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Math 1300: Section 8-3 Conditional Probability, Intersection, and Independence

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  • 1. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary Math 1300 Finite MathematicsSection 8-3: Conditional Probability, Intersection, and Independence Jason Aubrey Department of Mathematics University of Missouri Jason Aubrey Math 1300 Finite Mathematics
  • 2. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryConditional Probability Conditional probability is the probability of an event occuring given that another event has already occured. Jason Aubrey Math 1300 Finite Mathematics
  • 3. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryConditional Probability Conditional probability is the probability of an event occuring given that another event has already occured. The conditional probability of event A occuring, given that event B has occured is denoted P(A|B), and is read as “probability of A, given B.” Jason Aubrey Math 1300 Finite Mathematics
  • 4. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryDefinitionFor events A and B in a arbitrary sample space S, we definethe conditional probability of A given B by P(A ∩ B) P(A|B) = P(B) = 0 P(B) Jason Aubrey Math 1300 Finite Mathematics
  • 5. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary 0.25A B.35 0.15 .25 Jason Aubrey Math 1300 Finite Mathematics
  • 6. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary 0.25A B.35 0.15 .25P(A ∩ B) = 0.15 Jason Aubrey Math 1300 Finite Mathematics
  • 7. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary 0.25A B .35 0.15 .25P(A ∩ B) = 0.15“out of all outcomes in thesample space, 15% ofthem are in A ∩ B. Jason Aubrey Math 1300 Finite Mathematics
  • 8. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary P(A|B) asks: out of all 0.25 outcomes in B, whatA B proportion of them are also in A. .35 0.15 .25P(A ∩ B) = 0.15“out of all outcomes in the P(A ∩ B) P(A|B) =sample space, 15% of P(B)them are in A ∩ B. Jason Aubrey Math 1300 Finite Mathematics
  • 9. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary P(A|B) asks: out of all 0.25 outcomes in B, whatA B proportion of them are also in A. .35 0.15 .25 So, we can think of the conditional probability as restricting the sample space.P(A ∩ B) = 0.15“out of all outcomes in the P(A ∩ B) P(A|B) =sample space, 15% of P(B)them are in A ∩ B. Jason Aubrey Math 1300 Finite Mathematics
  • 10. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary P(A|B) asks: out of all 0.25 outcomes in B, whatA B proportion of them are also in A. .35 0.15 .25 So, we can think of the conditional probability as restricting the sample space.P(A ∩ B) = 0.15“out of all outcomes in the P(A ∩ B) P(A|B) =sample space, 15% of P(B)them are in A ∩ B. 0.15 = = 0.375 0.4 Jason Aubrey Math 1300 Finite Mathematics
  • 11. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: Suppose events A, B, D and E have probabilities asgiven in the table. A B Totals D 0.2 0.4 0.6 E 0.28 0.12 0.4 Totals 0.48 0.52 1 Jason Aubrey Math 1300 Finite Mathematics
  • 12. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: Suppose events A, B, D and E have probabilities asgiven in the table. A B Totals D 0.2 0.4 0.6 E 0.28 0.12 0.4 Totals 0.48 0.52 1 P(B ∩ D) P(B|D) = P(D) Jason Aubrey Math 1300 Finite Mathematics
  • 13. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: Suppose events A, B, D and E have probabilities asgiven in the table. A B Totals D 0.2 0.4 0.6 E 0.28 0.12 0.4 Totals 0.48 0.52 1 P(B ∩ D) .4 2 P(B|D) = = = P(D) .6 3 Jason Aubrey Math 1300 Finite Mathematics
  • 14. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: Suppose events A, B, D and E have probabilities asgiven in the table. A B Totals D 0.2 0.4 0.6 E 0.28 0.12 0.4 Totals 0.48 0.52 1 P(B ∩ D) .4 2 P(B|D) = = = P(D) .6 3 P(A|E) Jason Aubrey Math 1300 Finite Mathematics
  • 15. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: Suppose events A, B, D and E have probabilities asgiven in the table. A B Totals D 0.2 0.4 0.6 E 0.28 0.12 0.4 Totals 0.48 0.52 1 P(B ∩ D) .4 2 P(B|D) = = = P(D) .6 3 P(A ∩ E) 0.28 P(A|E) = = = .7 P(E) 0.4 Jason Aubrey Math 1300 Finite Mathematics
  • 16. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: In a survey of 255 people, the following data wereobtained relating gender to political orientation: Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118 Female (F) 40 77 20 137 Total 119 99 37 255A person is randomly selected. What is the probability that: Jason Aubrey Math 1300 Finite Mathematics
  • 17. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: In a survey of 255 people, the following data wereobtained relating gender to political orientation: Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118 Female (F) 40 77 20 137 Total 119 99 37 255A person is randomly selected. What is the probability that: The person is male? Jason Aubrey Math 1300 Finite Mathematics
  • 18. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: In a survey of 255 people, the following data wereobtained relating gender to political orientation: Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118 Female (F) 40 77 20 137 Total 119 99 37 255A person is randomly selected. What is the probability that: The person is male? n(M) 118 P(M) = = ≈ .463 n(S) 255 Jason Aubrey Math 1300 Finite Mathematics
  • 19. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: In a survey of 255 people, the following data wereobtained relating gender to political orientation: Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118 Female (F) 40 77 20 137 Total 119 99 37 255A person is randomly selected. What is the probability that: The person is male? n(M) 118 P(M) = = ≈ .463 n(S) 255 The person is male and a Democrat? Jason Aubrey Math 1300 Finite Mathematics
  • 20. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: In a survey of 255 people, the following data wereobtained relating gender to political orientation: Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118 Female (F) 40 77 20 137 Total 119 99 37 255A person is randomly selected. What is the probability that: The person is male? n(M) 118 P(M) = = ≈ .463 n(S) 255 The person is male and a Democrat? n(M ∩ D) 22 P(M ∩ D) = = ≈ .086 n(S) 255 Jason Aubrey Math 1300 Finite Mathematics
  • 21. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118Female (F) 40 77 20 137 Total 119 99 37 255 P(M) = .46 P(M ∩ D) = 0.086 Jason Aubrey Math 1300 Finite Mathematics
  • 22. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118Female (F) 40 77 20 137 Total 119 99 37 255 P(M) = .46 P(M ∩ D) = 0.086 a Democrat? Jason Aubrey Math 1300 Finite Mathematics
  • 23. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118Female (F) 40 77 20 137 Total 119 99 37 255 P(M) = .46 P(M ∩ D) = 0.086 a Democrat? n(D) 99 P(D) = = ≈ .39 n(S) 255 Jason Aubrey Math 1300 Finite Mathematics
  • 24. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118Female (F) 40 77 20 137 Total 119 99 37 255 P(M) = .46 P(M ∩ D) = 0.086 a Democrat? n(D) 99 P(D) = = ≈ .39 n(S) 255 Male, given that the person is a Democrat? Jason Aubrey Math 1300 Finite Mathematics
  • 25. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary Republican (R) Democrat (D) Ind. (I) Total Male (M) 79 22 17 118Female (F) 40 77 20 137 Total 119 99 37 255 P(M) = .46 P(M ∩ D) = 0.086 a Democrat? n(D) 99 P(D) = = ≈ .39 n(S) 255 Male, given that the person is a Democrat? P(M ∩ D) 0.086 P(M|D) = = ≈ .22 P(D) .39 Jason Aubrey Math 1300 Finite Mathematics
  • 26. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryNoteNote that for sample spaces with equally likely outcomes n(A ∩ B) P(A|B) = n(B) Jason Aubrey Math 1300 Finite Mathematics
  • 27. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryNoteNote that for sample spaces with equally likely outcomes n(A ∩ B) P(A|B) = n(B)So, for example, from the previous problem we could have done n(M ∩ D) 22 P(M|D) = = ≈ .22 n(D) 99 Jason Aubrey Math 1300 Finite Mathematics
  • 28. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIntersection of Events: Product Rule In the previous section we sometimes used the addition principle P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find P(A ∩ B). Jason Aubrey Math 1300 Finite Mathematics
  • 29. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIntersection of Events: Product Rule In the previous section we sometimes used the addition principle P(A ∪ B) = P(A) + P(B) − P(A ∩ B) to find P(A ∩ B). The conditional probability formula gives another method for finding P(A ∩ B) when appropriate, called the Product Rule. Jason Aubrey Math 1300 Finite Mathematics
  • 30. Conditional ProbabilityIntersection of Events: Product Rule Probability Trees Independent Events Summary P(A ∩ B) P(A|B) = P(B) Jason Aubrey Math 1300 Finite Mathematics
  • 31. Conditional ProbabilityIntersection of Events: Product Rule Probability Trees Independent Events Summary P(A ∩ B) P(A|B) = P(B) P(A ∩ B) P(B)P(A|B) = P(B) P(B) Jason Aubrey Math 1300 Finite Mathematics
  • 32. Conditional ProbabilityIntersection of Events: Product Rule Probability Trees Independent Events Summary P(A ∩ B) P(A|B) = P(B) P(A ∩ B) P(B)P(A|B) = P(B) P(B) P(B)P(A|B) = P(A ∩ B) Jason Aubrey Math 1300 Finite Mathematics
  • 33. Conditional ProbabilityIntersection of Events: Product Rule Probability Trees Independent Events Summary P(A ∩ B) P(A|B) = P(B) P(A ∩ B) P(B)P(A|B) = P(B) P(B) P(B)P(A|B) = P(A ∩ B) P(A ∩ B) = P(A|B)P(B) Jason Aubrey Math 1300 Finite Mathematics
  • 34. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary P(A ∩ B) P(A|B) = P(B) P(A ∩ B) P(B)P(A|B) = P(B) P(B) P(B)P(A|B) = P(A ∩ B) P(A ∩ B) = P(A|B)P(B)Similarly, the formula P(B ∩ A) P(B|A) = P(A)gives P(A ∩ B) = P(B|A)P(A). Jason Aubrey Math 1300 Finite Mathematics
  • 35. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryTheoremFor events A and B with nonzero probabilities in a samplespace S, P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B)and we can use either P(A)P(B|A) or P(B)P(A|B) to computeP(A ∩ B). Jason Aubrey Math 1300 Finite Mathematics
  • 36. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: A computer chip manufacturer produces 45% of itschips at Plant A and at and the remainder at Plant B. However,1% of the chips produced at Plant A are defective, while 0.5%of the chips produced at Plant B are defective. What is theprobability that a randomly chosen computer chip produced bythis manufactuer is defective and was produced at Plant A? Jason Aubrey Math 1300 Finite Mathematics
  • 37. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: A computer chip manufacturer produces 45% of itschips at Plant A and at and the remainder at Plant B. However,1% of the chips produced at Plant A are defective, while 0.5%of the chips produced at Plant B are defective. What is theprobability that a randomly chosen computer chip produced bythis manufactuer is defective and was produced at Plant A?Let A be the event that a chip was produced at plant A. Let Drepresent the event that a chip is defective. Jason Aubrey Math 1300 Finite Mathematics
  • 38. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: A computer chip manufacturer produces 45% of itschips at Plant A and at and the remainder at Plant B. However,1% of the chips produced at Plant A are defective, while 0.5%of the chips produced at Plant B are defective. What is theprobability that a randomly chosen computer chip produced bythis manufactuer is defective and was produced at Plant A?Let A be the event that a chip was produced at plant A. Let Drepresent the event that a chip is defective. Then, P(A) = 0.45 and P(A ) = 0.55(Note that A represents the event that a chip was produced atplant B.) Jason Aubrey Math 1300 Finite Mathematics
  • 39. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThe statement “1% of the chips produced at Plant A aredefective, gives the conditional probability P(D|A) = 0.01 Jason Aubrey Math 1300 Finite Mathematics
  • 40. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThe statement “1% of the chips produced at Plant A aredefective, gives the conditional probability P(D|A) = 0.01Similarly, the statement ”0.5% of the chips produced at Plant Bare defective“ gives P(D|A ) = 0.005 Jason Aubrey Math 1300 Finite Mathematics
  • 41. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThe statement “1% of the chips produced at Plant A aredefective, gives the conditional probability P(D|A) = 0.01Similarly, the statement ”0.5% of the chips produced at Plant Bare defective“ gives P(D|A ) = 0.005The question What is the probability that a randomly chosen computer chip produced by this manufactuer is defective and was produced at Plant A? Jason Aubrey Math 1300 Finite Mathematics
  • 42. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThe statement “1% of the chips produced at Plant A aredefective, gives the conditional probability P(D|A) = 0.01Similarly, the statement ”0.5% of the chips produced at Plant Bare defective“ gives P(D|A ) = 0.005The question What is the probability that a randomly chosen computer chip produced by this manufactuer is defective and was produced at Plant A?is asking for P(D ∩ A). Jason Aubrey Math 1300 Finite Mathematics
  • 43. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryBy the product rule P(D ∩ A) = P(D|A)P(A) Jason Aubrey Math 1300 Finite Mathematics
  • 44. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryBy the product rule P(D ∩ A) = P(D|A)P(A) P(D ∩ A) = (0.01)(0.45) = 0.0045 Jason Aubrey Math 1300 Finite Mathematics
  • 45. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryBy the product rule P(D ∩ A) = P(D|A)P(A) P(D ∩ A) = (0.01)(0.45) = 0.0045So, the probability that a randomly chosen computer chipproduced by this manufactuer is defective and was produced atPlant A is 0.45%. Jason Aubrey Math 1300 Finite Mathematics
  • 46. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryProbability Trees Probability Trees We used tree diagrams in the previous chapter to help us count the number of combined outcomes in a sequence of experiments. Jason Aubrey Math 1300 Finite Mathematics
  • 47. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryProbability Trees Probability Trees We used tree diagrams in the previous chapter to help us count the number of combined outcomes in a sequence of experiments. Similarly, we can use tree diagrams to help us compute the probabilities of combined outcomes in a sequence of experiments. Jason Aubrey Math 1300 Finite Mathematics
  • 48. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryProbability Trees Probability Trees We used tree diagrams in the previous chapter to help us count the number of combined outcomes in a sequence of experiments. Similarly, we can use tree diagrams to help us compute the probabilities of combined outcomes in a sequence of experiments. A seqence of experiments where the outcome of each experiment is not certain is called a stochastic process. Jason Aubrey Math 1300 Finite Mathematics
  • 49. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryConstructing Probability Trees Draw a tree diagram corresponding to all combined outcomes of the sequence of experiments. Label each node. Assign a probability to each tree branch. Use the results from the first two steps to answer various questions related to the sequence of experiements as a whole. Jason Aubrey Math 1300 Finite Mathematics
  • 50. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: Suppose that 5% of all adults over 40 have diabetes.A certain physician correctly diagnoses 90% of all adults over40 with diabetes as having the disease and incorrectlydiagnoses 3% of all adults over 40 without diabetes as havingthe disease. Jason Aubrey Math 1300 Finite Mathematics
  • 51. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: Suppose that 5% of all adults over 40 have diabetes.A certain physician correctly diagnoses 90% of all adults over40 with diabetes as having the disease and incorrectlydiagnoses 3% of all adults over 40 without diabetes as havingthe disease.(a) Let A = the event ”has diabetes", and let B = the event ”is diagnosed with diabetes."Draw a tree diagram which models this scenario and label itwith appropriate probabilities. Jason Aubrey Math 1300 Finite Mathematics
  • 52. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary A = ”has diabetes" D = ”is diagnosed with diabetes."”Suppose that 5% of alladults over 40 havediabetes.“ .05 A .95 A Jason Aubrey Math 1300 Finite Mathematics
  • 53. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary A = ”has diabetes" D = ”is diagnosed with diabetes."”Suppose that 5% of alladults over 40 have .90 Ddiabetes.“”A certain physician .10correctly diagnoses 90% of .05 Aall adults over 40 with Ddiabetes as having thedisease. . . “ .95 A Jason Aubrey Math 1300 Finite Mathematics
  • 54. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary A = ”has diabetes" D = ”is diagnosed with diabetes."”Suppose that 5% of alladults over 40 have .90 Ddiabetes.“”A certain physician .10correctly diagnoses 90% of .05 Aall adults over 40 with Ddiabetes as having thedisease. . . “ .95 .03 Dand incorrectly diagnoses3% of all adults over 40 A .97without diabetes as having Dthe disease. Jason Aubrey Math 1300 Finite Mathematics
  • 55. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(b) Find the probability that a randomlyselected adult over 40 does not have .90 Ddiabetes, and is diagnosed as havingdiabetes. .10 .05 A D .95 .03 D A .97 D Jason Aubrey Math 1300 Finite Mathematics
  • 56. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(b) Find the probability that a randomlyselected adult over 40 does not have .90 Ddiabetes, and is diagnosed as havingdiabetes. .10 .05 A D “does not have diabetes” (A ) .95 .03 D A .97 D Jason Aubrey Math 1300 Finite Mathematics
  • 57. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(b) Find the probability that a randomlyselected adult over 40 does not have .90 Ddiabetes, and is diagnosed as havingdiabetes. .10 .05 A D “does not have diabetes” (A ) “is diagnosed as having diabetes” .95 .03 D (D ) A .97 D Jason Aubrey Math 1300 Finite Mathematics
  • 58. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(b) Find the probability that a randomlyselected adult over 40 does not have .90 Ddiabetes, and is diagnosed as havingdiabetes. .10 .05 A D “does not have diabetes” (A ) “is diagnosed as having diabetes” .95 .03 D (D ) P(A ∩ D) = P(A )P(D|A )P(A ) = A .97 (0.95)(0.03) = .0285 D Jason Aubrey Math 1300 Finite Mathematics
  • 59. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(c) Find the probability that a randomlyselected adult over 40 is diagnosed asnot having diabetes. .90 D .10 .05 A D .95 .03 D A .97 D Jason Aubrey Math 1300 Finite Mathematics
  • 60. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(c) Find the probability that a randomlyselected adult over 40 is diagnosed asnot having diabetes. We are looking for P(D ). .90 D .10 .05 A D .95 .03 D A .97 D Jason Aubrey Math 1300 Finite Mathematics
  • 61. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(c) Find the probability that a randomlyselected adult over 40 is diagnosed asnot having diabetes. We are looking for P(D ).There are two cases. Such a personeither .90 D does not have it and is (correctly) .10 diagnosed as not having it, .05 A D P(A ∩ D ) = P(A )P(D |A ) .95 .03 D A .97 D Jason Aubrey Math 1300 Finite Mathematics
  • 62. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(c) Find the probability that a randomlyselected adult over 40 is diagnosed asnot having diabetes. We are looking for P(D ).There are two cases. Such a personeither .90 D does not have it and is (correctly) .10 diagnosed as not having it, .05 A D P(A ∩ D ) = P(A )P(D |A ) .95 .03 D has it, but is (incorrectly) diagnosed as not having it, A .97 P(A ∩ D ) = P(A)P(D |A) D Jason Aubrey Math 1300 Finite Mathematics
  • 63. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThen P(D ) = P(A ∩ D ) + P(A ∩ D ) Jason Aubrey Math 1300 Finite Mathematics
  • 64. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThen P(D ) = P(A ∩ D ) + P(A ∩ D ) = P(A )P(D |A ) + P(A)P(D |A) Jason Aubrey Math 1300 Finite Mathematics
  • 65. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThen P(D ) = P(A ∩ D ) + P(A ∩ D ) = P(A )P(D |A ) + P(A)P(D |A) = (0.95)(0.97) + (0.05)(0.1) Jason Aubrey Math 1300 Finite Mathematics
  • 66. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThen P(D ) = P(A ∩ D ) + P(A ∩ D ) = P(A )P(D |A ) + P(A)P(D |A) = (0.95)(0.97) + (0.05)(0.1) = 0.9215 + 0.005 = 0.9265 Jason Aubrey Math 1300 Finite Mathematics
  • 67. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryThen P(D ) = P(A ∩ D ) + P(A ∩ D ) = P(A )P(D |A ) + P(A)P(D |A) = (0.95)(0.97) + (0.05)(0.1) = 0.9215 + 0.005 = 0.9265So, 92.65% of adults over 40 who take this test will bediagnosed as not having diabetes. Jason Aubrey Math 1300 Finite Mathematics
  • 68. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(d) Find the probability that a randomly selected adult over 40actually has diabetes, given that he/she is diagnosed as nothaving diabetes. Jason Aubrey Math 1300 Finite Mathematics
  • 69. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(d) Find the probability that a randomly selected adult over 40actually has diabetes, given that he/she is diagnosed as nothaving diabetes. P(A ∩ D )Here we are looking for P(A|D ) = . P(D ) Jason Aubrey Math 1300 Finite Mathematics
  • 70. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(d) Find the probability that a randomly selected adult over 40actually has diabetes, given that he/she is diagnosed as nothaving diabetes. P(A ∩ D )Here we are looking for P(A|D ) = . P(D )From the previous part, P(A ∩ D ) = P(A)P(D |A) = (0.05)(0.1) = 0.005 Jason Aubrey Math 1300 Finite Mathematics
  • 71. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(d) Find the probability that a randomly selected adult over 40actually has diabetes, given that he/she is diagnosed as nothaving diabetes. P(A ∩ D )Here we are looking for P(A|D ) = . P(D )From the previous part, P(A ∩ D ) = P(A)P(D |A) = (0.05)(0.1) = 0.005And P(D ) = 0.9265. So, Jason Aubrey Math 1300 Finite Mathematics
  • 72. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(d) Find the probability that a randomly selected adult over 40actually has diabetes, given that he/she is diagnosed as nothaving diabetes. P(A ∩ D )Here we are looking for P(A|D ) = . P(D )From the previous part, P(A ∩ D ) = P(A)P(D |A) = (0.05)(0.1) = 0.005And P(D ) = 0.9265. So, P(A ∩ D ) 0.005 P(A|D ) = = ≈ 0.0054 P(D ) 0.9265 Jason Aubrey Math 1300 Finite Mathematics
  • 73. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events Summary(d) Find the probability that a randomly selected adult over 40actually has diabetes, given that he/she is diagnosed as nothaving diabetes. P(A ∩ D )Here we are looking for P(A|D ) = . P(D )From the previous part, P(A ∩ D ) = P(A)P(D |A) = (0.05)(0.1) = 0.005And P(D ) = 0.9265. So, P(A ∩ D ) 0.005 P(A|D ) = = ≈ 0.0054 P(D ) 0.9265So 0.54% of adults diagnosed as not having diabetes by thisdoctor actually do have the disease. Jason Aubrey Math 1300 Finite Mathematics
  • 74. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIndependent Events In probability, to say that two events are independent intuitively means that the occurence of one event makes it neither more nor less probable that the other occurs. Jason Aubrey Math 1300 Finite Mathematics
  • 75. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIndependent Events In probability, to say that two events are independent intuitively means that the occurence of one event makes it neither more nor less probable that the other occurs. For example, if two cards are drawn with replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are independent. Jason Aubrey Math 1300 Finite Mathematics
  • 76. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIndependent Events In probability, to say that two events are independent intuitively means that the occurence of one event makes it neither more nor less probable that the other occurs. For example, if two cards are drawn with replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are independent. By contrast, if two cards are drawn without replacement from a deck of cards, the event of drawing a red card on the first trial and that of drawing a red card on the second trial are not independent. Jason Aubrey Math 1300 Finite Mathematics
  • 77. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIndependent Events Caution: very often intuition is not a reliable guide to the notion of independence. Must use mathematics to verify independence, not intuition. Jason Aubrey Math 1300 Finite Mathematics
  • 78. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIndependent Events Caution: very often intuition is not a reliable guide to the notion of independence. Must use mathematics to verify independence, not intuition. In problems we will either Jason Aubrey Math 1300 Finite Mathematics
  • 79. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIndependent Events Caution: very often intuition is not a reliable guide to the notion of independence. Must use mathematics to verify independence, not intuition. In problems we will either be asked to determine whether or not two events are independent, or Jason Aubrey Math 1300 Finite Mathematics
  • 80. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIndependent Events Caution: very often intuition is not a reliable guide to the notion of independence. Must use mathematics to verify independence, not intuition. In problems we will either be asked to determine whether or not two events are independent, or be asked to assume that two events are independent and to use that assumption to solve the problem. Jason Aubrey Math 1300 Finite Mathematics
  • 81. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryDefinition (Independence)If A and B are any events in a sample space S, we say that Aand B are independent if and only if P(A ∩ B) = P(A)P(B)Otherwise, A and B are said to be dependent. Jason Aubrey Math 1300 Finite Mathematics
  • 82. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 arethe events A and B independent? Jason Aubrey Math 1300 Finite Mathematics
  • 83. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 arethe events A and B independent?We are asked to determine whether or not two events areindependent. To do this, we must determine whether or not P(A ∩ B) = P(A)P(B) Jason Aubrey Math 1300 Finite Mathematics
  • 84. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 arethe events A and B independent?We are asked to determine whether or not two events areindependent. To do this, we must determine whether or not P(A ∩ B) = P(A)P(B)We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B). Jason Aubrey Math 1300 Finite Mathematics
  • 85. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 arethe events A and B independent?We are asked to determine whether or not two events areindependent. To do this, we must determine whether or not P(A ∩ B) = P(A)P(B)We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B). This is aperfect place to use the addition principle: Jason Aubrey Math 1300 Finite Mathematics
  • 86. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 arethe events A and B independent?We are asked to determine whether or not two events areindependent. To do this, we must determine whether or not P(A ∩ B) = P(A)P(B)We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B). This is aperfect place to use the addition principle: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.8 + 0.5 − 0.9 = 0.4 Jason Aubrey Math 1300 Finite Mathematics
  • 87. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 arethe events A and B independent?We are asked to determine whether or not two events areindependent. To do this, we must determine whether or not P(A ∩ B) = P(A)P(B)We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B). This is aperfect place to use the addition principle: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.8 + 0.5 − 0.9 = 0.4We also have P(A)P(B) = (0.8)(0.5) = 0.4 Jason Aubrey Math 1300 Finite Mathematics
  • 88. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: If P(A ∪ B) = 0.9, P(A) = 0.8 and P(B) = 0.5 arethe events A and B independent?We are asked to determine whether or not two events areindependent. To do this, we must determine whether or not P(A ∩ B) = P(A)P(B)We know P(A), P(B) and P(A ∪ B), but not P(A ∩ B). This is aperfect place to use the addition principle: P(A ∩ B) = P(A) + P(B) − P(A ∪ B) = 0.8 + 0.5 − 0.9 = 0.4We also have P(A)P(B) = (0.8)(0.5) = 0.4Since P(A ∩ B) = P(A)P(B) we conclude that A and B areindependent events. Jason Aubrey Math 1300 Finite Mathematics
  • 89. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: On a stormy night, the probability that the electricitygoes out is 7% and the probability that the phone goes out is3%. Assuming that the two are independent, what is theprobability that neither the electricity nor the phone goes out ona stormy night? Jason Aubrey Math 1300 Finite Mathematics
  • 90. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: On a stormy night, the probability that the electricitygoes out is 7% and the probability that the phone goes out is3%. Assuming that the two are independent, what is theprobability that neither the electricity nor the phone goes out ona stormy night?Here we are asked to assume that two events are independent,and we will use this assumption to solve the problem Jason Aubrey Math 1300 Finite Mathematics
  • 91. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: On a stormy night, the probability that the electricitygoes out is 7% and the probability that the phone goes out is3%. Assuming that the two are independent, what is theprobability that neither the electricity nor the phone goes out ona stormy night?Here we are asked to assume that two events are independent,and we will use this assumption to solve the problem Let E represent the electricity going out. So P(E) = 0.07. Jason Aubrey Math 1300 Finite Mathematics
  • 92. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: On a stormy night, the probability that the electricitygoes out is 7% and the probability that the phone goes out is3%. Assuming that the two are independent, what is theprobability that neither the electricity nor the phone goes out ona stormy night?Here we are asked to assume that two events are independent,and we will use this assumption to solve the problem Let E represent the electricity going out. So P(E) = 0.07. Let F represent the phone going out. So P(F ) = 0.03. Jason Aubrey Math 1300 Finite Mathematics
  • 93. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: On a stormy night, the probability that the electricitygoes out is 7% and the probability that the phone goes out is3%. Assuming that the two are independent, what is theprobability that neither the electricity nor the phone goes out ona stormy night?Here we are asked to assume that two events are independent,and we will use this assumption to solve the problem Let E represent the electricity going out. So P(E) = 0.07. Let F represent the phone going out. So P(F ) = 0.03. The probability that neither the electricity nor the phone goes out is given by P(E ∩ F ) Jason Aubrey Math 1300 Finite Mathematics
  • 94. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIf E and F are independent, so are E and F . So P(E ∩ F ) = P(E )P(F ) Jason Aubrey Math 1300 Finite Mathematics
  • 95. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIf E and F are independent, so are E and F . So P(E ∩ F ) = P(E )P(F )Then to solve the problem: P(E ∩ F ) = P(E )P(F ) Jason Aubrey Math 1300 Finite Mathematics
  • 96. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIf E and F are independent, so are E and F . So P(E ∩ F ) = P(E )P(F )Then to solve the problem: P(E ∩ F ) = P(E )P(F ) = (1 − P(E))(1 − P(F )) Jason Aubrey Math 1300 Finite Mathematics
  • 97. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIf E and F are independent, so are E and F . So P(E ∩ F ) = P(E )P(F )Then to solve the problem: P(E ∩ F ) = P(E )P(F ) = (1 − P(E))(1 − P(F )) = (1 − 0.07)(1 − 0.03) = (0.93)(0.97) Jason Aubrey Math 1300 Finite Mathematics
  • 98. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIf E and F are independent, so are E and F . So P(E ∩ F ) = P(E )P(F )Then to solve the problem: P(E ∩ F ) = P(E )P(F ) = (1 − P(E))(1 − P(F )) = (1 − 0.07)(1 − 0.03) = (0.93)(0.97) = 0.9021 Jason Aubrey Math 1300 Finite Mathematics
  • 99. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryIf E and F are independent, so are E and F . So P(E ∩ F ) = P(E )P(F )Then to solve the problem: P(E ∩ F ) = P(E )P(F ) = (1 − P(E))(1 − P(F )) = (1 − 0.07)(1 − 0.03) = (0.93)(0.97) = 0.9021So, the probability that neither the electricity nor the phonegoes out is 90.21%. Jason Aubrey Math 1300 Finite Mathematics
  • 100. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryTheoremIf A and B are independent events with nonzero probabilities ina sample space S, then P(A|B) = P(B) and P(B|A) = P(B)If either equation holds, then A and B are independent. Jason Aubrey Math 1300 Finite Mathematics
  • 101. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: A card is drawn from a standard 52 card deck.Events M and N are M = the drawn card is a diamond (♦) N = the drawn card is even (face cards are not valued) Jason Aubrey Math 1300 Finite Mathematics
  • 102. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: A card is drawn from a standard 52 card deck.Events M and N are M = the drawn card is a diamond (♦) N = the drawn card is even (face cards are not valued)(a) Find P(N|M). Jason Aubrey Math 1300 Finite Mathematics
  • 103. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: A card is drawn from a standard 52 card deck.Events M and N are M = the drawn card is a diamond (♦) N = the drawn card is even (face cards are not valued)(a) Find P(N|M).Notice that N ∩ M is the set of all even diamonds. This is, N ∩ M = {2♦, 4♦, 6♦, 8♦, 10♦} Jason Aubrey Math 1300 Finite Mathematics
  • 104. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: A card is drawn from a standard 52 card deck.Events M and N are M = the drawn card is a diamond (♦) N = the drawn card is even (face cards are not valued)(a) Find P(N|M).Notice that N ∩ M is the set of all even diamonds. This is, N ∩ M = {2♦, 4♦, 6♦, 8♦, 10♦}So n(N ∩ M) = 5. Jason Aubrey Math 1300 Finite Mathematics
  • 105. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryM is the set of all diamonds, so n(M) = 13. Jason Aubrey Math 1300 Finite Mathematics
  • 106. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryM is the set of all diamonds, so n(M) = 13. n(N ∩ M) P(N|M) = n(M) Jason Aubrey Math 1300 Finite Mathematics
  • 107. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryM is the set of all diamonds, so n(M) = 13. n(N ∩ M) 5 P(N|M) = = n(M) 13 Jason Aubrey Math 1300 Finite Mathematics
  • 108. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryM is the set of all diamonds, so n(M) = 13. n(N ∩ M) 5 P(N|M) = = n(M) 13(b) Test M and N for independence. Jason Aubrey Math 1300 Finite Mathematics
  • 109. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryM is the set of all diamonds, so n(M) = 13. n(N ∩ M) 5 P(N|M) = = n(M) 13(b) Test M and N for independence.Here we could check whether P(M ∩ N) = P(M)P(N), but it isslightly more convenient to check whether P(N|M) = P(N). Jason Aubrey Math 1300 Finite Mathematics
  • 110. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryM is the set of all diamonds, so n(M) = 13. n(N ∩ M) 5 P(N|M) = = n(M) 13(b) Test M and N for independence.Here we could check whether P(M ∩ N) = P(M)P(N), but it isslightly more convenient to check whether P(N|M) = P(N).N is the set of all even cards in the deck. There are 5 evencards per suit and 4 suits, so n(N) = 20. Jason Aubrey Math 1300 Finite Mathematics
  • 111. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryM is the set of all diamonds, so n(M) = 13. n(N ∩ M) 5 P(N|M) = = n(M) 13(b) Test M and N for independence.Here we could check whether P(M ∩ N) = P(M)P(N), but it isslightly more convenient to check whether P(N|M) = P(N).N is the set of all even cards in the deck. There are 5 evencards per suit and 4 suits, so n(N) = 20. Then 20 5 P(N) = = 52 13 Jason Aubrey Math 1300 Finite Mathematics
  • 112. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryM is the set of all diamonds, so n(M) = 13. n(N ∩ M) 5 P(N|M) = = n(M) 13(b) Test M and N for independence.Here we could check whether P(M ∩ N) = P(M)P(N), but it isslightly more convenient to check whether P(N|M) = P(N).N is the set of all even cards in the deck. There are 5 evencards per suit and 4 suits, so n(N) = 20. Then 20 5 P(N) = = 52 13P(N|M) = P(N) so yes, N and M are independent. Jason Aubrey Math 1300 Finite Mathematics
  • 113. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryDefinition (Independent Set of Events)A set of events is said to be independent if for each finitesubset {E1 , E2 , . . . , Ek } P(E1 ∩ E2 ∩ · · · ∩ Ek ) = P(E1 )P(E2 ) · · · P(Ek ) Jason Aubrey Math 1300 Finite Mathematics
  • 114. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: One interpretation of a baseball player’s battingaverage is as the probability of getting a hit each time the playergoes to bat. Jason Aubrey Math 1300 Finite Mathematics
  • 115. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: One interpretation of a baseball player’s battingaverage is as the probability of getting a hit each time the playergoes to bat.For instance, a player with a .300 average has probability .3 ofgetting a hit. Jason Aubrey Math 1300 Finite Mathematics
  • 116. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryExample: One interpretation of a baseball player’s battingaverage is as the probability of getting a hit each time the playergoes to bat.For instance, a player with a .300 average has probability .3 ofgetting a hit.If a player with a .300 batting average bats four times in a gameand each at-bat is an independent event, what is the probabilityof the player getting at least one hit in the game? Jason Aubrey Math 1300 Finite Mathematics
  • 117. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryLet Hi be the probability that the player gets a hit at his i th timeat bat. Then Jason Aubrey Math 1300 Finite Mathematics
  • 118. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryLet Hi be the probability that the player gets a hit at his i th timeat bat. Then Jason Aubrey Math 1300 Finite Mathematics
  • 119. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryLet Hi be the probability that the player gets a hit at his i th timeat bat. Then P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4 ) = 1 − P(H1 ∩ H2 ∩ H3 ∩ H4 ) Jason Aubrey Math 1300 Finite Mathematics
  • 120. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryLet Hi be the probability that the player gets a hit at his i th timeat bat. Then P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4 ) = 1 − P(H1 ∩ H2 ∩ H3 ∩ H4 ) = 1 − P(H1 )P(H2 )P(H3 )P(H4 ) Jason Aubrey Math 1300 Finite Mathematics
  • 121. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryLet Hi be the probability that the player gets a hit at his i th timeat bat. Then P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4 ) = 1 − P(H1 ∩ H2 ∩ H3 ∩ H4 ) = 1 − P(H1 )P(H2 )P(H3 )P(H4 ) = 1 − (0.7)4 = 0.7599 Jason Aubrey Math 1300 Finite Mathematics
  • 122. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummaryLet Hi be the probability that the player gets a hit at his i th timeat bat. Then P(H1 ∪ H2 ∪ H3 ∪ H3 ∪ H4 ) = 1 − P(H1 ∩ H2 ∩ H3 ∩ H4 ) = 1 − P(H1 )P(H2 )P(H3 )P(H4 ) = 1 − (0.7)4 = 0.7599So if a player with a batting average of .300 bats four times in agame, then there is about a 76% chance of that player getting ahit. Jason Aubrey Math 1300 Finite Mathematics
  • 123. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummarySummary Conditional Probability P(A ∩ B) P(B ∩ A) P(A|B) = P(B|A) = P(B) P(A) Note: P(A|B) is a probability based on the new sample space B, while P(A ∩ B) is based on the original sample space S. Product Rule P(A ∩ B) = P(B|A)P(A) = P(A|B)P(B) Jason Aubrey Math 1300 Finite Mathematics
  • 124. Conditional Probability Intersection of Events: Product Rule Probability Trees Independent Events SummarySummary Independent Events A and B are independent if and only if P(A ∩ B) = P(A)P(B) If A and B are independent events with nonzero probabilities, then P(A|B) = P(A) and P(B|A) = P(A) If A and B are independent events with nonzero probabilities and either P(A|B) = P(A) or P(B|A) = P(B), then A and B are independent. If E1 , E2 , . . . , En are independent, then P(E1 ∩ E2 ∩ . . . ∩ En ) = P(E1 )P(E2 ) · · · P(En ) Jason Aubrey Math 1300 Finite Mathematics