefollowing characteristics: 1. Each trial has only two possible outcomes. (Ex: head or tail, yes or no, success or failure) 2. The probability of success on any trial remains fixed or constant from trial to trial. 3. Each trial is statistically independent, that is, the outcome of one trial could not affect the outcomes of the other 46 trials.
  47. 47. Binomial FormulaThe formula for determining the probability of a designatednumber of X success in n independent trials in a binomialdistribution is: 1/26/2013 P(x) = C(n,x) px (1-p)n-x Prepared by: SB SatorreWhere: x = 0, 1, 2, 3, …, n p = constant probability of success for each trial C(n,x) = combination of n trials taken r at a time 1 – p = probability of failure n = number of trials 47
  48. 48. Poisson DistributionThe Poisson Distribution was named after themathematician and physicist Simeon Poisson (1781-1840).The Poisson distribution can be used to determine the 1/26/2013probability of a designated number of successes when theevents occur continuously. It is similar to the Bernoulli Prepared by: SB Satorreprocess except that the events occurs continuously and noton fixed trials or observations.For instance, the number of cars arriving at a gasolinestation for refill in a given 10-minute period and thenumber of commuters arriving at a bus waiting stationwithin a given minute can be assumed to be Poisson 48variables.
  49. 49. Poisson Distribution (cont.)A Poisson experiment gives the number ofoutcomes occurring during a time interval 1/26/2013independent of the number of outcomes.In determining the probability of a designated Prepared by: SB Satorrenumber of successes x in a Poisson process, theonly value needed is the average number ofsuccesses for the specific time or space which isrepresented by λ(Greek letter lambda). 49
  50. 50. The Poisson Formula: P(x,λ) = λx * e –λ 1/26/2013 x! Prepared by: SB SatorreWhere:e = 2.7183 (base of natural logarithm)λ = (Greek letter lambda) is the average or mean of thedistributionx = 0, 1, 2, 3,…,n 50
  51. 51. When the number of trials n is large and p is small, binomialprobabilities are approximated by means of the Poissondistribution changing λ into np (λ = np). 1/26/2013 P(x) = (np)x e –np x! Prepared by: SB SatorreWhere: n = total number of trials 51
  52. 52. Hypergeometric DistributionThe Hypergeometric Distribution is the appropriatedistribution when sampling without replacement is used in 1/26/2013a situation that could qualify as a Bernoulli process. Prepared by: SB SatorreThe probability distribution of the hypergeometric variablex is the number of successes in a random sample size nselected from N items of which k are classified assuccesses, and N-k as failures. Its variables are denoted byP(x; N, n, k). 52
  53. 53. In a hypergeometric distribution, the formula fordetermining the probability of a designated number ofsuccesses x is: 1/26/2013 P(x;N,n,k) = C(k,x) C(N-k,n-x) Prepared by: SB Satorre C(N,n) 53
  54. 54. Trinomial DistributionA trinomial distribution is one in which: 1/26/2013 • A certain basic action is performed with n trials or repetitions. Prepared by: SB Satorre • Each repetition involves three possible outcomes. A, B, and C. • Each of the three outcomes has a fixed probability. • p = P(A) q = P(B) r = P(C) • The individual trials are independent. 54
  55. 55. The trinomial formula: P(x,y,z) = [n!/(x!y!z!)] [pxqyrz] 1/26/2013Where: Prepared by: SB Satorrex+y+z = n and p+q+r = 1The term, n!/(x!y!z!) is called a trinomial coefficient. 55
  56. 56. Thank you. Prepared by: SB Satorre 1/26/201356