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Rules of Probability
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1. Rules of Probability

2. The additive rule P[A  B] = P[A] + P[B] – P[A  B] and if P[A  B] = f P[A  B] = P[A] + P[B]

3. The additive rule for more than two events and if Ai  Aj = f for all i ≠ j. then

4. for any event E The Rule for complements

5. The multiplicative rule of probability and if A and B areindependent. This is the definition of independent

6. The multiplicative rule for more than two events

7. Proof and continuing we obtain

8. Example What is the probability that a poker hand is a royal flush i.e. • 10 , J , Q , K ,A • 10 , J , Q , K ,A • 10 , J , Q , K ,A • 10 , J , Q , K ,A

9. 10 , J , Q , K ,A • 10 , J , Q , K ,A • 10 , J , Q , K ,A • 10 , J , Q , K ,A Solution Let A1 = the event that the first card is a “royal flush” card. Let Ai = the event that the ith card is a “royal flush” card. i = 2, 3, 4, 5.

10. Another solution is by counting The same result

11. Independencefor more than 2 events

12. Definition: The set of k events A1, A2, … , Akare called mutually independent if: P[Ai1∩ Ai2∩… ∩ Aim] = P[Ai1] P[Ai2] …P[Aim] For every subset {i1, i2, … , im } of {1, 2, …, k } i.e.for k = 3 A1, A2, … , Akare mutually independentif: P[A1∩ A2] = P[A1] P[A2], P[A1∩ A3] = P[A1] P[A3], P[A2∩ A3] = P[A2] P[A3], P[A1∩ A2∩ A3] = P[A1] P[A2] P[A3]

13. P[A1] = .4, P[A2] = .5 , P[A3] = .6 P[A1∩A2] = (0.4)(0.5) = 0.20 P[A1∩ A3] = (0.4)(0.6) = 0.24 P[A2∩ A3] = (0.5)(0.6) = 0.30 P[A1∩ A2∩ A3] = (0.4)(0.5)(0.6) = 0.12 A1 A2 0.08 0.08 0.12 0.12 0.12 0.18 0.12 0.18 A3

14. Definition: The set of k events A1, A2, … , Akare called pairwise independent if: P[Ai∩ Aj] = P[Ai] P[Aj] for all i and j. i.e.for k = 3 A1, A2, … , Akare pairwise independentif: P[A1∩ A2] = P[A1] P[A2], P[A1∩ A3] = P[A1] P[A3], P[A2∩ A3] = P[A2] P[A3], It is not necessarily true that P[A1∩ A2∩ A3] = P[A1] P[A2] P[A3]

15. P[A1] = .4, P[A2] = .5 , P[A3] = .6 P[A1∩A2] = (0.4)(0.5) = 0.20 P[A1∩ A3] = (0.4)(0.6) = 0.24 P[A2∩ A3] = (0.5)(0.6) = 0.30 P[A1∩ A2∩ A3] = 0.14 ≠ (0.4)(0.5)(0.6) = 0.12 A1 A2 0.06 0.10 0.14 0.14 0.10 0.16 0.10 0.20 A3

16. Bayes Rule • Due to the reverend T. Bayes • Picture found on website: Portraits of Statisticians • http://www.york.ac.uk/depts/maths/histstat/people/welcome.htm#h

17. Proof:

18. Example: We have two urns. Urn 1 contains 14red balls and 12black balls. Urn 2 contains 6red balls and 20black balls. An Urn is selected at random and a ball is selected from that urn. Urn 1 Urn 2 If the ball turns out to be red what is the probability that it came from the first urn?

19. Solution: Let A = the event that we select urn 1 = the event that we select urn 2 Let B = the event that we select a red ball Note: the desired conditional probability is in the reverse direction of the given conditional probabilities. This is the case when Bayes rule should be used

20. Bayes rule states

21. Example: Testing for a disease Suppose that 0.1% of the population have a certain genetic disease. A test is available the detect the disease. If a person has the disease, the test concludes that he has the disease 96% of the time. It the person doesn’t have the disease the test states that he has the disease 2% of the time. Two properties of a medical test Sensitivity = P[ test is positive | disease] = 0.96 Specificity = P[ test is negative | disease] = 1 – 0.02 = 0.98 A person takes the test and the test is positive, what is the probability that he (or she) has the disease?

22. Solution: Let A = the event that the person has the disease = the event that the person doesn’t have the disease Let B = the event that the test is positive. Note: Again the desired conditional probability is in the reverse direction of the given conditional probabilities.

23. Bayes rule states Thus if the test turns out to be positive the chance of having the disease is still small (4.58%).Compare this to (.1%), the chance of having the disease without the positive test result.

24. An generlization of Bayes Rule Let A1, A2 , …, Ak denote a set of events such that for all i and j. Then

25. If A1, A2 , …, Ak denote a set of events such that for all i and j. Then A1, A2 , …, Ak is called a partition of S. S A2 Ak A1 …

26. Proof for all i and j. A2 Ak A1 B Then

27. and

28. Example: We have three urns. Urn 1 contains 14red balls and 12black balls. Urn 2 contains 6red balls and 20black balls. Urn 3 contains 3red balls and 23black balls. An Urn is selected at random and a ball is selected from that urn. Urn 1 Urn 3 Urn 2 If the ball turns out to be red what is the probability that it came from the first urn? second urn? third Urn?

29. Solution: Let Ai= the event that we select urn i Let B = the event that we select a red ball Note: the desired conditional probability is in the reverse direction of the given conditional probabilities. This is the case when Bayes rule should be used

30. Bayes rule states

31. Example: Suppose that an electronic device is manufactured by a company. During a period of a week • 15% of this product is manufactured on Monday, • 23% on Tuesday, • 26% on Wednesday , • 24% on Thursday and • 12% on Friday.

32. Also during a period of a week • 5% of the product is manufactured on Monday is defective • 3 % of the product is manufactured on Tuesday is defective, • 1 % of the product is manufactured on Wednesday is defective , • 2 % of the product is manufactured on Thursday is defective and • 6 % of the product is manufactured on Friday is defective. If the electronic device manufactured by this plant turns out to be defective, what is the probability that is as manufactured on Monday, Tuesday, Wednesday, Thursday or Friday?

33. A1 = the event that the product is manufactured on Monday A2 = the event that the product is manufactured on Tuesday A3 = the event that the product is manufactured on Wednesday A4 = the event that the product is manufactured on Thursday A5 = the event that the product is manufactured on Friday Solution: Let Let B = the event that the product is defective

34. Now P[A1]= 0.15, P[A2]= 0.23, P[A3]= 0.26, P[A4]= 0.24 and P[A5]= 0.12 Also P[B|A1]= 0.05, P[B|A2]= 0.03, P[B|A3]= 0.01, P[B|A4]= 0.02 and P[B|A5]= 0.06 We want to find P[A1|B], P[A2|B], P[A3|B], P[A4|B]and P[A5|B]. We will apply Bayes Rule

35. The sure thing principle and Simpson’s paradox

36. The sure thing principle Suppose Example – to illustrate Let A = the event that horse A wins the race. B = the event that horse B wins the race. C = the event that the track is dry = the event that the track is muddy

37. Proof:

38. Simpson’s Paradox Does Example to illustrate D = death due to lung cancer S = smoker C = lives in city, = lives in country

39. Solution similarly

40. whether is greater than depends also on the values of

41. whether than and