2. Probability Chapter 2 Towards a theory of probability… Chapter 2B Will the apple fall? It is a random event.

3. Today’s Menu 2-0 Complementary Rule 2-2 Axioms – one more time

4. 2-2.2 Axioms of Probability

5. Mutually Exclusive Events A Generalization A collection of events is mutually exclusive if for all pairs i and k, E i E k = For a collection of mutually exclusive events, We often rely on mutually exclusive events in a sample space They partition the space They make many computations feasible and simple P(E 1 E 2... E k ) = P(E 1 ) + P(E 2 ) +... + P(E k )

6. The Complimentary Rule Either an event will occur or it will not occur. That is P(E) + P(E c ) = 1 Or P(E) = 1 – P(E c ) It follows from E  E c = S and E, E c are mutually exclusive P(E) + P(E c ) = P(S) = 1 I knew that.

7. The Complimentary Rule Four members of the ENM faculty agree to go to one of 6 Los Vegas gambling casinos. What is the probability that at least two will end up in the same casino? Let E = the event, at least two meet, then E c = the event, none meet P(at least two meet) = 1 – P(none meet) P(E) = 1 - P(E c )

8. Picking your Casino Sampling with replacement I have 6 choices 6 x 6 x 6 x6 = 6 4

9. Picking your Casino Sampling without replacement I have 6 choices

10. Picking your Casino Sampling without replacement I choose Now I only have 5 choices

11. Picking your Casino Sampling without replacement I choose and I now have 4 choices

12. Picking your Casino Sampling without replacement I choose and I now have 3 choices 6 x 5 x 4 x 3 = P(6,4)

13. The Complimentary Rule P(at least two meet) = 1 – P(none meet) P(E) = 1 - P(E c )

14. The Birthday Problem The Enlightened Company gives each of its employees the day off on their birthday. What is the probability that at least two workers in the same office will be off on their birthdays the same day. Assume every day of the year is equally likely. Let n = the number of workers in an office and A = the event at least two people have a common birthday; then A c is the event that none do. P(A)

15. More Birthday Problems How many people are needed in order to have a better than 50% chance that two people have a birthday within k days of each other? within k days# people required 023 114 211 39 48 57 76

16. 2-3 The Addition Rule In general, P(A  B) = P(A) + P(B) – P(A  B) A B S If A & B do not intersect, i.e., P(A  B) = 0 P(A  B) = P(A) + P(B)  A & B are mutually exclusive Basically, we avoid double counting the overlap area

17. The Addition Rule – Example 1 Random process: draw a card at random from a deck of 52 playing cards Let A = the event, draw a spade: P(A) = 13/52 =1/4 let B = the event draw an ace: P(B) = 4/52 = 1/13 P(A  B) = 1/52 P(A  B) = P(A) + P(B) - P(A  B) = 13/52 + 4/52 – 1/52 = 16/52 =.3077

18. The Addition Rule – Example 2 In studying the cause of power failures, these data have been gathered: 20% are due to transformer damage (event A) 70% are due to line damage (event B) 2% involve both problems What is the probability that a given power failure involves transformer or line damage? P(A  B) = P(A) + P(B) – P(A  B) =.20 +.70 -.02 =.88

19. The three-event case AB C S

20. A three-event example The Pandemonium Company (TPC) produces powdered laundry detergent (SOPE). During final inspection a box of sope may be found to be underweight (Event A), not correctly labeled (Event B) or not properly sealed (Event C). Quality Control has provided the following data: 1.(easy) Underweight or incorrectly labeled boxes are rejected.* What fraction falls In this category? 2. (harder) What fraction of the boxes have no problems? *unsealed boxes are resealed and shipped to the customers. Event Probability A5% B7% C12% A and B3% A and C4% B and C5% A and B and C1%

21. A three-event example solved 2. P(A c  B c  C c ) = 1 - P(A c  B c  C c ) c =1 – P(A  B  C) = 1 – [ P(A) + P(B) + P(C) – P(A  B) – P(A  C) – P(B  C) + P(A  B  C)] = 1 – [.05 +.07 +.12 -.03 -.04 -.05 +.1] = 1 -.13 =.87 1. P(A  B) = P(A) + P(B) - P(A  B) =.05 +.07 -.03 =.09 1.(easy) Underweight or incorrectly labeled boxes are rejected. What fraction falls In this category? 2. (harder) What fraction of the boxes have no problems? Event Probability A5% B7% C12% A and B3% A and C4% B and C5% A and B and C1%

22. 2-4 Conditional Probability Conditional Probabilities are based upon a reduced sample space P(B|A) is the probability B occurs given that A has occurred. A B S

23. More Conditional Probability AB NTNT NBNB NANA N AB P(A) = N A /N T P(A B) = N AB /N T P(B|A) = P(A B) / P(A) = (N AB /N T )/(N A /N T ) = N AB / N A Let N i = the number of equally likely outcomes in event I N T = total number of outcomes in the sample space

24. A Conditional Example In a study of waters near power plants and other industrial plants, 5% showed signs of both chemical and thermal pollution. 40% showed signs of chemical pollution (event A) 30% showed signs of thermal pollution. (event B) What is the probability that a stream showing chemical pollution will show signs of thermal pollution? Desired: P(B|A) Given: P(A) =.4, P(B) =.3, P(A  B) = P(B  A) =.05 P(B|A) = P(B  A)/ P(A) =.05 /.40 = 1/8 =.125

25. Complementary Law A B S (A  B) (A  B c )

26. More of A Conditional Example In a study of waters near power plants and other industrial plants, 5% showed signs of chemical and thermal pollution. 40% showed signs of chemical pollution (event A) 30% showed signs of thermal pollution. (event B) What is the probability that a stream showing chemical pollution will not show signs of thermal pollution? Desired: P(B c |A) P(B C |A) = 1 - P(B|A) = 1-.125 =.875

27. A another conditional probability A company has 150 employees that are categorized according to their education and work assignment. An employee is selected at random: 1. What is the probability that the employee has a Master’s degree? 2. What is the probability that the employee works in sales given that the employee has a Master’s degree? 3. Given that the employee works in the office, what is probability the employee has a 2-yr technical degree?

28. 2-5 The Multiplication Rule P(A  B) = P(A|B) P(B) Since P(A|B) = P(B  A) / P(B) And P(B  A) = P(A  B)

29. Multiplication Rule Example 1 A parts bin contains 5 defective items from among 20 items in the bin. If 2 are selected at random, what is the probability that both are defective? Let A i = the event, the i th item selected is defective. Desired P(A 1  A 2 ) Since P(A 1  A 2 ) = P(A 2  A 1 ) = P(A 2 |A 1 ) P(A 1 ) P(A 1 ) = 5/20 and P(A 2 |A 1 ) = 4/19 Therefore P(A 1  A 2 ) = P(A 2 |A 1 ) P(A 1 ) = (5/20) (4/19) =.0526 I selected poorly. Is there an alternate approach to this problem?

30. An Alternate Approach A parts bin contains 5 defective items from among 20 items in the bin. If 2 are selected at random, what is the probability that both are defective? Let A i = the event, the i th item selected is defective. Desired P(A 1  A 2 ) P(A 1  A 2 ) = P(A 2 |A 1 ) P(A 1 ) = (5/20) (4/19) =.0526

31. Multiplication Rule Example 2 Following aircraft accidents, a detailed investigation is conducted. The probability that an accident due to structural failure is correctly identified is.9. If 25% of all accidents are due to structural failures, find the probability that an aircraft accident is due to structural failure and is diagnosed as due to structural failure. Define: A = the event, structural failure and B = the event, diagnosed correctly Given: P(B|A) =.9 and P(A) =.25 Required: P(A  B )= P(B|A) P(A) = (.9) (.25) =.225

32. Multiplication Rule Example 3 In a machine shop, there are four automatic screw machines. Past inspection reports yields the following data: If a screw is randomly picked from inventory, what is the probability that It will be from machine 3 and be defective? I’m a non- defective machine 3 screw. Let E i = the event screw came from machine i D = the event, screw is defective Given: P(E 3 ) =.20 and P(D | E 3 ) =.05 Required: P(E 3  D) = (.05) (.20) =.01

33. A harder question In a machine shop, there are four automatic screw machines. Past inspection reports yields the following data: If a screw is randomly picked and found to be defective, what is the probability that it was produced by machine 3? I’m a non- defective machine 3 screw. Let E i = the event screw came from machine i D = the event, screw is defective Given: P(E i ) and P(D | E i ) Required: P(E 3 |D) = P(E 3  D) / P(D) need P(D)

34. Continuing with a harder question Let D = (E 1  D)  (E 2  D)  (E 3  D)  (E 4  D) Then P(D) = P(E 1  D) + P(E 2  D) + P(E 3  D) + P(E 4  D) = P(D|E 1 ) P(E 1 ) + P(D|E 2 ) P(E 2 ) + P(D|E 3 ) P(E 3 ) + P(D|E 4 ) P(E 4 ) = (.04)(.15) + (.03)(.30) + (.05)(.20) + (.02)(.35) =.032 Required: P(E 3 |D) = P(E 3  D) / P(D) =.01/.032 =.3125

35. 2-5 Multiplication and Total Probability Rules We must use this power wisely! This is pretty powerful stuff!

36. Total Probability Rule (two events)

37. Total Probability Rule (multiple events) The events E i, i = 1,…,k, form a partition

38. Example Blood type distribution in the United States is Type A 41%, Type B 9%, Type AB 4%, and Type O 46%. 4% of the Army recruits with Type O are classified as Type A, 4% with Type B are classified as Type A, 10% with Type AB are classified as Type A, and 88% with Type A are classified as Type A Define the relevant events, express the above as probabilities of these events, and find the probability a recruit is classified as type A. Let A = event, blood type A B = event, blood type B AB = event, blood typeA B O = event, blood type O T = event, classified as type A P(A) =.41, P(B) =.09, P(AB) =.04, P(O) =.46 P(T|A) =.88, P(T|O) =.04 P(T|AB) =.10, P(T|B) =.04 P(T) = (.88)(.41) + (.04)(.46) + (.04)(.09) + (.10) (.04) =.3868

39. The Multiplication Rule applied to the Addition Rule P(A  B) = P(A) + P(B) – P(A  B) = P(A) + P(B) – P(A|B) P(B) = P(A) + P(B) – P(B|A) P(A) Gosh, both rules at once.

40. But if A and B are mutually exclusive… P(A  B) = P(A) + P(B) – P(A  B) Where P(A  B) =  and P(B|A) = P(A  B) / P(A) =  AB Note that A c and B c are not mutually exclusive.

41. A Two Rule Example The 17 th National Bank & Trust Company has observed that customers who have sufficient funds in their checking accounts postdate their checks, by mistake, once every 200 times. Customers who write checks on insufficient funds postdate them 95% of the time. Seven percent of the customers have insufficient funds. What is the probability of the bank receiving a postdated check or a check from an account having insufficient funds? Ye Olde Solution Box Let A = the event, check is postdated B = the event, customer has insufficient funds Given: P(A|B c ) =.005, P(A|B) =.95 P(B) =.07 Required: P(A  B) = P(A) + P(B) – P(A|B) P(B) P(A) = P(A|B c ) P(B c ) + P(A|B)P(B) = (.005)(.93) + (.95)(.07) =.07115 P(A  B) =.07115 +.07 – (.95)(.07) =.07465 17 th P(A  B) = P(A) + P(B) – P(B|A) P(A)

42. One Last Example Twenty percent of a shipment of preformatted CD’s contain defects. Twenty-five percent of the defective CD’s also contain a virus. Thirty-three percent of the non-defective disks also contain a virus. What is the probability of selecting at random a non-defective disk without a virus?

43. One Last Example Twenty percent of a shipment of formatted CD’s contain defects. Twenty-five percent of the defective CD’s also contain a virus. Thirty-three percent of the non-defective disks also contain a virus. What is the probability of selecting at random a non-defective disk without a virus? Ye Olde Solution Box: Let A = the event, CD has a defect B = the event, CD contains a virus Given: P(A) =.2, P(B|A) =.25, P(B|A c ) =.33 Required: P(A c  B c ) = 1 – P(A  B) = 1 – [P(A) + P(B) – P(B|A)P(A)] P(B) = P(B|A)P(A) + P(B|A c )P(A c ) = (.25)(.2) + (.33)(.8) =.314 P(A c  B c ) = 1 – [.2 +.314 – (.25)(.2)] = 1 -.464 =.536

44. Towards a Theory of Probability… This has been great! What do we have to look forward to next time? And many more examples!