1. 機率 SOME BASIC PROBABILITY CONCEPTS 台北醫學大學公共衛生學系 葉錦瑩

2. 機率的基本觀念 集合 (set;S) 是由許多元素 (elements; s) 構成。 – 若 x ＝ X (s) S ＝ {1,2,3,4,5} A = {1,2,3} B = {2,3,4} – 則 x : random variable 5 ⊂ S: 唸成 '5 belong to S' 表示 5 是 S 的元素 (element) 。 A ⊂ S: 唸成 'A include in S' 表示 A 是 S 的部份集合 (subset) 。 A ＝ {4,5} 表示是 A 的餘集合 (complement; A* ) 。 C ＝ {} 表示 C 是空集合 (empty set; null set;φ) 。 A ∪ B ＝ {1,2,3,4} : 唸成 'A union B' 表示 A 與 B 的聯集合。 A ∩ B ＝ {2,3} : 唸成 'A intersection B' 表示 A 與 B 的交集合。

3. 機率的基本觀念 實驗 (Random Experiment):any planned process of data collection, which consists of a number of independent trials under the same condition. 樣本空間 (Sample Space; Outcome Space; N):the collection of all possible different outcomes in a experiment. 樣本點 (Sample Point): any possible outcome in the sample space. 事件 (Event):a single outcome 或 a set of outcomes. 機率 (Probability):sample points of an event (A) in a sample space (N).

4. Definition If an event can occur in N mutually exclusive and equally likely ways, and if m of these possess a trait, E, the probability of the occurrence of E is equal to m/N. P(E) = m / N

5. The subjects in the study by Erickson and Murray consisted of a sample of 75 men and 36 women. Table 3.4.1 shows the lifetime frequency of cocaine use and the gender of these subjects.

6. P(M) = number of males / total number of subjects = 75 / 111 =.6757 What is the probability that this person will be a male?

7. 機率的基本觀念 機率的獲得 – 理論機率 (Theoretical Probability; 事先機 率,Priority Probability): the frequency distribution for all the measurements in the entire population. – 經驗機率 (Empirical Probability): a frequency distribution is tabulated from a set of sample measurement. – 主觀機率 (Subjective Probability)

8. 機率運算法則 加法法則 (Addition rule) –Pr(A or B) ＝ Pr(A) ＋ Pr(B) － Pr(A and B) i.e. Pr(A ∪ B) ＝ Pr(A) ＋ Pr(B) － Pr(A ∩ B) if A and B are mutually exclusive （互斥） then : Pr(A ∪ B) ＝ Pr(A) ＋ Pr(B) 乘法法則 (Multiplication rule) – 兩個或兩個以上事件聯合發生時，稱為組合機率 (joint probability or compound probability) 。 Pr(A ∩ B) ＝ Pr(A) ×Pr(B │ A) ＝ Pr(B) ×Pr(A │ B) but A and B are independent （獨立） if and only if : Pr(A ∩ B) ＝ Pr(A) ×Pr(B)

9. Example 3.4.2 Suppose we pick a subject at random from the 111 subjects and find that he is a male (M). What is the probability that this male will be one who has used cocaine 100 times or more during his lifetime (C)? P(C\M) = 25 / 75 =.33

10. Example 3.4.3 Let us refer again to Table 3.4.1. What is the probability that a person picked at random from the 111 subjects will be a male (M) and be a person who has used cocaine 100 times or more during his lifetime (C)? P (M∩C) = 25 / 111 =.2252

11. Example 3.4.4 We wish to compute the joint probability of male (M) and a lifetime frequency of cocaine use of 100 times or more (C) from a knowledge of an appropriate marginal probability and an appropriate conditional probability. P(M) = 75 / 111 =.6757P(C \ M) = 25 / 75 =.3333 P(M ∩ C) = P(M) P(C \ M) = (.6757)(.3333) =.2252

12. 條件機率 (Conditional Probability; Posterior Probability) Pr(B │ A) ＝ [N (A and B) /N]/[N (A) /N] ＝ Pr(A ∩ B)/Pr(A) ＝ [Pr(A │ B) ×Pr(B)]/Pr(A) Where: Pr(A│B) ≧ 0,Pr(B│B) ＝ 1 Pr(A1 ∪ A2 ∪ …│B) ＝ Pr(A1│B) ＋ Pr(A2│B) ＋ … [ Ai∩Aj ＝ φ,i≠j ] but Pr(A) = ΣPr(A│B j ) × Pr(B j ) so Pr(B i │A) = [Pr(A│B i ) × Pr(B i )]∕[ΣPr(A│B j ) ×Pr(B j )] That is the “ Bayes' theorem".

13. Definition The conditional probability of A given B is equal to the probability of A ∩ B divided by the probability of B, provided the probability of B is not zero. P(A\B) = P(A∩B) / P(B), where P(B)≠0

14. Example 3.4.5 We wish to use the following equation to find the conditional probability, P(C │ M). P(C\M) = P(C∩M) / P(M) = (25/111) / (75/111) = 25/ 75

15. Definition Given two events A and B, the probability that event A, or event B, or both occur is equal to the probability that event A occurs, plus the probability that event B occurs, minus the probability that the events occur simultaneously. P(A ∪ B) = P(A) ＋ P(B) － P(A∩B)

16. P(M ∪ C) = P(M) ＋ P(C) － P(M∩C) P(M) = 75 / 111 =.6757 P(M∩C) = 25 / 111 =.2252 P(C) = 34 / 111 =.3063 P(M ∪ C) =.6757 ＋.3063 －.2252 =.7568 If we select a person at random from the 111 subjects, what is the probability that this person will be a male (M) or will have used cocaine 100 times or more during his lifetime (C) or both ？

17. In a certain high school class, consisting of 60 girls and 40 boys, it is observed that 24 girls and 16 boys wear eyeglasses. If a student is picked at random from this class, the probability that the student wears eyeglasses, P(E), is 40/100, or.4. –What is the probability that a student picked at random wears eyeglasses, given that the student is a boy? P(E\B) ＝ P(E∩B)/P(B) ＝ (16/100)/(40/100) ＝.4 P(E\B) ＝ P(E∩B)/P(B) ＝ (24/100)/(60/100) ＝ 24/60 ＝.4 Example 3.4.7 eyeglassgirlboytotal yes241640 no362460 total6040100

18. Example 3.4.7 What is the probability of the joint occurrence of the events of wearing eyeglasses and being a boy? P(E∩B) ＝ P(B)P(E\B) P(E∩B) ＝ P(B)P(E) ＝ (40/100)(40/100) ＝.16 eyeglassgirlboytotal yes241640 no362460 total6040100

19. Example 3.4.8 Suppose that of 1200 admissions to a general hospital during a certain period of time, 750 are private admissions. If we designate these as set A, then A is equal to 1200 minus 750, or 450. We may compute P(A) ＝ 750/1200 ＝.625 P(A) ＝ 450/1200 ＝.375 P(A) ＝ 1 － P(A).375 ＝ 1 －.625.375 ＝.375

20. Definition Given some variable that can be broken down into m categories designated by A 1, A 2, …, A i, …, A m and another jointly occurring variable that is broken down into n categories designated by B 1, B 2, …, B j, …, B n, the marginal probability of A i, P(A i ), is equal to the sum of the joint probabilities of A i with all the categories of B. That is, P(A i ) ＝ Σ P (A i ∩ B j ), for all values of j

21. Example 3.4.9 We wish to use the following equation and the data in Table 3.4.1 to compute the marginal probability P(M). P(M) ＝ P(M∩A)+ P(M∩B) +P(M∩C)

22. P(M∩A) ＝ 32/111 ＝.2883 P(M∩B) ＝ 18/111 ＝.1622 P(M∩C) ＝ 25/111 ＝.2252 P(M) ＝ P(M∩A) ＋ P(M∩B) ＋ P(M∩C) ＝.2883 ＋.1622 ＋.2252 ＝.6757

23. Definitions A false positive results when a test indicates a positive status when the true status is negative. A false negative results when a test indicates a negative status when the true status is positive.

25. Definition The sensitivity of a test (or symptom) is the probability of a positive test result (or presence of the symptom) given the presence of the disease. P(T\D) ＝ a/(a ＋ c)

26. Definition The specificity of a test (or symptom) is the probability of a negative test result (or absence of the symptom) given the absence of the disease. P(T \ D) ＝ d/(b ＋ d)

27. Definition The predictive value positive of a screening test (or symptom) is the probability that a subject has the disease given that the subject has a positive screening test result (or has the symptom). P(D \ T) ＝ P(T \ D) P(D) P(T \ D) P(D) ＋ P(T \ D) P(D)

28. Definition The predictive value negative of a screening test (or symptom) is the probability that a subject does not have the disease, given that the subject has a negative screening test result (or does not have the symptom). P(T│D) P(D) ＋ P(T│D) P(D) P(D │T) ＝ P(T│D) P(D)

30. sensitivity ＝ P(T \ D) ＝ 436/450 ＝.9689 specificity ＝ P(T \ D) ＝ 495/500 ＝.99 P(T \ D) ＝ 5/500 ＝.01 P(D) ＝ 450/950 ＝. 47 P (D \ T ) ＝ P(D) P (T\D)/ [P (D) P (T\D) ＋ P(D) P(T\D)] ＝ (.47) (.9689) /[(.47)(.9689) ＋ (1-.47)(1 －.99)] ＝.98 DDtotal T4365441 T14495509 total450500950

31. 期望值 (expected value; expectation) 隨機變數之平均值稱之 E(X) ＝ μ ＝ ΣX i Pr(X i ) E(X － μ) 2 ＝ σ 2 ＝ Σ(X i － μ) 2 Pr(X i ) ＝ ΣX i 2 Pr(X i ) － μ 2 ΣPr(X i ) ＝ ΣX i 2 Pr(X i ) － μ 2 E(c) ＝ c where c is constant E(cx) ＝ cμ E(c ＋ x) ＝ c +μ E(Σx) ＝ Nμ E(X 2 ) ＝ μ 2 ＋ σ 2 E(ΣX) 2 ＝ N 2 μ 2 ＋ Nσ 2

32. Thanks for your attention To be continued …..