1. 1 Chapter 3 Probability 3-1 Fundamentals 3-2 Addition Rule 3-3 Multiplication Rule: Basics 3-4 Multiplication Rule: Complements and Conditional Probability 3-5Counting Techniques

2. 2 Objectives  develop sound understanding of probability values used in subsequent chapters  develop basic skills necessary to solve simple probability problems

3. 3 General Comments  This chapter tends to be the most difficult one encountered in the course  Homework note:  Show setup of problem even if using calculator

4. 4  Experiment – Action being performed (book uses the word procedure)  Event (E) – A particular observation within the experiment  Sample space (S) - all possible events within the experiment 3-1 Fundamentals Definitions

5. 5 Notation P - denotes a probability A, B,... - denote specific events P (A) - denotes the probability of event A occurring

6. 6 Basic Rules for Computing Probability Let A equal an Event P(A) = number of outcomes favorable to event “A” total possible experimental outcomes(sample space)

7. 7 Probability Limits  The probability of an impossible event is 0.  The probability of an event that is certain to occur is 1. 0  P(A)  1 Impossible to occur Certain to occur

8. 8 Possible Values for Probabilities Certain Likely 50-50 Chance Unlikely Impossible 1 0.5 0

9. 9 Unlikely Probabilities Examples: Winning the lottery Being struck by lightning 0.0000035892 1 / 727,235 Typically any probability less than 0.05 is considered unlikely.

10. 10 Example: Roll a die and observe a 4? Find the probability. What is the experiment? Roll a die What is the event A? Observe a 4 What is the sample space? 1,2,3,4,5,6 Number of outcomes favorable to A is 1. Number of total outcomes is 6. What is P(A)? P(A) = 1 / 6 = 0.167 Similar to #4 on hw

11. 11 Example: Toss a coin 3 times and observe exactly 2 heads? Experiment: toss a coin 3 times Event (A): observe exactly 2 heads Sample Space: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Note there are 3 outcomes favorable to the event and 8 total outcomes P(A) = 3 / 8 = 0.375 Test problem Similar to #6 on HW

12. 12 Law of Large Numbers As a procedure is repeated again and again, the probability of an event tends to approach the actual probability. This is the reason to quit while your ahead when gambling!

13. 13 P(A) Complementary Events The complement of event A, denoted by A, consists of all outcomes in which event A does not occur. P(A) (read “not A”) or P(A c )

14. 14 P(A) = 1 – P(A c ) Complementary Events Property of complementary events Example: If the probability of something occurring is 1/6 what is the probability that it won’t occur?

15. 15 Example: A study of randomly selected American Airlines flights showed that 344 arrived on time and 56 arrived late, What is the probability of a flight arriving late? Let A = late flight A c = on time flight P(A c ) = 344 /(344 + 56) = 344/400 =.86 P(A) = 1 – P(A c ) = 1 -.86 = 0.14 Similar to number 11 & 12 on hw

16. 16 Rounding Off Probabilities  give the exact fraction or decimal or  round off the final result to three significant digits Examples: 1/3 is exact and could be left as a fraction or rounded to.333 0.00038795 would be rounded to 0.000388

17. 17  Compound Event – Any event combining 2 or more events  Notation – P(A or B) = P (event A occurs or event B occurs or they both occur)  General Rule – add the total ways A can occur and the total way B can occur but don’t double count 3-2 Addition Rule Definitions

18. 18 Formal Addition Rule P(A or B) = P(A) + P(B) - P(A and B) where P(A and B) denotes the probability that A and B both occur at the same time. Alternate form P(A B) = P(A) + P(B) – P(A B) Compound Event

19. 19 Definition Events A and B are mutually exclusive if they cannot occur simultaneously.

20. 20 Definition Total Area = 1 P(A) P(B) P(A and B) Non-overlapping Events Overlapping Events Not Mutually Exclusive P(A or B) = P(A) + P(B) – P(A and B) Mutually Exclusive P(A or B) = P(A) + P(B)

21. 21 Applying the Addition Rule P(A or B) Addition Rule Are A and B mutually exclusive ? P(A or B) = P(A)+ P(B) - P(A and B) P(A or B) = P(A) + P(B) Yes No

22. 22 Mutually Exclusive Example: P(A) = 2/7 and P(B) = 3/7, P(A or B) = 5/7, are A and B mutually exclusive? Why? Test question

23. 23 Example: You have an URN with 2 green marbles, 3 red marbles and 4 white marbles Let A = choose red marble and B = choose white marble 1.What is the probability of choosing a red marble? P(A) = 3/9 2.What is the probability of choosing a white marble? P(B) = 4/9 3.What is the probability of choosing a red or a white? P(B or A) = 3/9 + 4/9 = 7/9 Test Questions Why are event A and B mutually exclusive events?

24. 24 Example: A card is drawn from a deck of cards. 1.What is the probability that the card is an ace or jack? P(ace) + P(jack) = 4/52 + 4/52 = 8/52 2.What is the probability that the card is an ace or heart? P(ace) + P(heart) – P(ace of hearts) = 4/52 + 13/52 – 1/52 = 16/52

25. 25 Example: Toss a coin 3 times and observe all possibilities of the number of heads Experiment: toss a coin 3 times Events (A): observe exactly 0 heads (B): observe exactly 1 head (C): observe exactly 2 heads (D):observe exactly 3 heads Sample Space: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Find the P(A) + P(B) + P(C) + P(D)

26. 26 Example: Toss a coin 3 times and observe all possibilities of the number of heads EventP(event) A = 0 heads1/8 B = 1 head3/8 C = 2 heads3/8 D = 3 heads1/8 Total1 Probability distribution: Table of all possible events along with the probability of each event. The sum of all probabilities must sum to ONE. Note: Events are mutually exclusive

27. 27 Example: Roll 2 dice and observe the sum Experiment: roll 2 dice Event (F): observe sum of 5 Sample Space: 36 elements One 2 Two 3’s Three 4’s Four 5’s Five 6’s, One 12 Two 11’s Three 10’s Four 9’s Five 8’s Six 7’s Find the P(F)

28. 28 Example: Roll 2 dice and observe the sum EventP(event) A: Sum = 21/36 B: Sum = 32/36 And so on……. Construct a probability distribution Let’s Try #8 From the HW

29. 29 Let A = select a man Let B = select a girl P(A or B) =+ = = 0.781 Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table (Titanic Mortality) * Mutually Exclusive * 1692 2223 45 2223 1737 2223

30. 30 Let A = select a woman Let B = select someone who died. P (A or B) = (422 + 1517 - 104) / 2223 = 1835 / 2223 = 0.825 Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table (Titanic Mortality) * NOT Mutually Exclusive * Very similar to test problem

31. 31 How could you define a probability distribution for this data? Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table (Titanic Mortality)

32. 32 Complementary Events P(A) & P(A c )  P(A) and P(A c ) are mutually exclusive  P(A) + P(A c ) = 1 (this has to be true)  P(A)= 1 - P(A c )  P(A c ) = 1 – P(A)

33. 33 Venn Diagram for the Complement of Event A Total Area = 1 P (A) P (A) = 1 - P (A)

34. 34  Notation: P(A and B) = P(event A occurs in a first trial and event B occurs in a second trial)  Formal Rule  P(A and B) = P(A) P(B) if independent (with replacement)  P(A and B) = P(A) P(B A) if dependent (without replacement) 3-3 Multiplication Rule Definitions will define later

35. 35 Definitions  Independent Events Two events A and B are independent if the occurrence of one does not affect the probability of the occurrence of the other.  Dependent Events If A and B are not independent, they are said to be dependent.

36. 36 Ta Tb Tc Td Te Fa Fb Fc Fd Fe abcdeabcdeabcdeabcde TFTF P(T) = P(c) = P(T and c) = Tree Diagram of Test Answers 1 2 1 5 1 10

37. 37 P (both correct) = P (T and c) 1 10 1 2 1 5 = Multiplication Rule INDEPENDENT EVENTS

38. 38  Choose 2 marbles from an URN with 3 red marbles and 3 white marbles  Dependent – choose the 1 st marble then choose the 2 nd marble  Independent – choose the 1 st marble, replace it, then choose the 2 nd marble Independence vs. Dependence

39. 39 P(B A) represents the probability of event B occurring after it is assumed that event A has already occurred (read B A as “B given A”). = given Notation for Conditional Probability

40. 40 Example: You have an URN with 3 red marbles and 4 white marbles Let A = choose red marble and B = choose white marble 1.What is the probability of choosing a red marble? P(A) 2.What is the probability of choosing a white marble? P(B) 3.If two are chosen find the probability of choosing a white on the a second trial given a red marble was chosen 1 st. P(B A) a)Assume the 1 st marble is replaced {independent} b)Assume the 1 st marble is not replaced {dependent} Test Questions

41. 41 Example: You have an URN with 3 red marbles and 4 white marbles Let’s change things a bit…. If two are chosen and we want to find the probability of choosing a white then choosing a red marble. P(A and B) So if we let: A = choose red 1 st and B = choose white 2 nd then we need to find P(A and B) The problem here is that calculating this probability depends on what happens on the first draw. We need a rule that helps us with this.

42. 42 Formal Multiplication Rule  P(A and B) = P(A) P(B A)  If A and B are independent events, P(B A) is really the same as P(B). Will see this in the next section.

43. 43 Applying the Multiplication Rule P(A and B) Multiplication Rule Are A and B independent ? P(A and B) = P(A) P(B A) P(A and B) = P(A) P(B) Yes No

44. 44 Example: You have an URN with 3 red marbles and 4 white marbles Let A = choose red marble and B = choose white marble If two are chosen find the probability of choosing a red then choosing a white marble. In other words find P(A and B) a)Assume the 1 st marble is replaced {independent} P(A and B) = P(A) P(B) b)Assume the 1 st marble is not replaced {dependent} P(A and B) = P(A) P(B A) Use as an example for #6 Test Questions

45. 45 Class Assignment – Part I You have an URN with 3 red marbles, 7 white marbles, and 1 green marble Let A = choose red B = choose white C = choose green Find the following: 1.P(A c ), that is find P(not red) 2.P(A or B) 3.If two marbles chosen what is the probability that you choose a white marble 2 nd when a red marble was chosen first. This is, find P(B A) 4.If two marbles are chosen, find P(A and C) with replacement 5.If two marbles are chosen, find P(A and C) without replacement Very Similar to test question #1

46. 46 Class Assignment – Part II

47. 47 Mutually Exclusive vs. Independent Events  Mutually Exclusive Events P(A or B) = P(A) + P(B) Independent Events P(A and B) = P(A) P(B) Example: if P(A) =.3, P(B)=.4, P(A or B)=.7, and P(A and B) =.12, what can you say about A and B? Note: Test Question

48. 48 Select two –Find P(2 women) = 422/2223 x 421/2222 –Find P(2 that died) = 1517/2223 x 1516/2222 Select one –Find P(woman and died) = 104/2223 –Find P(Boy and survived) = 29 / 2223 Men Women Boys Girls Totals Survived 332 31829 27 706 Died 1360 10435 18 1517 Total 1692 422 64 45 2223 Contingency Table (Titanic Mortality)

49. 49 3-4 Topics  Probability of “at least one”  More on conditional probability  Test for independence

50. 50 Probability of ‘At Least One’  ‘At least one’ is equivalent to ‘one or more’.  The complement of getting at least one item of a particular type is that you get no items of that type. IfP(A) = P(getting at least one), then P(A) = 1 - P(A c ) where P(A c ) = P(getting none)

51. 51 Probability of ‘At Least One’  Find the probability of a getting at least 1 head if you toss a coin 4 times. P(A) = 1 - P(A c ) where P(A) is P(no heads) P(A c ) = (0.5)(0.5)(0.5 )(0.5) = 0.0625 P(A) = 1 - 0.0625 = 0.9375

52. 52 Conditional Probability P(A and B) = P(A) P(B | A)  Divide both sides by P(A)  Formal definition for conditional probability P(B | A) = P(A and B) P(A)

53. 53 If P(B | A) = P(B) then the occurrence of A has no effect on the probability of event B; that is, A and B are independent events. or If P(A and B) = P(A) P(B) then A and B are independent events. (with replacement) Testing for Independence Example: if A and B are independent, find P(A and B) if P(A) = 0.3 and P(B) = 0.6 (test question)

54. 54 3-5 Counting  fundamental counting rule  two events (“mn” rule)  multiple events (n r rule)  permutations  factorial rule  different items  not all items different  combinations

55. 55 Fundamental Counting Rule (‘mn” rule) If one event can occur m ways and the second event can occur n ways, the events together can occur a total of m n ways. Example 1: Example 1: How many ways can your order a meal with 3 main course choices and 4 deserts? First list then use rule. Main Courses Tacos (T) Pasta (P) Liver & Onions (LO) Deserts Ice Cream (IC) Jello (J) Cake © Fruit (F)

56. 56 Fundamental Counting Rule (n r rule) If one event that can occur n ways is repeated r times, the events together can occur a total of n r ways. Example 2: Example 2: How many outcomes are possible when tossing a coin 3 times? First list then use rule. Example 3: Example 3: How many outcomes are possible when tossing a coin 20 times? Would you care to list all the outcomes this time?

57. 57  The factorial symbol ! denotes the product of decreasing positive whole numbers.  n ! = n (n- 1 ) (n-2) (n- 3 ) (3) (2) (1)  Special Definition: 0 ! = 1  Find the ! key on your calculator Notation

58. 58 A collection of n different items can be arranged in order n ! different ways. Factorial Rule Example: Example: How many ways can you order the letters A, B, C? List first then use rule. Note: actually a special type of permutation, will define next

59. 59 Compare the N R and Factorial Rule Example: Example: How many ways can you order the letters A, B, C? a)N R _______ _________ ________ (with replacement) b)N! _______ _________ ________ (without replacement)

60. 60  n is the number of available items (without replacement)  r is the number of items to be selected  the number of permutations (or sequences) is Permutations Rule (when items are all different)  Order matters P n r = ( n - r ) ! n!

61. 61  Example: Eight men enter a race. In how many ways can the first 4 positions be determined? Permutations Rule (when items are all different)

62. 62 Permutations Rule ( when some items are identical )  If there are n items with n 1 alike, n 2 alike,.... n k alike, the number of permutations is n 1 !. n 2 !........ n k ! n!n!

63. 63 Permutations Rule ( when some items are identical ) 1.How many ways can you arrange the word statistics? Or Mississippi? Examples: 2.How many ways can you arrange 3 green marbles and 4 red marbles? Test Question

64. 64 Permutations Rule  Factorial rule is special case n! = n P Can you show this is true?

65. 65  n different items  r items to be selected  different orders of the same items are not counted (order doesn’t matter)  the number of combinations is (n - r )! r! n! n C r = Combinations Rule

66. 66 TI-83 Calculator Calculate n!, n P r, n C r 1.Enter the value for n 2.Press Math 3.Cursor over to Prb 4.Choose 2: n P r or 3: n C r or 4: n! as required 5a. Press Enter for the n! case 5b. Enter the value for “r” for the n P r and n C r cases

67. 67 Pick five numbers from 1 to 47 and a MEGA number from 1 to 27 Pick five numbers from 1 to 56 and a MEGA number from 1 to 46 Note: game has 2 separate sets of numbers

68. 68  Example: Find the probability of winning the Pennsylvania Super 6 lotto. Select 6 numbers from 69.  What’s the probably of getting 5 of 6? 4 of 6?, etc. (see lottery handout)  What’s the probability if you have to get all 6 numbers in a specified order? Combinations Rule

69. 69 Recall previous example? Experiment: Toss a coin 3 times Event (A): Observe 2 heads Sample Space: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT Note there are 3 outcomes favorable to the event and 8 total outcomes P(A) = 3 / 8 = 0.375 Test problem

70. 70 Let’s take a different approach Experiment: Toss a coin 3 times Event (A): Observe 2 heads There are 2 3 possible outcomes and 3 C 2 ways to get 2 heads P(A) = 3 C 2 / 2 3 = 3 / 8 = 0.375 Test problem

71. 71 Example: Experiment: Toss a coin 6 times Event (A): Observe 4 heads There are 2 6 possible outcomes and 6 C 4 ways to get 4 heads P(A) = 6 C 4 / 2 6 = 15 / 64 = 0.234 Test problem