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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. Conditional Probability,Independence andThe Multiplicative Rue

  6. Then the conditional probability of A given B is defined to be:

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

  8. The multiplicative rule for more than two events

  9. Proof and continuing we obtain

  10. 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

  11. 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.

  12. Another solution is by counting The same result

  13. Independencefor more than 2 events

  14. 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]

  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] = (0.4)(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

  16. 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]

  17. 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

  18. 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

  19. Proof:

  20. 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?

  21. 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

  22. Bayes rule states

  23. 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?

  24. 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.

  25. 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.

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

  27. 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 …

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

  29. and

  30. 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?

  31. 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

  32. Bayes rule states

  33. 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.

  34. 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?

  35. 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

  36. 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

  37. The sure thing principle and Simpson’s paradox

  38. 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

  39. Proof:

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

  41. Solution similarly

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

  43. whether than and