1. Data Analysis (Quantitative Methods) Lecture 2 Probability

2. zAn experiment is an act or process of observation that leads to a single outcome that cannot be predicted with certainty. zA sample point is the most basic outcome of an experiment. Ob 1, Ob 2, …, Ob n.

3. Sample Space & Venn Diagram zThe sample space of an experiment is the collection of all its sample points. S: {Ob 1, Ob 2, …, Ob n } zVenn diagrams. Graphical representations. Ob 1 Ob 2 Ob n S

4. Examples Experiment: Observe the up face on a coin Sample Space: 1. Observe a head H 2. Observe a tail T S={ H, T} H T S

5. Examples Experiment: Observe the up face on a die. Sample Space: 1. Observe a 1. 2. Observe a 2. 3. Observe a 3. 4. Observe a 4. 5. Observe a 5. 6. Observe a 6. S= {1,2,3,4,5,6} 1 2 3 4 5 6 s

6. Examples Experiment: Observe the up face on two coins Sample Space: 1. Observe HH 2. Observe HT 3. Observe TH 4. Observe TT S: {HH,HT,TH,TT} HH HT TH TT S

7. Probability Rules for Sample Points z All sample point probabilities must lie between 0 and 1. zThe probabilities of all the sample points within a sample space must sum to 1.

8. Probability An Event is a specific collection of Sample points. Example: Consider the experiment of tossing two balanced coins. Events: A: {Observe exactly one head} B: { Observe at least one head}

9. Probability Sample point Probability HH 1/4 HT 1/4 TH 1/4 TT 1/4 P(A)=P(HT)+P(TH)=1/2 P(B)=P(HH)+P(TH)+P(HT)=3/4

10. Probability of an Event The probability of an event A is calculated by summing the probabilities of the sample points in the sample space for A.

11. Steps for Calculating Probabilities of Events zDefine the experiment. zList the sample points. Ob 1,Ob 2,…,Ob n zAssign probabilities to the sample points. P(Ob 1 ), …, P(Ob n ). zDetermine the collection of sample points contained in the event of interest. zSum the sample point probabilities to get the event probability.

12. Steps for Calculating Probabilities of Events Example: Experiment: Observe the up face on a die. Find the probability of event of sum of two throws is equal to 6. Solution: Sample points 36 (6 by 6). A: sum of two throws is equal to 6. (1,5),(2,4),(3,3),(4,2),(5,1) Pr(A)=5/36

13. Unions and intersections zThe Union of two events A and B is the event that occurs if either A or B or both occur on a single performance of the experiment, denoted as the symbol

14. Unions and intersections zThe intersection of two events A and B is the event that occurs if both A and B occur on a single performance of the experiment, denoted as the symbol P(A  B) zEvents A and B are mutually exclusive if A  B contains no sample points, i.e. if A and B have no sample points in comon.

15. Unions and intersections A A B

16. Example 1. Consider the die-toss experiment. Define the following events: A: {Toss an even number} B: {Toss a number less than or equal to 3} Find:

17. Unions and intersections

18. Complementary Events zThe complement of an event A is the event that A does not occur -- that is, the event consisting of all sample points that are not in event A and denoted as symbol A c zP(A)+P(A c )=1

19. zAdditive Rule of Probability Probability

20. Conditional Probability zTo find the conditional probability that event A occurs given that event B occurs, divide the probability that both A and B occur by the probability that B occurs, that is,

21. Probability zTree diagram H H T T H T HH HT TH TT

22. Independent Events zEvents A and B are independent events if the occurrence of B does not alter the probability that A has occurred; that is, events A and B are independent if P(A|B)=P(A) zWhen events A and B are independent, it is also true that P(B|A)=P(B) zEvents that are not independent are said to be dependent.

23. Probability zProbability of Intersection of Two independent Events zIf events A and B are independent, the probability of the intersection of A and B equals the product of the probabilities of A and B; that is P(A  B)=P(A) P(B) zThe converse is also true: if P(A  B)=P(A) P(B), then A and B are independent.

24. Example zUse tree-diagram to obtain the Sample space of an experiment that consists of a “fair” coin being tossed three times. Consider the following events: zA=“All three results are the same”. zB=“exactly one Head occurs”. zC=“at least two Heads occur”. zFind: P(A),P(B),P(C), P(A)+P(B)+P(C), P(A  C), P(A  B), P(A c ),P(A|C)  Hence, explain if all the events A,B and C are not mutually exclusive and independent as well.

25. Solution: zP(A)=1/4;P(B)=3/8;P(C)=1/2; P(A)+P(B)+P(C)=1, P(A  C)=1/8, P(A  B)=P(A)+P(B)-P(A  B)=1/2. zP(A c )=1-1/4=3/4,P(A|C)= P(A  C)/P(C)=1/4  P(A)*P(B)=3/32  P(A  B) (P(A  B)= 0)  P(A)*P(C)=1/8 = P(A  C)  P(B)*P(C)=3/16  P(B  C) (P(B  C)= 1/8)

26. Exercise zCalculate the mode, mean, and median of the following data (1) 12, 13, 15, 18, 12, 56, 13, 17, 19, 20, 35, 36 (2) 35, 23, 18, 26, 35, 23, 39, 45, 47, 37, 23, 35, 19

27. Answer: (1): Mode: 12,13 Mean: 22.17 Median 17.5 (2): Mode: 23, 35 Mean: 31.15 Median 35

28. Excercise zCalculate the range, variance and standard deviation of the following data (1) 2, 3, 1, 6, 8, 5, 9, 4, 5 (2) 2, 0, 8, 4, 7, 5, 3, 2, 100

29. Answer: (1): range: 8 sample variance: 6.94 sample standard deviation: 2.64 (2): range: 98 sample variance: 1159.32 sample standard deviation: 34.05

30. Excercise zTwo fair coins are tossed and the following events are defined: A:{Observed at least one head} B:{Observed exactly one head} C:{Observed exactly one tail} D:{Observed at most one head} Find: P(A), P(B  D), P(A|D)

31. Answer: Pr(A)=3/4 Pr(B  D)=1/2 Pr(A|D)=2/3

32. Exercise Use tree-diagram to obtain the Sample space of an experiment that consists of a “fair” coin being tossed four times. Consider the following events: A=“All four results are the same”. B=“exactly one Head occurs”. C=“at least two Heads occur”. Find: P(A),P(B),P(C), P(A)+P(B)+P(C), P(A  C), P(A  B) Hence, explain why all the events A,B and C are not mutually exclusive.

33. Answer: Pr(A)=1/8, Pr(B)=1/4,Pr(C)=11/16 Pr(A)+Pr(B)+Pr(C)=17/16 Pr(A  C)=1/16 Pr(A  C)=3/4 Since Pr(A  C) is not equal to 0, so A,B,C are not mutually exclusive. But Pr(A  B)=Pr(B  C)=0, A,B and B,C are mutually exclusive respectively.

34. Exercise Let P(A)=0.7, P(B)=0.5 and P(A  B)= 0.8. Find: (1) P(A  B) (2) P(B|A) (3) Is event A independent of event B?

35. Answer: Pr(A  B)=Pr(A)+Pr(B)-Pr(A  B)=0.4 Pr(B|A)= Pr(A  B)/Pr(A)=4/7 Since Pr(B|A) is not equal to Pr(B), so event B is not independent of A.

36. References zStatistics, 8th Edition MaClave and Sincich Prentice Hall, 2000.