introduction of probabilityChapter5.pptx

for health care students to do research

1. 1 Amanuel Ayele [BSc, MPH in Epidemiology] Hawassa College of Health Sciences Oct , 2024 Introduction to Probability

2. Learning Objectives At the end of this chapter, the student will be able to: Understand the concepts and characteristics of probabilities and probability distributions Compute probabilities of events and conditional probabilities Differentiate between the binomial and normal distributions Understand the concepts and uses of the standard normal distribution 2

3. Introduction Definition In general, there is no completely satisfactory definition of probability The probability that something occurs is the proportion of times it occurs when exactly the same experiment is repeated a very large (preferably infinite!) number of times in independent trials. 3 3

4. Probability theory developed from the study of games of chance like dice and cards.  A process like tossing or flipping a coin, rolling a die, drawing a card from a deck or drawing marbles from bag are probability experiments. Introduction... 4

5. Why probability in statistics? Results are not certain To evaluate how accurate our results are: Given how our data were collected, are our results accurate? Given the level of accuracy needed, how many observations need to be collected? 5

6. 6 Why probability in medicine? Because medicine is an inexact science, physicians seldom predict an outcome with absolute certainty. E.g ., to formulate a diagnosis, a physician must rely on available diagnostic information about a patient. History, PE, Laboratory tests, X-ray, ECG findings, etc.

7. More importantly probability theory is used to understand: About probability distributions: Binomial Distributions, Gaussian Distributions (Normal Distribution), Poisson Distributions Sampling and sampling distributions Estimation and Hypothesis testing Advanced statistical analysis Why probability in medicine? ... 7

8. Terms commonly encountered in probability: Experiment : anything/process that results in a count or a measurement of uncertain outcome. Sample space : the set of all possible outcomes of an experiment. E.g.,: (H,T); (1,2,3,4,5, 6) of die; ( 52 cards) . Event : any subset of the sample space H or T . Events are represented by UPPERCASE letters (A, B, C,). 8 Definitions of Terms 8

9. 9 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 Ace which is also a red card from a deck of cards. Definitions of Terms

10. Mutually Exclusive Events & The Additive Rule Two events A and B are mutually exclusive if they have no elements in common. P (A ∩ B) = 0 If A and B are outcomes of an experiment they cannot both happen at the same time. That is, the occurrence of A precludes the occurrence of B and vice versa. 10 10

11. Mutually Exclusive….. Example: In the toss of a coin, the event A (it lands heads) and event B ( it lands tails) are mutually exclusive. The Additive Law The additive law, when applied to two mutually exclusive events, states that: - the probability of either of the two events occurring is obtained by adding the probabilities of each event. 11

12. Thus, if A and B are mutually exclusive events, Pr (A or B) = Pr (A) + Pr (B). Example 1 A thrown die may show a one or a two, but not both. The probability that it shows a one or a two is _ Pr (1 or 2) = Pr (1) + Pr (2) Pr (1 or 2) = Pr (1/6) + Pr (1/6) = 2/6 12 Mutually exclusive events and…… 12

13. Extension of the additive law to more than two events indicates that if A, B, C … are mutually exclusive events, Pr (A or B or C or…) = Pr(A) + Pr(B)+ Pr(C) + … When A and B are not mutually exclusive , Pr (A or B) = Pr (A) + Pr(B) – Pr(A and B) 13 Mutually exclusive events and…… 13

14. Eg 2: One die is rolled and sample space = S = (1, 2, 3, 4, 5, 6) Let A = the event an odd number turns up, A = (1,3,5) Let B = the event a 1, 2 or 3 turns up; B = (1, 2, 3) Let C = the event a 2 turns up, C= (2) 14 Mutually exclusive events and……

15. Solution Find Pr (A); Pr (B) and Pr (C) Pr (A) = Pr (1) + Pr (3) + Pr (5) = 1/6 + 1/6 + 1/6 = 3/6 = ½=0.5 Pr (B) = Pr (1) + pr(2) + Pr (3) = 1/6 + 1/6 + 1/6 = 3/6 = ½=0.5 Pr (C ) = Pr (2) = 1/6=0.167 15 Mutually exclusive events and……

16. Are A and B; A and C; B and C mutually exclusive? A and B are not mutually exclusive . Because they have the elements 1 and 3 in common. B and C are not mutually exclusive . They have the element 2 in common. A and C are mutually exclusive. They don’t have any element in common. 16 Mutually exclusive events and……

17. Eg 3: Of 200 seniors at a certain college, 98 are women (W), 34 are majoring in Biology (B), and 20 Biology majors are women (BW). If one student is chosen at random from the senior class, what is the probability that the choice will be either a Biology major or a women? Solution: Pr (Biology major or Woman) = = Pr (B)+ Pr (W)– Pr (BW)= 34/200 + 98/200 – 20/200 = 112/200 = 0.56 17 Mutually exclusive events and… 17

18. Conditional Probability Definition Refers to the probability of an event, given that another event is known to have occurred. “What happened first is assumed” Hint - When thinking about conditional probabilities, think in stages. Think of the two events A and B occurring chronologically, one after the other, either in time or space.

19. The conditional probability that event B has occurred given that event A has already occurred is denoted P(B|A) and is defined provided that P(A) ≠ 0. The notation is Pr(B/A ), which is read as “the probability event B occurs given that event A has already occurred ”. Conditional Probability…

20. Eg : A study investigating the effect of prolonged exposure to bright light on retina damage in premature infants. Retinopathy YES Retinopathy NO TOTAL Bright light Reduced light 18 21 3 18 21 39 TOTAL 39 21 60 Conditional Probability…

21. The probability of developing retinopathy is: P (Retinopathy) = No. of infants with retinopathy Total No. of infants = (18+21)/(21+39) = 39/60 = 0.65 Conditional Probability….

22. We want to compare the probability of retinopathy , given that the infant was exposed to bright light , with that the infant was exposed to reduced light . Exposure to bright light and exposure to reduced light are conditioning events , events we want to take into account when calculating conditional probabilities . Conditional Probability….

23. The conditional probability of retinopathy, given exposure to bright light, is: P(Retinopathy/exposure to bright light) = No. of infants with retinopathy exposed to bright light No. of infants exposed to bright light = 18/21 = 0.86 Conditional Probability….

24. P(Retinopathy/exposure to reduced light) = # of infants with retinopathy exposed to reduced light No. of infants exposed to reduced light = 21/39 = 0.54 The conditional probabilities suggest that premature infants exposed to bright light have a higher risk of retinopathy than premature infants exposed to reduced light. Conditional Probability…

25. Often there are two events such that the occurrence or non-occurrence of one does not in any way affect the occurrence or non occurrence of the other. Eg : A classic example is n tosses of a coin and the chances that on each toss it lands heads. These are independent events . 25 Independent Events and Multiplicative Rule

26. Independent Events…. The chance of heads on any one toss is independent of the number of previous heads. No matter how many heads have already been observed, the chance of heads on the next toss is ½ or 0.5. 26

27. With independent events, the multiplicative law becomes: Pr(A and B) = P (A ∩ B ) = Pr(A) * Pr (B) Hence, Pr(A) = Pr(A and B) / Pr(B),… where Pr(B)  Pr (B) = Pr (A and B) / Pr (A),… where Pr (A)  Independent events & Multiplicative …. 27

28. Example: Suppose we toss a coin twice, and the probability of two heads occurring is the product of their probabilities, that is_ Pr (two heads) = Pr (1/2)* Pr (1/2) = 1/4 = 0.25 28 Independent events & Multiplicative ….

29.  For independent events A and B P(A/B) = P(A).  For non-independent events A and B P(A and B) = P(A/B) P(B) (General Multiplication Rule) Independent events & Multiplicative….

30. Basic Properties of Probability Probabilities are real numbers on the interval from 0 to 1, inclusive; i.e. , 0  Pr(E)  1 If an event is certain to occur, its probability is 1, and if the event is certain not to occur, its probability is 0. If two events A and B, are mutually exclusive (disjoint), the probability that one or the other will occur equals the sum of the probabilities; Pr(A or B) = P(A U B) = Pr(A) + Pr(B) 30 30

31. If A and B are two events, not mutually exclusive, then: Pr (A or B) = Pr (A) + Pr (B) – Pr (A and B) The sum of the probabilities that an event will occur and that it will not occur is equal to 1; hence, P(A ’ ) = 1 – P(A) (Complementary events) If A and B are two independent events, then Pr (A and B)= P(A∩B )=P(A) x P(B) (Independent events) P(A∩B) ≠ P(A) x P(B) (Dependent events) 31 31 Basic properties of probability…

32. The term probability distribution or just distribution refers to the collection of all possible outcomes along with their probabilities. Listing of the possible values that a random variable may assume along with their probabilities. A probability distribution of a random variable can be displayed by a table or a graph or a mathematical formula. The Probability Distribution 32

33. Example Toss a coin 2 times. Let, x be the number of heads obtained. Find the probability distribution of x . Pr (X = x i ), i = 0, 1, 2, Pr(x = 0) = 1/4 ……………………. TT Pr(x = 1) = 1/2 .………………….... HT TH Pr(x = 2) = 1/4 .…………………… HH 33 33 Probability Distribution….

34. 34 Probability distribution of X ( i.e. probability distribution of heads) . X = x i 1 2 Pr(X=x i ) 1/4 1/2 1/4 Probability Distribution…..

35. 35 Types of Probability Distributions The are different types of probability distributions. However, certain types of distributions give good approximations to the distributions of many random variables. Important types of discrete distributions include: Binomial , Poisson, Multinomial, Hyper geometric, etc … 35

36. Important types of continuous distributions include: The Normal (Gaussian ), Exponential, Gamma, Lognormal, etc … 36 Types of Probability Distributions

37. 1. Binomial Distribution It is one of the most widely encountered discrete probability distributions. It is a type of distribution that has TWO possible outcomes, & derived from a process known as Bernoulli Trial . Bernoulli Trial is a trial which can result in only one of two mutually exclusive outcomes in a single experiment. 37 37

38. It is simply the probability of a SUCCESS or FAILURE outcome in an experiment that is repeated multiple times. Many instances you can think of that can only be a success or failure can be found in a real life, represented by a binomial distribution . 38 1. Binomial Distribution

39. Binomial distribution.... Example : A coin toss has only two possible outcomes: (Head or Tail) If a new drug is introduced to cure a disease, it is either cures the disease (success) or it doesn’t cure the disease (failure). If you purchase a lottery ticket, you are either going to win money, or you are not. 39

40. Consider other dichotomous (binary) random variables Patient Survives or Dies A Specimen is Positive or Negative A Child has Vaccinated or Not Vaccinated 40 Binomial distribution....

41. 41 Characteristics of Binomial Distribution The data arise from a sequence of “n” independent trials. At each trial there are only two possible outcomes, conventionally called success and failure . The probability of success, “P” , is the same in each trial. If P is the probability of success in one trial , then 1 – p is the probability of failure . Note that: p ( success) + q ( failure) = 1 . 41

42. The random variable of interest is the number of successes, “r” , in the “n” trials. “n” and “P” are the two parameters that determine the function of Binomial Distribution . The Mean ,  = nP The Variance ,  2 = nP (1- P) 42 Characteristics of Binomial Distribution

43. 43 Application of Binomial Distribution It is appropriate when: Sampling is from infinite population. “n” is small relative to “N” (“N” is at least ten times as large as “n” ). If the binomial assumptions are satisfied, the probability of ‘ r ’ successes in ‘n‘ trials computed as: r = 0,1,2…n

44. 44 OR r = 0,1,2…n r = 0,1,2…n Application of binomial distribution...

45. The general formula for the binomial coefficients 45 is n = denotes the number of fixed trials r = denotes the number of successes in the n trials n! = n factorial ( the product of an integer & all the integers below it) p = denotes the probability of success q = denotes the probability of failure (1- p ) Application of binomial distribution... = n C r is the number of possible ways to choose r success from n observations . 45

46. Example: the number of ways to achieve 2 heads in a set of four tosses is “ 4 choose 2”, or 4!/ 2!2! = 6, the possibilities are: {HHTT, HTHT, HTTH, TTHH, THTH, THHT} Suppose we know that 40% of a certain population are cigarette smokers. If we take a random sample of 10 people from this population, what is the probability that we will have exactly 4 smokers in our sample? 46 Application of binomial distribution... 46

47. Solutions If the probability that any individual in the population is a smoker to be P=0.40, then the probability that r = 4 smokers out of n = 10 subjects selected is: P(X=4) = 10 C 4 (0.4) 4 (1- 0.4) 10-4 = 10 C 4 (0.4) 4 (0.6) 6 = 210(0.0256)(0.04666) = 0.25 The probability of obtaining exactly 4 smokers in the sample is about 0.25 . 47 Application of binomial distribution... 47

48. Exercise 1: The probability of obtaining HIV exposed neonate among mothers who took delivery service in Hawassa millennium HC per a year is 10% . If we take a random sample of 5 neonates delivered in that HC, what is the probability that we will have exactly 2 neonate in the sample who is exposed for HIV? 48 Application of binomial distribution... 48

49. Solutions: If the probability of obtaining HIV exposed neonate from the total mothers who took delivery service to be P = 0.10 , then the probability that r = 2 HIV exposed out of n=5 neonates delivered in that HC selected is: P(X=2) = 5 C 2 (0.1) 2 (1- 0.1) 5-2 = 5 C 2 (0.1) 2 (0.9) 3 = 10x0.01x0.729 = 0.0729 The probability of obtaining exactly 2 HIV exposed neonate in the sample is about = 0.0729 49 49 Application of binomial distribution...

50. If the true proportion of events of interest is P , then in a sample of size n The Mean of the binomial distribution is: µ = np The Standard deviation is: σ = 50 Application of binomial distribution... 50

51. Example: About 70% of a certain population has been immunized for polio. If a sample of size 50 is taken, what is the “ expected total number ”, in the sample who have been immunized? µ = np = 50(0.70) = 35 This tells us that “ on the average” we expect to see 35 immunized subjects in a sample of 50 from this population. 51 51 Application of binomial distribution...

52. Class-Exercise 2: The probability of obtaining HIV exposed neonate among mothers who took delivery service in Alamura HC per a year is 10%. If we take a random sample of 5 neonates delivered in that HC, what is the “ expected total number ”, in the sample who are exposed for HIV? µ = np = 5(0.10) = 0.5 This tells us that “on the average” we expect to see 1 HIV exposed neonate in a sample of 5 from the total delivered. 52 52 Application of binomial distribution...

53. 2. The Normal Distribution The normal distribution is an example of probability distribution for continuous random variables . The most important probability distribution in statistics. Frequently called the “Gaussian distribution” or Bell-shape curve [ Named after Carl Friedrich Gauss (1777–1855)] 53 53

54. 54 Variables such as, Blood Pressure, Weight, Height, Serum Cholesterol Level, and IQ score are approximately normally distributed. A random variable is said to have a normal distribution if it has a probability distribution that is symmetric and bell-shaped. 2. The Normal Distribution

55. 55 Normal distribution.... 55

56. Normal distribution.... A random variable x is said to follow normal distribution if and only if, its probability density function (PDF) is: where, -  < x <  56

57. 57 Normal distribution.... Probability density function (PDF) is the total area bounded by its curve and the x-axis is equal to 1. Sub area under the curve bounded, x-axis and the perpendiculars erected at any two points give the probability that x is between a and b . Any Binomial distribution may be approximated by the normal distribution of the same mean and variance provided n is large enough.

58. Characteristics of the Normal Distribution The important characteristics of the normal distribution are: It is unimodal, bell-shaped and symmetrical about mean ( µ ). The two important parameters are the mean and the standard deviation. We have different normal distributions depending on the values of μ and  σ 2 . As the values of σ 2 increases, the curve becomes more and more flat and vise versa. 58

59. The mean µ tells you about location – Increase µ - Location shifts right Decrease µ – Location shifts left Shape is unchanged The variance σ 2 tells you about narrowness or flatness of the bell – Increase σ 2 - Bell flattens. Extreme values are more likely Decrease σ 2 - Bell narrows. Extreme values are less likely Location is unchanged 59 The important characteristics of ND....

60. The important characteristics .... The mean, median, and mode are all equal. The curve never touches the x-axis. The total area under the curve about the x-axis is 1 square unit. It is a probability distribution of a continuous variable. It extends from minus infinity (-  ) to plus infinity (+  ). Tabulated normal probability calculations are available only for the Normal Distribution with µ = 0 and σ =1. 60

61. Standard Normal Distribution It is a normal distribution that has a mean equal to and a SD equal to 1 , and is denoted by N(0, 1). Normal distribution can be transformed to standard normal distribution by Z-transformation defined by: this process is known as Standardization and it gives you the position on a normal curve  = 0 and  = 1 i.e. the Standard Normal Distribution, Z . 61

62. These Z-scores can then be used to find the area (and thus the probability) under the normal curve. The standard normal distribution has mean 0 and variance 1. 62 Standard Normal Distribution

63. Properties of the Standard Normal Distribution The cumulative area is close to 0 for z -scores close to z = 3.49. The cumulative area increases as the z -scores increase. The cumulative area for z = 0 is 0.5000. The cumulative area is close to 1 for z -scores close to z = 3.49 z = 3.49 Area is close to 0. z = Area is 0.5000. z = 3.49 Area is close to 1. z 3 1 2  1 2 3 Standard Normal Distribution

64. If a random variable X~N(  ,  ) then we can transform it to a SND with the help of Z-transformation . SND = Z score = Z represents the Z-score for a given x value 64 Standard normal distribution…. 64

65. Area Under any Normal Curve To find the area under a normal curve (with mean  and standard deviation  ) between x = a and x = b , find the Z scores corresponding to a and b (call them Z 1 and Z 2 ) and, then find the area under the standard normal curve between Z 1 and Z 2 from the published table. Z-table 65

66. Using a SND Table Procedure: How to use a Standard Normal Distribution Table For areas to the left of a specified z value, use the table entry directly. For areas to the right of a specified z value, look up the table entry for z and subtract the area from 1 . 66

67. Note : another way to find the same area is to use the symmetry of the normal curve and look up the table entry for –z . For area between two z values, z 1 and z 2 (where z 2 > z 1 ), subtract the table area for z 1 from the table area for z 2 . Some Useful Tips: 67 Using a SND Table

68. Example What is the probability that Z < -1.96? Sketch a normal curve Draw a perpendicular line for Z = -1.96 Find the area in the table The answer is the area to the left of the line P(Z < -1.96) = 0.0250; ( i.e. 1 - 0.9750 = 0.0250) 68

69. What is the probability that (-1.96 < Z < 1.96) ? 69 Example….. 69 The area between the values P(-1.96 < Z< 1.96) P(z= -1.96) = 0.0250, and P(z<1.96) = 0.9750 P(-1.96 < Z< 1.96) = 0.9750 – 0.0250 = 0.9500

70. What is the probability that Z > 1.96? 70 70 The answer is the area to the right of the line P(Z > 1.96) = 0.0250 (i.e. 1 – 0.9750 = 0.0250) N.B: From the symmetry properties of the standard normal distribution , P(Z  -x) = P(Z  x)

71. Exercise 3: Compute P(-1 ≤ Z ≤ 1.5) Compute P( -1.66 < Z < 2.85) Find the area under SND for P(0.83 < Z < 1.25) Find the area for Z < 1.96 71 71 Answer: 0.9493 (i.e. P( Z > -1.66) = 0.0485, and P(Z< 2.85) = 0.9978) ( P(-1.66 ≤ Z ≤ 2.85) = 0.9978 - 0.0485 = 0.9493) Answer: 0.7745 (i.e. P(Z  -1) = 0.1587, and P(Z ≤ 1.5) = 0.9332) ( P(-1 ≤ Z ≤ 1.5) = 0.9332 - 0.1587 = 0.7745) Answer: 0.8944  0.7967 = 0.0977 P(0.83 < Z < 1.25) = 0.8944 – 0.7967 = 0.0977) Answer: 0.9750

72. Example The height of adult men in United Kingdom, which is approximately normal with mean  = 171.5cm & standard deviation  = 6.5cm . What is the probability that a randomly selected men has a height taller than 180cm? What is the probability that a randomly selected men has a height shorter than 160cm? What is the probability that a randomly selected men has a height between 165cm and 175cm? 72 72

73. What is the probability that a randomly selected men has a height taller than 180cm? First find the corresponding SND = Z scores P(Z > 1.31) = 0.9049 Answer = 0.0951 (i.e. 1 – 0.9049 = 0.0951) Or equivalently 9.51% of adult men are taller than 180cm. 73 Solutions 73

74. What is the probability that a randomly selected men has a height shorter than 160cm? P( Z < -1.77) = 0.0384 Thus, 3.84% of men are shorter than 160cm . 74 Solutions 74

75. What is the probability that a randomly selected men has a height between 165cm and 175cm? SND corresponding to 165cm P( Z < -1.0) = 0.1587 proportion below this height (165cm) is 0.1587 75 Solutions 75

76. SND corresponding to 175cm P( Z > 0.54) = 0.7054 (1-0.2946) Proportion of men with height between 165cm & 175cm P(-1 ≤ Z ≤ 0.54) = 0.7054 – 0.1587 = 0.5467 = 54.67%. OR proportion below this height (165cm) is 0.1587 proportion above this height (175cm) is 0.2946 Solutions

77. Proportion of men with height between 165cm & 175cm. = 1- proportion below 165cm - proportion above 175cm = 1- 0.1587- 0.2946 = 0.5467 = 54.67%. 77 Solutions…..

78. Exercise 4: The diastolic blood pressures of males 35 – 44 years of age are normally distributed with µ = 80 mmHg and σ = 12mmHg. Let individuals with BP above 95mmHg are considered to be hypertensive. What is the probability that a randomly selected male has a BP above 95 mm Hg? What is the probability that a randomly selected male has a DBP above 110 mm Hg? What is the probability that a randomly selected male has a DBP below 60 mm Hg? 78 78

79. What is the probability that a randomly selected male has a BP above 95 mm Hg? Answer: P(z >1.25) = 0.1056 (i.e., 1- 0.8944= 0.1056) Approximately 10.6% of this population would be classified as hypertensive. What is the probability that a randomly selected male has a DBP above 110 mm Hg? Answer: P(z >2.50) = 0.0062 Approximately 0.6% of the population has a DBP above 110 mm Hg. 79 Solutions 79

80. What is the probability that a randomly selected male has a DBP below 60 mm Hg? Answer. P (Z < -1.67) = 0.0475 Approximately 4.8% of the population has a DBP below 60 mmHg. 80 Solutions 80

81. Other distributions 81 Student t-distribution F- Distribution  2 –Distribution The Poisson Distribution

82. 82