2. 6 Probability

3. Chapter6 p126 6.1 Operations on events and probability An event is the basic element to which probability can be applied. Notations Event: A, B A∩B  both A and B A ∪ B  either A or B

4. Chapter6 p126 A∩B  both A and B A ∪ B  either A or B A c  not A, complement of A

5. Chapter6 p128 Definition of probability -Many definitions, text used the frequentist definition Probability of A P(A) = m / n m and n denote the frequency of occurrence of A and the total number of repeated experiments Example – Table 5.1 P(a child survives his or her first year) = 99149 / 100000 = 0.99149 P(A ∪ A c ) = 1 P(A ∩ A c ) = 0 P(A c ) = (n – m) / n The probability that a newborn does not survive the first year of life is 1- 0.99149 = 0.00851

6. Chapter6 p128 Mutually exclusively or disjoint event A ∩ B = Ø P( A ∩ B) = 0 When two events are mutually exclusively, the additive rule of probability applied,  P( A ∪ B) = P(A) + P(B) Example the probability a newborn’s birth weight is under 2000 grams is 0.025, and the probability that it is between 2000 and 2499 grams is 0.043  P( baby weight is under 2500 grams) = 0.025 + 0.043 = 0.068 n mutually exclusively events, A 1 ∩A 2 = Ø, ……. A n-1 ∩A n = Ø  P(A 1 ∪ A 2 …… ∪ A n ) = P(A 1 ) + P(A 2 ) + ….. P(A n ) When two events are NOT mutually exclusively,  P( A ∪ B) = P(A) + P(B) – P(A ∩ B) Figure 6.2 Venn diagram representing two mutually exclusively events.

7. Chapter6 p128 6.2 Conditional probability The probability that an event B will occur given that we know the outcome of event A, does the prior occurrence of a cause the probability of B to change ? Example Find the probability that a person will live to the age of 65 given that the person has reached the age of 60  conditional probability, P(B|A) represent the probability of the event B given that the probability that event A has occurred. Multiplicative rule of probability The probability that two events A and B will both occur = probability of A * probability of B given that A has already occurred  P( A ∩ B) = P(A) P(B|A) since is arbitrary which event we call A which we call B, we can also write  P( A ∩ B) = P(B) P(A|B) Example Event A = a person is alive at age 60, Event B = a person survives to 65 A ∩ B = the event that the person is alive at age 60 and also at 65. What are P(A), P(A ∩ B) and P(B|A) ? P(A) = 85993 / 100000 = 0.85993 P(A ∩ B) = 80145 / 100000 = 0.80145 P(B|A) = 0.80145 / 0.85993 = 0.9320 If a person lives to be 60, this person chance of surviving to age 65 is greater than it was at birth, P(B), i.e. 0.9320 > 0.80145

8. Chapter6 p128 6.2 Conditional probability When we are concerned with two events such that the outcome of one event has no effect on the occurrence or nonoccurrence of the other, the events are said to be independent. If A and B are independent event, P(A|B) = P(A) and P(B|A) = P(B)  P(A ∩ B) = P(A) P(B) Independent and mutually exclusive DO NOT mean the same thing  If A and B are mutually exclusive, and event A occurs, event B cannot occur. By definition, P(B|A) = 0

9. 6.3 Bayes’ Theorem

10. 3 Defects 7 Good Given 10 films, 3 of them are defected. What is the probability two successive films are defective?

11. 6.3 Bayes’ Theorem Loyalty of managers to their employer.

12. 6.3 Bayes’ Theorem Probability of new employee loyalty

13. 6.3 Bayes’ Theorem Probability (over 10 year and loyal) = ? Probability (less than 1 year or loyal) = ?

14. Chapter6 p128 6.3 Bayes’ Theorem Let E 1, E 2 and E 3 = a person is currently employed, unemployed, and not in the labor force respectively P(E 1 ) = 98917 / 163157 = 0.6063 P(E 2 ) = 7462 / / 163157 = 0.0457 P(E 3 ) = 56778 / 163157 = 0.3480 Let H = a person has a hearing impairment due to injury, what are P(H), P(H|E 1 ), P(H|E 2 ) and P(H|E 3 ) ? P(H) = 947 / 163157 = 0.0058 P(H|E 1 ) = 552 / 98917 = 0.0056 P(H|E 2 ) = 27 / 7462 = 0.0036 P(H|E 3 ) = 368 / 56778 = 0.0065 Employment statusPopulationImpairments Currently employed98917552 Currently unemployed746227 Not in the labor force56778368 Total163157947

15. Chapter6 p128 6.3 Bayes’ Theorem H = a person has a hearing impairment due to injury, What is P(H)? May be expressed as the union of three mutually exclusively events, i.e. E 1 ∩H, E 2 ∩H, and E 3 ∩ H H = (E 1 ∩H) ∪ (E 2 ∩H) ∪ (E 3 ∩ H) Apply the additive rule P(H) = P(E 1 ∩H) + P(E 2 ∩H) + P(E 3 ∩ H) Apply the Bayer’ theorem P(H) = P(E 1 ) P(H|E 1 ) + P(E 2 ) P(H|E 2 ) + P(E 3 ) P(H|E 3 ) EventP(E i )P(H | E i )P(E i ) P(H | E i ) E1E1 0.60630.00560.0034 E2E2 0.04570.00360.0002 E3E3 0.34800.00650.0023 P(H)0.0059

16. Chapter6 p128 6.3 Bayes’ Theorem The more complicate method P(H) = P(E 1 ) P(H|E 1 ) + P(E 2 ) P(H|E 2 ) + P(E 3 ) P(H|E 3 ) ………………. (1) is useful when we are unable to calculate P(H) directly. How about we want to compute P(E 1 |H) ? The probability that a person is currently employed given that he or she has a hearing impairment. The multiplicative rule of probability states that P(E 1 ∩H) = P(H) P(E 1 | H)  P(E 1 | H) = P(E 1 ∩ H) / P(H) Apply the multiplicative rule to numerator, we have P(E 1 | H) = P(E 1 ) P(H | E 1 ) / P(H) ……………………………………..(2) Substitute (1) into (2), we have the expression for Bayes’ Theorem

17. Chapter6 p128 6.3 Bayes’ Theorem EventP(E i )P(H | E i )P(E i ) P(H | E i ) E1E1 0.60630.00560.0034 E2E2 0.04570.00360.0002 E3E3 0.34800.00650.0023 P(H)0.0059 = (0.6063)(0.0056) / [(0.6063)(0.0056)+(0.0457)(0.0036)+(0.3480)(0.0065)] = 0.583 P(E 1 | H) = 552 /947 = 0.583 Employment statusPopulationImpairments Currently employed98917552 Currently unemployed746227 Not in the labor force56778368 Total163157947

18. 6.3 Bayes’ Theorem Exercise A box contains 10 balls, 3 black in color and 7 white in color.. Balls are drawn from the box without return. (a)Calculate the probability that the second ball is black in color. (b)Given that the second ball is black in color, determine the probability that the first is also black in color, that is compute P (first black | second black).

19. Chapter6 p138 6.4 Diagnostic test Screening is the application of a test to individuals who have not yet exhibited any clinical symptoms in order to classify them with respect to their probability of having a particular disease. Test positive  likely to have the disease  further diagnostic Bayes’ theorem allows us to use probability to evaluate the associated uncertainties

20. Probability of false negative P(test negative | cancer) Probability of true positive P(test positive | cancer) Probability of false positive P(test positive | no cancer) 6.4.1 Sensitivity and Specificity

22. Correlation coefficient (CC) SpSp Correlation coefficient (CC) ranges from -1 to 1, where a value of 1 means prefect prediction, a value of -1 indicates zero correction prediction.

23. 6.4.2 Applications of Bayes’ Theorem Notation : D 1 is the event that an individual has a particular disease D 2 is the event that an individual does not has a particular disease T + denotes a positive screening test result T - denotes a negative screening test result Example Cervical cancer, Pap smear ( 子宮頸抹片 ) screening testPap smear ( 子宮頸抹片 ) screening test Probability of false negative P(test negative | cancer) = P(T - | D 1 )= 0.1625 Probability of true positive P(test positive | cancer) P(T + | D 1 )= 1-0.1625 = 0.8375 Probability of false positive P(test positive | no cancer) = P(T + | D 2 )= 0.1864 P(D 1 ) = 0.000083, rate of cervical cancer in 1983-1984 was 8.3 per 100,000  prevalence of the disease P(D 2 ) = 1 - P(D 1 ) P(D 1 |T + ) is called the p redictive value of a positive test For every 1,000,000 women with positive Pap smear, only 373 represent true cases of cervical cancer.

24. Chapter6 p138 6.4.2 Applications of Bayes’ Theorem Calculate the predictive value of a negative test. For every 1,000,000 women with negative Pap smear, 999,983 do not have cervical cancer

25. Figure 6.3 illustrates the results of the entire diagnostic testing process. All numbers have been rounded to the nearest integer.

26. 6.4.3 Receiver Operation Characteristic (ROC) curve - Sensitivity and Specificity

28. Chapter6 p126 6.4.3 ROC curve 2.9 mg% as an indicator of imminent rejection, the test has a sensitivity of 0.303 and a specificity of 0.909. To increase the sensitivity, we could lower the arbitrary cutoff poing that distinguishes a positive test result from a negative one, if we use 1.2 mg%, for example, a much greater proportion of the results will reject the organ. At the same time we would increase the probability of a FP result, thereby decreasing the specificity.

30. 6.5 The relative risk and the odds ratio Relative Risk (RR) – want to compare the probabilities of disease in two different groups. Example – breast cancer study Exposed – a woman first gave birth at age 25 of older Sample – 4540 (1628) women who gave birth to their first child before the age of 25 (or older), 65 (31) developed breast cancer Women who first gave birth at a later age are 33% more likely to develop breast cancer. In chapter 15, we will explain how to determine whether this is an important difference.

31. 6.5 The relative risk and the odds ratio Odds Ratio ( 機會, 勝算 ), or relative odds (OR) – measure of the relative probabilities of disease. If an event takes place with probability p, the odds in favor of the event are p/(1– p) to 1. If p = ½, the odds are (1/2)/(1/2) = 1  the event is equally likely either to occur or not to occur. For every 100,000 individuals there are 9/3 cases of tuberculosis, the odds of randomly selected person’s having the disease are OR is defined as the odds of disease among exposed individuals divided by the odds of disease among the unexposed, or

32. 6.5 The relative risk and the odds ratio Risk factors for breast cancer - use of oral contraceptives A case-control study – examine the effects of the use of oral contraceptives. Determine whether the exposure in question was present or absent for each individual 989 women who has breast cancer, 273 had used oral contraceptives and 716 had not 9901 women who did not have breast cancer, 2641 had used oral contraceptives and 7260 had not Subjects with and without the disease are chosen, therefore, the probability of disease in the exposed and unexposed group cannot be determined. We can determine the probability of exposure for both cases and controls Women who have used oral contraceptives have an odds of developing breast cancer that is only 1.05 times the odds of nonusers. Chapter 15 will interpret this result.

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36. Chapter6 p126