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Conditional Probability Two-Way Tables

Conditional Probability Two-Way Tables. Problem: Is ABS useful?

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Conditional Probability Two-Way Tables

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  1. Conditional Probability Two-Way Tables

  2. Problem: Is ABS useful? In a survey involving 100 cars, each vehicle was classifies according to whether or not it has antilock brakes system(ABS) and whether or not it has been involved in an accident in the past year. Suppose that any one of these cars is randomly selected for inspection. ABS No ABS Accident 3 12 No Accident 40 45 Transform this table into a probability table first.

  3. Transform this table into a probability table first. ABS No ABS Accident 3 12 No Accident 40 45 15 85 43 57 100

  4. Transform this table into a probability table first. ABS No ABS Accident 3 12 No Accident 40 45 15 Accident 15 85 43 57 100 Questions: 1) What is probability that the car has been involved in an accident in the past year? Total accidents 15 3 P(accident) = = = Total inspected 100 20

  5. Transform this table into a probability table first. ABS No ABS Accident 3 12 No Accident 40 45 ABS 15 85 43 43 57 100 Questions: 2) What is probability that the car has ABS Total with ABS 43 P(accident) = = Total inspected 100

  6. Transform this table into a probability table first. ABS No ABS Accident 3 12 No Accident 40 45 ABS 15 No Accident 40 85 43 57 100 Questions: 3) What is probability that the car has not been in an accident and has ABS ABS and No Accident 40 2 P(accident) = = = Total inspected 100 5

  7. Transform this table into a probability table first. ABS No ABS Accident 3 12 No Accident 40 45 ABS ABS No ABS Accident 3 3 12 15 15 No Accident 40 45 85 85 43 57 100 43 57 100 • Questions: • Given that the car has been involved in an accident, what is the probability that it has ABS? ABS and Accident 3 1 P(ABS|accident) = = = Total accident 15 5

  8. Transform this table into a probability table first. ABS No ABS Accident 3 12 No Accid 40 45 ABS No ABS Accident Accident 3 3 12 15 15 No Accident 40 45 85 85 43 43 57 100 57 100 Questions: 5) Given that the car has ABS, what is the prob that it has been involved in an accident? Accident and ABS 3 P(accident|ABS) = = Total ABS 43

  9. Transform this table into a probability table first. ABS No ABS Accident 3 12 No Accid 40 45 ABS No ABS Accident Accident 3 12 12 15 15 No Accident 40 45 85 85 43 43 57 100 57 100 Questions: 6) Given that the car does not have ABS, what is the prob that it has been involved in an accident? Accident and No ABS 12 4 P(acci|No ABS) = = = Total No ABS 57 19

  10. p203 #34-36

  11. Counting Principle 1. A car dealer offers a choice of 6 vinyl top colors, 18 body colors, and 7 upholstery colors. How many color combinations are possible? 6 * 18 * 7 = 756

  12. Counting Principle 2. How many ways can six prizes be awarded to 6 children (assuming each child only gets one prize)? What are the possibilities for the first prize? Any of the 6 What are the possibilities for the second prize? Any of the remaining 5 What are the possibilities for the third prize? Any of the 4

  13. Counting Principle 2. How many ways can six prizes be awarded to 6 children (assuming each child only gets one prize)? 6*5*4*3*2*1 = 6! 6! = 720

  14. Counting Principle 2. How many ways can six prizes be awarded to 6 children (assuming each prize can be awarded to any of the 6 children)? What are the possibilities for the first prize? Any of the 6 What are the possibilities for the second prize? Any of the 6 What are the possibilities for the third prize? Any of the 6

  15. Counting Principle 2. How many ways can six prizes be awarded to 6 children (assuming each prize can be awarded to any of the 6 children)? 6*6*6*6*6*6 = 66 66 = 46,656

  16. Counting Principle 3. The letters a, b, c, d, and e are used to form 5-letter patterns. How many patterns can be formed if repetitions are allowed? What are the possibilities for the first letter? Any of the 5 What are the possibilities for the second letter? Any of the 5 What are the possibilities for the third letter? Any of the 5

  17. Counting Principle 3. The letters a, b, c, d, and e are used to form 5-letter patterns. How many patterns can be formed if repetitions are allowed? 5*5*5*5*5 = 55 55 = 3125

  18. Counting Principle 3. The letters a, b, c, d, and e are used to form 5-letter patterns. How many patterns can be formed if repetitions are NOT allowed? What are the possibilities for the first letter? Any of the 5 What are the possibilities for the second letter? Any of the remaining 4 What are the possibilities for the third letter? Any of the remaining 3

  19. Counting Principle 3. The letters a, b, c, d, and e are used to form 5-letter patterns. How many patterns can be formed if repetitions are NOT allowed? 5*4*3*2*1 = 5! 5! = 120

  20. Factorial Notation product of all the integers from 1 to n We write "n factorial" with an exclamation mark. Symbol: n! *(Multiply n times every number between 1 and n)

  21. Try some on your own: a. 9! b. 1! c. _8!_ 6! 4. _19!_ 13! =(9)(8)(7)(6)(5)(4)(3)(2)(1) = 362,880 =(1) = 1 = (8)(7)(6)(5)(4)(3)(2)(1) (6)(5)(4)(3)(2)(1) = 56 = (19)(18)(17)(16)(15)(14)13! 13! = 19,535,040

  22. Permutation An arrangement of n objects in a specific order *ORDER MATTERS! n- total number r- number you are choosing

  23. Permutation Rule Taking n objects in a specific order using r objects at a time Symbol: nPr Formula: ___n!___ ((n-r)!) nPr =

  24. Permutation Rule ___n!___ ((n-r)!) Using : nPr = Find the following: 1. 12P5 2. 20P4 ___12!___ ((12-5)!) ___20!___ ((20-4)!) 95,040 116,280

  25. Permutation Rule Example 1 There are 8 stories for the local news tonight. One will be used as the lead story, one will be the second story, and one will be the closing story. If the director has a total of 8 stories to choose from, how many possible ways can the program be set up? Is there any significance to the order of the 3 stories selected? Lead, 2nd, closing YES! This is why it is a PERMUTATION n= 8 possible stories r = choosing 3 stories (lead, 2nd, closing)

  26. Permutation Rule Example 1 There are 8 stories for the local news tonight. One will be used as the lead story, one will be the second story, and one will be the closing story. If the director has a total of 8 stories to choose from, how many possible ways can the program be set up? _8*7*6*5!_ 5! ___8!___ (8-3)! 336 nPr = 8P3 = = =

  27. Permutation Rule Example 2 How many ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available? Is there any significance to the order of the 2 scientists selected? Chairperson, Assistant Chairperson YES! This is why it is a PERMUTATION n= 7 possible scientists r = choosing 2 scientists (chair, assist. chair)

  28. Permutation Rule Example 2 How many ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available? ___7!___ (7-2)! _7*6*5!_ 5! nPr = 7P2 = 42 = =

  29. Combination A selection of distinct objects without regard to order *ORDER does NOT matter

  30. Combination Rule Symbol: nCr Formula: ___n!___ ((n-r)!r!) nCr =

  31. Combination Rule ___n!___ ((n-r)!r!) Using : nCr = Find the following: 1. 13C8 2. 9C7 ___13!___ ((13-8)!8!) ___9!___ ((9-7)!7!) 1,287 36

  32. Combination Rule Example 1 A bicycle shop owner has 12 mountain bicycles in the showroom. The owner wishes to select 5 of them to display at a bicycle show. How many different ways can a group of 5 be selected? Is there any significance to the order of the 5 bicycles selected? NO! This is why it is a COMBINATION n= 12 possible bicycles r = choosing 5 bicycles

  33. Combination Rule Example 1 A bicycle shop owner has 12 mountain bicycles in the showroom. The owner wishes to select 5 of them to display at a bicycle show. How many different ways can a group of 5 be selected? ___12!___ ((12-5)!5!) _12!_ ((7!)5!) nCr = 12C5 = = _12*11*10*9*8*7!_ ((7!)5!) _12*11*10*9*8_ 5*4*3*2*1 = =

  34. Combination Rule Example 1 3 2 _12*11*10*9*8_ 5*4*3*2*1 _12*11*9*8_ 4*3*1 = 5*2 = 10 _12*11*3*2_ 1 792 = =

  35. Combination Rule Example 2 All of the 15 students in a class draw pictures. Only 9 can be displayed at a time. How many different combinations of pictures are there? Is there any significance to the order of the 9 pictures selected? NO! This is why it is a COMBINATION n= 15 possible pictures r = choosing 9 pictures

  36. Combination Rule Example 2 All of the 15 students in a class draw pictures. Only 9 can be displayed at a time. How many different combinations of pictures are there? ___15!___ ((15-9)!9!) _15!_ ((6!)9!) nCr = 15C9 = = _15*14*13*12*11*10*9!_ (6!)9! =

  37. Combination Rule Example 2 _15*14*13*12*11*10_ 6*5*4*3*2*1 = _14*13*11*10_ 4*1 = 5,005 =

  38. Combination or Permutation? A teacher has 5 different roles to be assigned to the students each day. If there are 18 students in the class, how many possible ways can the 5 roles be assigned? Ask yourself, does order matter or are you just choosing 5 students regardless of order? Order matters, therefore it is a PERMUTATION

  39. Combination or Permutation? A teacher has 5 different roles to be assigned to the students each day. If there are 18 students in the class, how many possible ways can the 5 roles be assigned? Since it is a permutation we use nPr nPr = 18P5 __n!__ ((n-r)!) __18!__ ((18-5)!) _18!_ 13! 1,028,160 18P5 = = = =

  40. Combination or Permutation? In a club with 12 members, they need 6 members to represent the club at the schools open house. How many ways can they come up with the 6 students? Ask yourself, does order matter or are you just choosing 6 members regardless of order? Order does not matter, therefore it is a COMBINATION

  41. Combination or Permutation? In a club with 12 members, they need 6 members to represent the club at the schools open house. How many ways can they come up with the 6 students? Since it is a combination we use nCr nCr = 12C6 __n!__((n-r)!r!) __12!___ ((12-6)!6!) _12!_ 6!6! 924 12C6 = = = =

  42. p211 #1, 5, 8, 10, 13, 18, 23, 24, 28

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