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Chapter 5

A Survey of. Probability. Concepts. Chapter 5. 1. 2. 3. 4. Chapter Goals. When you have completed this chapter, you will be able to:. Explain the terms random experiment , outcome , sample space , permutations , and combinations. Define probability.

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Chapter 5

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  1. A Survey of Probability Concepts Chapter 5

  2. 1 2 3 4 Chapter Goals When you have completed this chapter, you will be able to: Explain the termsrandom experiment, outcome,sample space,permutations,andcombinations. Define probability. Describetheclassical,empirical,andsubjectiveapproachesto probability. Explainandcalculateconditional probabilityand joint probability. and...

  3. 5 6 7 Chapter Goals Calculate probability using the rules of addition andrules of multiplication. Usea tree diagram to organize and compute probabilities. Calculate a probabilityusing Bayes’ theorem.

  4. Descriptive Inferential Methods of…collectingorganizingpresenting and analyzingdata Science of…making inferences about a population, based onsample information. Types of Statistics Recall Emphasis now to be on this!

  5. Terminology Probability …is a measure of the likelihood that an event in the future will happen! • It can only assume a value between 0 and 1. • A value near zero means the event is not likely happen;nearone means it is likely.. • There are three definitions of probability: classical, empirical, and subjective More

  6. Terminology Random Experiment …is a process repetitive in nature the outcome of any trial is uncertain well-defined set of possible outcomes each outcome hasa probabilityassociated with it More

  7. Terminology Outcome Sample space Event …is a particularresult of a random experiment. ... is the collection or set of all the possible outcomes of a random experiment. …is the collection of one or moreoutcomes of an experiment. More

  8. Approaches to Assigning Probability NUMBER of favourableoutcomes Probability of an Event = TotalNUMBERofpossibleoutcomes Examples Subjective …probability is based on whatever information is available Objective Classical Probability Empirical Probability … is based on the assumption that the outcomes of an experiment are equally likely … applies when the number of timesthe event happens is divided by the number of observations

  9. S ubjective Probability Example …. refers to the chance of occurrence assigned to an event by a particular individual It is not computed objectively, i.e., notfrom prior knowledge or from actual data… …that the Toronto Maple Leafs will win the Stanley Cup next season! …that you will arrive to class on time tomorrow!

  10. E mpirical Probability A Students measure the contents of their soft drink cans… 10 cans are underfilled, 32are filled correctly and 8 are overfilled When the contents of the next can is measured, what is the probability that it is… (a) filled correctly? P(C) = 32 / 50 = 64% …(b) notfilled correctly? P(~C) = 1 – P(C) = 1 - .64 = 36% This is called theComplement of C

  11. Random Experiment The experiment is rolling the die...once! The possible outcomes are the numbers… 1 2 3 4 5 6 An event is the occurrence of an even number i.e. we collect the outcomes2, 4, and 6.

  12. Tree Diagrams This is a useful device to show all the possible outcomes of the experiment and their corresponding probabilities Consider the random experiment of flipping a coin twice.

  13. Tree Diagrams First Flip SecondFlip New 1.00 Origin Expressed as: P(HH)= 0.25 HH H H P(HT)= 0.25 HT T Simple Events H TH P(TH)= 0.25 T T TT P(TT)= 0.25

  14. Tree Diagrams Pie Menu Ice Cream Beef Appetizer: Soup or Juice Entrée: Beef Turkey Fish Dessert: Pie Ice Cream Pie Turkey Ice Cream Pie Fish Ice Cream Beef Pie Ice Cream Turkey Pie Fish Ice Cream Pie Ice Cream Origin Appetizer Entrée Dessert Soup Juice

  15. Tree Diagrams 12 How many complete dinners are there?

  16. Tree Diagrams How many dinnersinclude beef? 1. 2. 4 3. 4.

  17. Tree Diagrams Juice? Turkey? Both beef and soup? What is the probability that a complete dinner will include… 6/12 4/12 See next slide… 2/12

  18. The M *NRule  If one thing can be done in M ways, and if after this is done, something else can be done in N ways, then both things can be done in a total of M*N different ways in that stated order! Legend: Appetizer Entrée Dessert Refer back to tree diagram example:  # different meals = 2 * 3 * 2 = 12  # meals with beef = 2 * 1 * 2 = 4  # meals with juice = 1 * 3 * 2 = 6

  19. A Example When getting dressed, you have a choice between wearing one of: 3 pairs of shoes 2 pairs of pants 5 shirts Find the number of different “outfits” possible 3 * 2 * 5 = 30

  20. P robability What is the probability of drawing ared Ace from a deck of well-shuffled cards? P( Red Ace) = 2/52

  21. P Deck = 52 Cards robability 4 Suits Spades Clubs Diamonds Hearts 13 cards in each P robability Using Analysis Key steps 1. Determine….the Outcomes that Meet Our Condition 2. List….all Possible Outcomes

  22. P robability …of getting four(4) aces Deck 4 Suits Data Spades Clubs Hearts Diamonds P = probability = 52 Cards(the Population) 4 Suits x 13 cards 13 cards in each

  23. P = 52 Cards robability 4 Suits(13 cards in each) Spades Clubs Scenarios Deck ‘Honours’ cards Hearts Diamonds Each Suit has a…….

  24. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck 2 52 26 52 Condition Outcomes All Possible Outcomes Condition Outcomes All Possible Outcomes Condition Outcomes All Possible Outcomes Hearts Diamonds 4 1. Draw an Ace 52 2. Draw a Black Ace 3. Draw a Red Card

  25. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck 28 52 28 52 26 52 2 52 Condition Outcomes All Possible Outcomes = 26 + 4 - 2 52 Hearts Diamonds 4. Drawing…aRed Card or a Queen + = -or-P(Red) + P(Queen) - P (Red Queen) =

  26. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck Hearts Diamonds What is the probability of drawing a Jack or a King from a deck of well-shuffled cards? = 4/52 = 4/52 P( Jack or King) = 4/52 + 4/52 = 8/52

  27. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck These are MUTUALLY EXCLUSIVE events, i.e.they can’t both happen at the same time! Note Hearts Diamonds What is the probability of drawing one card that is both a Jack and a King from a deck of well-shuffled cards? P( Jack and King) = 0

  28. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck Hearts Diamonds What is the probability of drawing one card that is both BLACK and a King from a deck of well-shuffled cards? P( Black and King) =2/52

  29. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck Formula P(A or B) = Note This is called the Addition Rule Hearts Diamonds What is the probability of drawing a card that is either BLACK or a King from a deck of well-shuffled cards? P (A) + P(B) –P(Both) P( Black or King) = 26/52 + 4/52 - 2/52 = 28/52

  30. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck This is called a CONDITIONALprobability Note Hearts Diamonds Alternate solution What is the probability of drawing a King given that you have drawn a BLACK card? Our sample space is now just theBLACKcards = 2/26 P(King|Black )

  31. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck P (Both) Hearts Diamonds What is the probability of drawing a King given that you have drawn a BLACK card? Formula P(A|B) = P(Given) = (2/52) / (26/52) = (2/52) * (52/26) = 2/26

  32. = 52 Cards = 52 Cards 4 Suits(13 cards in each) 4 Suits(13 cards in each) Scenarios Scenarios Spades Spades Clubs Clubs Deck Deck P (Both) P(A|B) = P(Given) Hearts Hearts Diamonds Diamonds What is the probability of drawing a King of Clubs given that you have drawn a BLACK card? Formula = (1/52) / (26/52) P(King of Clubs|Black ) = (1/52) * (52/26) = 1/26

  33. = 52 Cards 4 Suits(13 cards in each) Scenarios Spades Clubs Deck Hearts Diamonds What is the probability of drawing a King of Clubs given that you have drawn a CLUB? P(King of Clubs given Club) = P(King of Clubs|Club) = P(1/52) / (13/52) = (1/52) * (52/13) = 1/13

  34. ReadingProbabilitiesfrom a Table

  35. ReadingProbabilitiesfrom a Table A More A survey of undergraduate students in the School of Business Management at Eton College revealed the following regarding the gender and majors of the students: Gender AccountingInternationalHR TOTAL What is the Probability of selecting a female student? 400/750 = 53.33%

  36. ReadingProbabilitiesfrom a Table A More What is the Probability of selecting a Human Resources or Internationalmajor? Gender AccountingInternationalHR TOTAL P(HR or I) = P(HR) + P(I) = 115/750 + 310/750 = 425/750 = 56.67%

  37. ReadingProbabilitiesfrom a Table A More What is the Probability of selecting a Female or Internationalmajor? Gender AccountingInternationalHR TOTAL P(F or I) = P(F) + P(I) – P(F and I) – 160/750 = 400/750 + 310/750 = 550/750 = 73.33%

  38. ReadingProbabilitiesfrom a Table A More What is the Probability of selecting a FemaleAccounting student? Gender AccountingInternationalHR TOTAL P(F and A) = 175/750 = 23.33%

  39. ReadingProbabilitiesfrom a Table A Alternative Solution What is the Probability of selecting a Female, given that the person selectedis anInternational major? Gender AccountingInternationalHR TOTAL 160/310 = 51.6% P(F|I) =

  40. ReadingProbabilitiesfrom a Table A Formula P(A|B) = P(Both) P(Given) What is the Probability of selecting a Female, given that the person selectedis anInternational major? P(F|I) = P(F and I) / P(I) = (160/750) / (310/750) = 160/310 = 51.6%

  41. ReadingProbabilitiesfrom a Table A More What is the Probability of selecting an International major, given that the person selectedis a Female? Gender AccountingInternationalHR TOTAL 160/400 = 40% P(I|F) =

  42. ReadingProbabilitiesfrom a Table and …betweenF given I …I given F! Notice the significant difference:

  43. Terminology A Independent Events Events are independentif the occurrence ofone eventdoes not affectthe probability of the other Each flip is independent of the other! Find the probability of flipping 2 Heads in a row Flip once Flip twice P(2H) = .5*.5 = .25 or 25% Consider the random experiment of flipping a coin twice.

  44. Terminology Independent Events Draw three cards withreplacementi.e.,draw one card, look at it, put it back, and repeat twice more. Each draw is independent of the other Find the probability of drawing 3 Queens in a row: = 0.00046=most unlikely! P(3Q) = 4/52 * 4/52 *4/52

  45. Note Independent Events • Consider 2 events: • Drawing a RED card from a deck of cards • Drawing a HEART from a deck of cards Are these two events considered to be independent? If two events, A and B are independent, then P(A|B) = P(A) P(Red) = 26/52 = 1/2 P(Red|Heart) = 13/13 = 1 Therefore these are NOT independent events!

  46. B T ayes’ heorem

  47. B ayes’ T heorem Example …isamethod for revising a probability given additional information! Formula P(B|A1 ) P(A1 ) P(A1|B) = P(A1 ) P(A2 ) P(B|A1)+ P(B|A2 )

  48. B ayes’ T heorem Duff Cola Company recently received several complaints that their bottles are under-filled. A complaint was received today but the production manager is unable to identify which of the two Springfield plants (A or B) filled this bottle. What is the probability that the under-filled bottle came from plant A?

  49. B ayes’ T heorem 1 2 What is the probability that the under-filled bottle came from plant A? List theProbabilities given P(plant A) = .55 P(plant B) = .45 P(Underfilled -A) = .03 P(Underfilled -B) = .04 Input values into formula and compute

  50. B ayes’ T heorem 1 2 P(B|A1 ) P(A1 ) P(A1 |B) = P(B|A1 )+ P(B|A2 ) P(A1 ) P(A2 ) .55(.03) What is the probability that the under-filled bottle came from plant A? List theProbabilities given P(plant A) = .55 P(plant B) = .45 P(Underfilled/A) = .03 P(Underfilled/B) = .04 Input values into formula and compute The likelihood that the underfilled bottle came from Plant A has been reduced from 55% to 47.83% = = .4783 .55(.03) + .45(.04)

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