1. Introduction to Probability

2. A probability experiment (or trial) is any process with a result determined by chance. Each individual result that is possible for a probability experiment is an outcome. The sample space is the set of all possible outcomes for a given probability experiment. An event is a subset of outcomes from the sample space.

3. Identifying Outcomes in a Sample Space or Event Consider an experiment in which a coin is tossed and then a six-sided die is rolled. a. List the outcomes in the sample space for the experiment.

4. Identifying Outcomes in a Sample Space or Event (cont.) Solution a. Each outcome consists of a coin toss and a die roll. For example, heads and a 3 could be denoted as H3. Using this notation, the sample space can be written as follows.

5. Identifying Outcomes in a Sample Space or Event (cont.) Consider an experiment in which a coin is tossed and then a six-sided die is rolled. b. List the outcomes in the event “tossing a tail then rolling an odd number.”

6. Identifying Outcomes in a Sample Space or Event (cont.) b. Choosing the members of the sample space that fit the event “tossing a tail then rolling an odd number” gives the following:

7. Identifying Outcomes in a Sample Space Consider the experiment in which two die are rolled together. List the outcomes in the sample space..

8. Using a Pattern to List All Outcomes in a Sample Space The previous question has the same answer as the following: Consider the experiment in which a red six- sided die and a blue six-sided die are rolled together. a. Use a pattern to help list the outcomes in the sample space.

9. Using a Pattern to List All Outcomes in a Sample Space (cont.) Solution a. Although there are many patterns that could help us list all outcomes in the sample space, one pattern will be to start with keeping the red die at 1, and allowing the blue die to vary from 1 to 6.

10. Using a Pattern to List All Outcomes in a Sample Space (cont.) There are six of these outcomes: and. Next we list all outcomes in which a 2 is rolled on the red die. Again, there are six outcomes of this form. The pattern continues until all 36 outcomes are listed.

11. Using a Pattern to List All Outcomes in a Sample Space (cont.) Sample space =

12. Using a Pattern to List All Outcomes in a Sample Space (cont.) Note that rolling a 1 on the red die and a 2 on the blue die, denoted, is a different outcome than rolling a 2 on the red die and a 1 on the blue die, denoted.

13. Using a Pattern to List All Outcomes in a Sample Space (cont.) Consider the experiment in which a red six-sided die and a blue six-sided die are rolled together. b. List the outcomes in the event “the sum of the numbers rolled on the two dice equals 6.”

14. Using a Pattern to List All Outcomes in a Sample Space (cont.) b. Note that 1 + 5 = 6, so the outcome of rolling a 1 on the red die and a 5 on the blue die is in the event. The reverse would be rolling a 5 on the red die and a 1 on the blue die. These are two separate possible outcomes that are both in the event. The event consists of the following outcomes.

15. Using a Tree Diagram to List All Outcomes in a Sample Space Consider a family with three children. Use a tree diagram to find the sample space for the gender of each child in regard to birth order.

16. Using a Tree Diagram to List All Outcomes in a Sample Space Consider a family with three children. Use a tree diagram to find the sample space for the gender of each child in regard to birth order. Solution The tree begins with the two possibilities for the first child—girl or boy. It then branches for each of the other two births in the family as shown.

17. Using a Tree Diagram to List All Outcomes in a Sample Space (cont.) Key: The notation GBG indicates the family shown in red with a girl, then a boy, and finally a girl.

18. Using a Tree Diagram to List All Outcomes in a Sample Space (cont.) Using the tree diagram as a guide, the sample space can be written as follows.

19. Types of Probability Experimental probability (or empirical probability) uses the outcomes obtained by repeatedly performing an experiment to calculate the probability. Classical probability (or theoretical probability) is the most precise type of probability and can only be calculated when all possible outcomes in the sample space are known and equally likely to occur.

20. Introduction to Probability Experimental Probability In experimental probability, if E is an event, then P(E), read “the probability that E occurs,” is given by where f is the frequency of event E and n is the total number of times the experiment is performed.

21. Introduction to Probability Classical Probability In classical probability, if all outcomes are equally likely to occur, then P(E), read “the probability that E occurs,” is given by where n(E) is the number of outcomes in the event and n(S) is the number of outcomes in the sample space.

22. Calculating Classical Probability Beck is allergic to peanuts. At a large dinner party one evening, he notices that the cheesecake options on the dessert table contain the following flavors: 10 slices of chocolate, 12 slices of caramel, 12 slices of peanut butter chocolate, and 8 slices of strawberry. Assume that the desserts are served to guests at random. a. What is the probability that Beck’s cheesecake contains peanuts?

23. Calculating Classical Probability (cont.) Solution a. There are 12 slices of chocolate peanut butter cheesecake, and 10 + 12 + 12 + 8 = 42 pieces of cheesecake total. The probability is then calculated as follows.

24. Calculating Classical Probability (cont.) Beck is allergic to peanuts. At a large dinner party one evening, he notices that the cheesecake options on the dessert table contain the following flavors: 10 slices of chocolate, 12 slices of caramel, 12 slices of peanut butter chocolate, and 8 slices of strawberry. Assume that the desserts are served to guests at random. b. What is the probability that Beck’s dessert does not contain chocolate?

25. Calculating Classical Probability (cont.) b. There are 12 + 8 = 20 slices of cheesecake that do not contain chocolate. The probability of being served one of these desserts is calculated as follows.

26. Calculating Classical Probability Consider a beginning archer who only manages to hit the target 50% of the time. What is the probability that in three shots, the archer will hit the target all three times?

27. Calculating Classical Probability (cont.) Let’s use a tree diagram.

28. Calculating Classical Probability (cont.) Notice that this gives 8 possible outcomes for the three shots, only 1 of which consists of hitting the target all three times. Thus, the probability of the novice archer hitting the target three times in a row is calculated as follows.

29. Calculating Classical Probability Consider a family with six boys. What is the probability that the seventh child will also be a boy?

30. Example 4.7: Calculating Classical Probability (cont.) The only two outcomes left in the sample space would be BBBBBBG and BBBBBBB. The probability of baby number seven being a boy is then just as with any other pregnancy!

31. Calculating Classical Probability In biology, we learn that many diseases are genetic. One example of such a disease is Huntington’s disease, which causes neurological disorders as a person ages. Each person has two huntington genes—one inherited from each parent. If an individual inherits a mutated huntington gene from either of his or her parents, he or she will develop the disease. On the TV show House, the character who Dr. House calls “13” inherited this disease from her mother. Assume for a moment that “13” has a child with a person who has two healthy huntington genes. What is the probability that her child will develop Huntington’s disease?

32. Calculating Classical Probability (cont.) So the four possibilities for the child’s genes are {Hh, Hh, hh, hh}. Someone only has to have one mutated gene to develop the disease, so that means two of the four combinations would result in a child having the disease, {Hh, Hh}. Thus, if E = the event of the child inheriting a mutated huntington gene, then the probability that E occurs is calculated as follows.

33. Calculating Classical Probability (cont.) This means that “13” has a 50% chance of passing on the disease to a child. This is true for anyone with Huntington’s disease. The probability goes up if the other parent also has the disease.

34. More Probability

35. Properties of Probability 1.For any event, E, 2.For any sample space S, P(S) = 1. 3.For the empty event ∅, P( ∅ ) = 0.

36. The Complement Complement Rule for Probability The sum of the probabilities of an event, E, and its complement, E c, is equal to one.

37. Using the Complement Rule for Probability (cont.) b. If there is a 5% chance that none of the items on a scratch-off lottery ticket will be a winner, what is the probability that at least one of the scratch-off items will win?

38. Using the Complement Rule for Probability (cont.) b. The complement to having none of something is having at least one of that thing. Thus, the probability is calculated as follows. Thus, there is a 95% chance of having at least one winning scratch-off item.

39. Using the Complement Rule for Probability Roll a pair of standard six-sided dice. What is the probability that neither die is a three?

40. Using the Complement Rule for Probability (continued) Solution We could list the outcomes in E, every combination of the dice that does not have a three. It is much easier to count the outcomes in the complement, E c. The complement of this event contains the outcomes in which either die is a three. (Check for yourself by making sure that adding the event and its complement covers the entire sample space.) Let’s list these outcomes.

41. Using the Complement Rule for Probability (cont.) Let’s list these outcomes. E c =

42. Using the Complement Rule for Probability (cont.) There are 11 outcomes where at least one of the dice is a three. Since we have already seen that there are 36 possible ways to roll a pair of dice we have that

43. Using the Complement Rule for Probability (cont.) Subtracting this probability from 1 gives us the following. Therefore, the probability that neither die is a three is approximately 0.6944.

44. Using the Addition Rule for Probability Teresa is looking for a new condo to rent. Teresa’s realtor has provided her with the following list of amenities for 17 available properties. The list contains the following. Close to the subway: 6 properties Low maintenance fee: 7 properties Green space: 5 properties Newly renovated: 2 properties Close to the subway and low maintenance fee: 2 properties Green space and newly renovated: 1 property

45. Using the Addition Rule for Probability (cont.) If Teresa’s realtor selects the first condo they visit at random, what is the probability that the property is either close to the subway or has a low maintenance fee?

46. Using the Addition Rule for Probability (cont.) Solution Before we calculate the probability, let’s verify that Teresa’s realtor has accurately counted the total number of properties. At first glance, it might seem that there are more than 17 properties if you simply add all the numbers in the list the realtor gave. However, there are 6 + 7 + 5 + 2 = 20 properties that have single characteristics, and 2 + 1 = 3 properties containing two characteristics each. So, there are in fact only 20  3 = 17 individual properties.

47. Using the Addition Rule for Probability (cont.) The probability that the first condo Teresa sees is either close to the subway or has a low maintenance fee is 11 out of 17 approximately 64.71%.

48. Addition Rules for Probability Addition Rule for Probability For two events, E and F, the probability that E or F occurs is given by the following formula.

49. Using the Addition Rule for Probability (cont.)

50. Using the Addition Rule for Probability Suppose that after a vote in the US Senate on a proposed health care bill, the following table shows the breakdown of the votes by party. If a lobbyist stops a random senator after the vote, what is the probability that this senator will either be a Republican or have voted against the bill? Votes on Health Care Bill Democrat Republican Independent Voted in FavorVoted Against 23 43 2 21 7 4

51. Using the Addition Rule for Probability (cont.) Solution There are a total of 100 US senators, all of whom voted on this bill according to the table. Of these senators, 43 + 7 = 50 are Republicans and 21 + 7 + 4 = 32 voted against the bill. However, 7 of the senators are both Republican and voted against the bill.

52. Using the Addition Rule for Probability (cont.) Thus, the Addition Rule would apply as follows. So the probability that the senator the lobbyist stops will either be a Republican or have voted against the bill is 75%.

53. Using the Addition Rule for Probability Roll a pair of dice. What is the probability of rolling either a total less than four or a total equal to ten? Solution

54. Using the Addition Rule for Probability (cont.) We need to determine the number of outcomes that give a total less than four and the number of outcomes that give a total of ten. Let’s list these outcomes in a table. Totals Less Than FourTotals of Ten

55. Using the Addition Rule for Probability (cont.) By counting the outcomes, we see that there are 3 outcomes that have totals less than four and 3 outcomes that have totals of exactly ten. Note that there are no outcomes that are both less than four and exactly ten.

56. Using the Addition Rule for Probability (cont.) Hence, we fill in the probabilities as follows.

57. Addition Rules for Probability Addition Rule for Probability of Mutually Exclusive Events If two events, E and F, are mutually exclusive, then the probability that E or F occurs is given by the following formula.

58. Using the Addition Rule for Probability of Mutually Exclusive Events At a major exit on the interstate, past experience tells us that the probabilities of a truck driver refueling at each of the five possible gas stations are given in the table below. Assuming that the truck driver will refuel at only one of the stations (thus making the events mutually exclusive), what is the probability that the driver will refuel at Shell, Exxon, or Chevron? Probabilities of Refueling Gas StationProbability BP Chevron Exxon Shell Texaco 0.0351 0.1539 0.2793 0.3207 0.2110

59. Using the Addition Rule for Probability of Mutually Exclusive Events (cont.) Solution Since these three events are mutually exclusive, we can add the individual probabilities together.

60. And more probability

61. Fundamental Counting Principle Bulloch Drug sells ice cream cones. They have three different types of cones (cake, waffle, and sugar) and 12 different types of ice creams. How many different combinations you could make if you must choose one type of cone and one type of ice cream?

62. Fundamental Counting Principle You could have: a sugar cone with each of the 12 types of ice cream. a waffle cone with each of the 12 types of ice cream. a cake cone with each of the 12 types of ice cream. 3 x 12 = 36

63. Using the Fundamental Counting Principle The governing board of the local charity, Mission Stateville, is electing a new vice president and secretary to replace outgoing board members. If the board consists of 11 members who don’t already hold an office, in how many different ways can the two positions be filled if no one may hold more than one office?

64. Using the Fundamental Counting Principle Solution There are two slots to fill in this example. Once someone is chosen for one office, they cannot be chosen for the other. The first slot may be filled with any of the 11 board members. The second position has one fewer to choose from, so there are only 10 choices for it. Using the Fundamental Counting Principle, we then multiply. There are 11 ⋅ 10 = 110 possible ways to elect the new officers.

65. Using the Fundamental Counting Principle to Calculate Probability Robin prepares an afternoon snack for her friends, Matthew and Lainey, every day. She wants to give each friend one item. She has the following snacks on hand: carrots, raisins, crackers, grapes, apples, yogurt, and granola bars. If she randomly chooses one snack for Matthew and one snack for Lainey, what is the probability that each friend gets the same snack as yesterday?

66. Using the Fundamental Counting Principle to Calculate Probability Solution To begin, we need to count the number of ways in which Robin can randomly choose a snack for her friends. To do this, think of there being two slots to fill—one for each friend. Because there is no requirement that they have different snacks or the same snack, there are 7 possibilities for each friend. Therefore, there are 7 ⋅ 7 = 49 possible ways she can prepare the snacks.

67. Using the Fundamental Counting Principle to Calculate Probability Putting this information together, we can calculate the probability. Thus, there is about a 2% chance that each child will eat the same thing two days in a row.

68. Probability of independent events Two out of three people can roll their tongue. Nine out of ten people can wink. What is the probability that and given person can roll their tongue and wink?

69. Probability of independent events

70. Multiplication Rules for Probability Multiplication Rule for Probability of Independent Events For two independent events E and F, the probability that E and F occur is given by the following formula.

71. Using the Multiplication Rule for Probability of Independent Events Choose two cards from a standard deck, with replacement. What is the probability of choosing a king and then a queen? Solution Because the cards are replaced after each draw, the probability of the second event occurring is not affected by the outcome of the first event. Thus, these two events are independent.

72. Using the Multiplication Rule for Probability of Independent Events Using the Multiplication Rule for Probability of Independent Events, we have the following.

73. Using the Extended Multiplication Rule for Probability of Independent Events Assume that a study by Human Resources has found that the probabilities of an employee being written up for the following infractions are the values shown in the following table. Assume that each infraction is independent of the others. What is the probability that a given employee will be written up for being late for work, taking unauthorized breaks, and leaving early?

74. Using the Extended Multiplication Rule for Probability of Independent Events Since these three events are independent, we can apply the Multiplication Rule for Probability of Independent Events to this situation. Probabilities of Being Written Up at Work InfractionProbability Insubordination Late for work Failure to show up for work Leaving early Taking unauthorized breaks 0.1356 0.2478 0.1026 0.1954 0.3186

75. Using the Extended Multiplication Rule for Probability of Independent Events

76. Calculating Probability What is the probability of drawing a king and then a queen from a standard deck if the cards are drawn without replacement?

77. Calculating Probability What is the probability of drawing a king and then a queen from a standard deck if the cards are drawn without replacement? Solution This situation is essentially the same as drawing two cards at the same time from a standard deck. We think of the experiment in two stages, though, for ease in calculation. Begin by determining the probability of drawing a king from a standard deck of cards.

78. Calculating Probability Since there are 4 kings in a deck of 52 cards, the probability is Next, assume that when you drew the first card, it was a king. What is the probability that you now draw a queen? Well, since we are holding a king in our hand, there are still 4 queens left in the deck of cards. But, there are no longer 52 cards total, only 51. Thus,

79. Calculating Probability We can now find the probability of the multistage experiment by multiplying the two individual probabilities.

80. Calculating Conditional Probability One card has already been chosen from a standard deck without replacement. What is the probability of now choosing a second card from the deck and it being red, given that the first card was a diamond?

81. Calculating Conditional Probability One card has already been chosen from a standard deck without replacement. What is the probability of now choosing a second card from the deck and it being red, given that the first card was a diamond? Solution First, we must determine how many red cards are left in the deck after the first pick. Because the first card was a diamond (which is a red card), there are only 25 red cards left in the deck instead of 26. There are also only 51 cards total left in the deck.

82. Calculating Conditional Probability (cont.) Thus, the conditional probability is calculated as follows.

83. Using the Multiplication Rule for Probability of Dependent Events What is the probability of choosing two face cards in a row? Assume that the cards are chosen without replacement.

84. Using the Multiplication Rule for Probability of Dependent Events Solution When the first card is picked, all 12 face cards are available out of 52 cards. When the second card is drawn, there are only 11 face cards left out of the 51 cards remaining in the deck.

85. Using the Multiplication Rule for Probability of Dependent Events Thus, we have the following.

86. Using the Multiplication Rule for Probability of Dependent Events Assume that there are 17 men and 24 women in the Rotary Club. Two members are chosen at random each year to serve on the scholarship committee. What is the probability of choosing two members at random and the first being a man and the second being a woman?

87. Using the Multiplication Rule for Probability of Dependent Events When the first member is picked, there are 17 men out of 41 members. When the second member is picked, we assume that we have already picked a man, so that leaves all 24 women, but only 40 remaining members.

88. Using the Multiplication Rule for Probability of Dependent Events The calculation is as follows.