This document provides examples and explanations of key concepts in probability, including: 1) Probability is a number between 0 and 1 that indicates the likelihood of an event. Experimental probability is calculated from observations, while theoretical probability uses the composition of a sample space. 2) Tree diagrams and the fundamental counting principle can be used to determine the number of possible outcomes and probabilities in multi-stage experiments. 3) Union, intersection, and complements of events are probability concepts used to calculate probabilities of combined events.

1. Chapter 10 Probability

2. Section 10.1 Simple Experiments Goals Study probability Experimental probability Theoretical probability Study probability properties Mutually exclusive events Unions and intersections of events Complements of events

3. Interpreting Probability Probability is the mathematics of chance. For example, the statement “The chances of winning the lottery game are 1 in 150,000” means that only 1 of every 150,000 lottery tickets printed is a winning ticket.

4. Probability Terminology Making an observation or taking a measurement is called an experiment . An outcome is one of the possible results of an experiment. The set of all possible outcomes is called the sample space . An event is any collection of possible outcomes.

5. Example 1 An experiment consists of rolling a standard six-sided die and recording the number of dots showing on the top face. List the sample space. List one possible event.

6. Example 1 The sample space contains 6 possible outcomes: {1, 2, 3, 4, 5, 6}. One possible event: The event of rolling an even number: {2,4,6}

7. Example 2 An experiment consists of tossing a coin 3 times and recording the results in order. List the sample space. List one possible event.

8. Example 2, cont’d Solution: The sample space contains 8 possible outcomes and can be written {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. One possible event is {HTH, HTT, TTH, TTT}, which is the event of getting a tail on the second coin toss.

9. Example 3 An experiment consists of spinning a spinner twice and recording the colors it lands on. List the sample space. List one possible event.

10. Example 3 Solution: The sample space contains 16 possible outcomes and can be written {RR, RY, RG, RB, YR, YY, YG, YB, GR, GY, GG, GB, BR, BY, BG, BB}. One possible event is {RR, YY, GG, BB}, which is the event of getting the same color on both spins.

11. Example 4 The experiment consists of rolling 2 standard dice and recording the number appearing on each die. List the sample space. List one possible event.

12. Example 4 Solution: The sample space contains 36 possible outcomes and can be written {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.

13. Example 4 Solution, cont’d: One possible event is {(6,1), (5,2), (4,3), (3,4), (2,5), (1,6)}, which is the event of getting a total of 7 dots on the two dice.

14. Probability, cont’d The probability of an event is a number from 0 to 1, and can be written as a fraction, decimal, or percent. The greater the probability, the more likely the event is to occur. An impossible event has probability 0. A certain event has probability 1.

15. Experimental Probability One way to find the probability of an event is to conduct a series of experiments.

16. Example 5 An experiment consisted of tossing 2 coins 500 times and recording the results Let E be the event of getting a head on the first coin Find the experimental probability of E .

17. Example 5 The even of getting a head on the first coin: {HH, HT}. occurred a total of 137 + 115 = 252 times out of 500. The experimental probability of E is

18. Theoretical Probability Another way to find the probability of an event is to use the theory of what “should” happen rather than conducting experiments.

19. Example 6 An experiment consists of tossing 2 fair coins. Find the theoretical probability of: Each outcome in the sample space. The event E of getting a head on the first coin. The event of getting at least one head.

20. Example 6, cont’d Solution: There are 4 outcomes in the sample space: {HH, HT, TH, TT}. Each outcome is equally likely to occur.

21. Example 6 The event E is {HH, HT} and the theoretical probability of E is the number of outcomes in E divided by the number of outcomes in the sample space.

22. Example 6 c) The event of getting at least one head is E = {HH, HT, TH}.

23. Example 7 An experiment consists of rolling 2 fair dice. Find the theoretical probability of: Event A : getting 7 dots. Event B : getting 8 dots. Event C : getting at least 4 dots.

24. Example 7 Solution: There are 36 outcomes in the sample space. (see pg 635 for a list of the sample space.) Event A : getting 7 dots contains 6 outcomes: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. Each outcome is equally likely to occur.

25. Example 7 Solution, cont’d: Event B : getting 8 dots contains 5 equally likely outcomes: {(2,6), (3,5), (4,4), (5,3), (6,2)}.

26. Example 7 Solution, cont’d: Event C : getting at least 4 dots (or a total of 4 or more) contains 33 equally likely outcomes: { (1,3), (1,4), (1,5), (1,6), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }.

27. Example 8 A jar contains four marbles: 1 red, 1 green, 1 yellow, and 1 white.

28. Example 8 If we draw 2 marbles in a row, without replacing the first one, find the probability of: Event A : One of the marbles is red. Event B : The first marble is red or yellow. Event C : The marbles are the same color. Event D : The first marble is not white. Event E : Neither marble is blue.

29. Example 8 If we draw 2 marbles in a row, without replacing the first one, find the probability of: Event A : One of the marbles is red. A = {RG, RY, RW, GR, YR, WR}.

30. Example 8 If we draw 2 marbles in a row, without replacing the first one, find the probability of: Event B : The first marble is red or yellow. B = {RG, RY, RW, YR, YG, YW}. Event C : The marbles are the same color.

31. Example 8 If we draw 2 marbles in a row, without replacing the first one, find the probability of: Event D : The first marble is not white. D = {RG, RY, RW, GR, GY, GW, YR, YG, YW}. e) Event E : Neither marble is blue.

32. Union and Intersection The union of two events, A U B , refers to all outcomes that are in one, the other, or both events. The intersection of two events, A ∩ B, refers to outcomes that are in both events.

33. Mutually Exclusive Events Events that have no outcomes in common are said to be mutually exclusive . If A and B are mutually exclusive events, then

34. Example 9 A card is drawn from a standard deck of cards. Let A be the event the card is a face card. Let B be the event the card is a black 5. Find and interpret P( A U B ).

35. Example 9 Solution: The sample space contains 52 equally likely outcomes. (this chart is on pg 639)

36. Example 9 Event A has 12 outcomes, one for each of the 3 face cards in each of the 4 suits. P( A ) = 12/52. Event B has 2 outcomes, because there are 2 black fives. P( B ) = 2/52. Events A and B are mutually exclusive because it is impossible for a 5 to be a face card. P( A U B ) = 12/52 + 2/52 = 14/52 = 7/26. This is the probability of drawing either a face card or a black 5.

37. Complement of an Event The set of outcomes in a sample space S , but not in an event E , is called the complement of the event E . The complement of E is written Ē . For example: The weather man says there is a 45% chance of rain. The complement of this event is the chance that it will not rain which is 55%.

38. Complement of an Event, cont’d The relationship between the probability of an event E and the probability of its complement Ē is given by:

39. Example 10 In a number matching game, First Carolan chooses a whole number from 1 to 4. Then Mary guesses a number from 1 to 4. What is the probability the numbers are equal? What is the probability the numbers are unequal?

40. Example 10 Solution: The sample space contains 16 outcomes: { (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4) }. Let E be the event the numbers are equal. P( E ) = ¼ Then Ē (complement of E) is the event the numbers are unequal. P( Ē ) = 1 – ¼ = ¾

41. Example 12 An experiment consists of spinning the spinner once and recording the number on which it lands.

42. Example 12 Define 4 events: A: an even number B: a number greater than 5 C: a number less than 3 D: a number other than 2 Find P( A ), P( B ), P( C ), and P( D ). Find and interpret P( A U B ) and P( A ∩ B ). Find and interpret P( B U C ) and P( B ∩ C ).

43. Example 12 Solution: The sample space has 8 equally likely outcomes: {1, 2, 3, 4, 5, 6, 7, 8}. Find P( A ),P( B ), P( C ), and P( D ). A = {2, 4, 6, 8}, so P( A ) = 4/8 = 1/2 B = {6, 7, 8}, so P( B ) = 3/8 C = {1, 2}, so P( C ) = 2/8 = 1/4 D = {1, 3, 4, 5, 6, 7, 8}, so P( D ) = 7/8.

44. Example12 Solution, cont’d: Find and interpret P( A U B ) and P( A ∩ B ). A and B are not mutually exclusive, so A ∩ B = {6, 8}, so P( A ∩ B ) = 2/8

45. Example 12 Solution, cont’d: Find and interpret P( B U C ) and P( B ∩ C ). B and C are mutually exclusive, so and

46. Section 10.2 Multistage Experiments Goals Study tree diagrams Study the fundamental counting principle Study probability tree diagrams Study the additive property Study the multiplicative property

47. 10.2 Initial Problem A friend who likes to gamble wagers that if you toss a coin repeatedly you will get 2 tails before you get 3 heads. Should you take the bet? The solution will be given at the end of the section.

48. Example 1 A tree diagram is drawn for the experiment of drawing 1 ball from a box containing 1 red ball, 1 white ball, and 1 blue ball.

49. Example 2 There are 12 possible outcomes. 12 is the product of 4 branches times 3 branches.

50. Fundamental Counting Principle The number of outcomes in an experiment can also be determined using the fundamental counting principle : If an event or action A can occur in r ways, and , for each of these r ways, an event or action B can occur in s ways, then the number of ways events or actions A and B can occur, in succession, is r times s . The principle can be extended to more than two events or actions.

51. Example 3 The options on a pizza are: Small, medium, or large White or wheat crust Sausage, pepperoni, bacon, onion, mushrooms How many different one-topping pizzas are possible?

52. Example 3 Solution: The first action is choosing 1 of 3 sizes. The second action is choosing 1 of 2 crusts. The third action is choosing 1 of 5 toppings. There are 3(2)(5) = 30 different one-topping pizzas possible.

53. Example 4 Find the probability of getting a sum of 11 when tossing a pair of fair dice.

54. Example 4 Solution: There are 36 equally likely outcomes in the sample space. 6 possible outcomes on the first roll 6 possible outcomes on the second roll 6(6) = 36 There are 2 ways of rolling a sum of 11: (5,6) and (6,5). The probability is 2/36 = 1/18.

55. Probability Tree Diagrams Tree diagrams can also be used to determine probabilities in multistage experiments. Tree diagrams that are labeled with the probabilities of events are called probability tree diagrams .

56. Probability Tree Diagrams

57. Example 6 Draw a probability tree diagram to represent the experiment of drawing one ball from a container holding 2 red balls and 3 white balls.

58. Example 6, cont’d Solution: The first tree has one branch for each ball. The second tree was simplified by combining branches.

59. Example 7 Draw a probability tree diagram to represent the experiment of spinning the spinner once. Find the probability of landing on white or on green.

60. Example 7 Solution: There are 4 outcomes in the sample space. They are not all equally likely. They are all mutually exclusive. (no overlapping) Use the central angles to find the probability of each outcome and draw the probability tree diagram.

61. Example 7 The probability of white or green is:

62. Example 8 A jar contains 3 marbles, 2 black and 1 red. A marble is draw and replaced, and then a second marble is drawn. What is the probability both marbles are black?

63. Example 8 Solution: Draw a probability tree diagram to represent the experiment.

64. Example 8 Assign a probability to the end of each secondary branch. (multiply fractions as you move down branches, add factions as you move across the end of branches.)

65. Example 8 Either tree diagram can be used to find that P( BB ) = 4/9.

66. Example 9 A jar contains 3 red balls and 2 green balls. First a coin is tossed. If the coin lands heads, a red ball is added to the jar. If the coin lands tails, a green ball is added to the jar. Second a ball is selected from the jar. What is the probability a red ball is chosen?

67. Example 9 Solution: A probability tree diagram is created.

68. Example 9 The probability of choosing a red ball is found by adding the probabilities at the end of the branches labeled R .

69. Example 10 A jar contains 3 marbles, 2 black and 1 red. A marble is drawn and not replaced before a second marble is drawn. What is the probability that both marbles were black?

70. Example 10 Solution: Create a probability tree diagram to represent the experiment.

71. Example 10 The probability of choosing 2 black marbles is:

72. Example 11 A jar contains 2 red gumballs and 2 green gumballs. An experiment consists of drawing gumballs one at a time from a jar until a red one is chosen. Find the probability of: A: only 1 draw is needed B: exactly 2 draws are needed C: exactly 3 draws are needed

73. Example 11 Solution: Create a probability tree diagram. The probabilities are:

74. 10.2 Initial Problem Solution A friend who likes to gamble wagers that if you toss a coin repeatedly you will get 2 tails before you get 3 heads. Should you take the bet? You can figure out what you should do by creating a probability tree diagram.

75. Initial Problem Solution, cont’d

76. Initial Problem Solution Add the probabilities of the outcomes that result in a win for you . You are much more likely to lose than to win, so you probably should not take the bet.

77. Section 10.3 Conditional Probability, Expected Value, and Odds Goals Study conditional probability Study independent events Study odds

78. 10.3 Initial Problem If you bet $100 on one number on the roulette wheel, what is your expected gain or loss? The solution will be given at the end of the section.

79. Conditional Probability In the experiment of tossing 3 fair coins suppose you know the first coin came up heads. The sample space is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}. Then the conditional sample space is {HHH, HHT, HTH, HTT} because the first coin was Head.

80. Conditional Probability The probability of A given B is called a conditional probability . The probability of A given B means the probability of event A occurring within the conditional sample space of event B . The probability of A given B is written P( A | B ),

81. Conditional Probability Suppose A and B are events in a sample space S and that the probability of B is not zero. The formula for conditional probability is

82. Example 1 One jar of marbles contains 2 white marbles and 1 black marble. Another jar of marbles contains 1 white marble and 2 black marbles.

83. Example 1 A coin is tossed. Heads: a marble is selected from the first jar. Tails: a marble is selected from the second jar. Find the probability that the coin landed heads up, given that a black marble was drawn.

84. Example 1 A coin is tossed. Heads: a marble is selected from the first jar. Tails: a marble is selected from the second jar. Find the probability that the coin landed heads up, given that a black marble was drawn. Solution: Create a probability tree diagram. The sample space is {HW, HB, TW, TB}. The probabilities are labeled on the diagram.

85. Example 1 Solution, cont’d:

86. Example 2 Suppose a test for a viral infection is not 100% accurate. Of the population, ¼ is infected and ¾ is not. Of those infected, 90% test positive. Of those not infected, 80% test negative. What is the probability the test is correct? Given that a person’s test is positive, what is the probability the person is infected?

87. Example 2 Solution: Create a probability tree diagram. Of the population, ¼ is infected and ¾ is not. Of those infected, 90% test positive. Of those not infected, 80% test negative.

88. Example 2 The probability of a correct test is the probability of a positive test for an infected person or a negative test for an uninfected person.

89. Example 2 Given that a person’s test is positive, what is the probability the person is infected? Find the conditional probability: P(positive) =

90. Example 2 Solution, cont’d: P(infected and positive) = P(infected|positive) =

91. Example 3 Results from an inspection of a candy company’s 2 production lines are shown in the table. If a customer find a sub-standard piece of candy, what is the probability it came from the Bay City factory?

92. Example 3 Solution: Use the numbers in the table to solve: P(Bay City and sub-standard) =

93. Example 3 Solution, cont’d: P(sub-standard) =

94. Example 3 Solution, cont’d: P(Bay City|sub-standard) =

95. Independent Events Two events are called independent if one event does not influence the other. When 2 events are independent, their probabilities follow the rule given below:

96. Example 5 A student’s name is chosen at random from the college enrollment list and the student is interviewed. Let A be the event the student regularly eats breakfast. Let B be the event the student has a 10:00 AM class. Explain in words what is meant by:

97. Example 5, Solution: : the probability the student regularly eats breakfast and has a 10:00 AM class. : the probability the student regularly eats breakfast given that the student has a 10:00 AM class : the probability the student does not regularly eat breakfast.

98. Expected Value The average numerical outcome for many repetitions of an experiment is called the expected value . If E = 0, the game is said to be fair .

99. Example 6 An experiment consists of rolling a fair die and noting the number on top of the die. Compute the expected value of one roll of the die.

100. Example 6 Solution: The calculations are shown below: Multiply 1(1/6) and 2(1/6) and 3(1/6) etc.

101. Example 7 How much should an insurance company charge as its average premium in order to break even?

102. Example 7 Solution: The calculations are shown above. The product row is the product of the 2 numbers directly above each value in the product row. The average premium should cost $760.

103. Odds The odds in favor of an event compare the number of favorable outcomes to the number of unfavorable outcomes. If the odds in favor are a : b , then

104. Odds The odds against an event compare the number of unfavorable outcomes to the number of favorable outcomes. The odds in favor of an event E are The odds against an event E are

105. Example 8 Suppose a card is randomly drawn from a standard deck. What are the odds in favor of drawing a face card?

106. Example 8 Solution: There are 12 face cards in the deck, and there are 40 other cards. The odds in favor are 12:40, which simplifies to 3:10.

107. Example 9 Find P( E ) given the following odds: The odds in favor of E are 3:7. The odds against E are 5:13.

108. Example 9, cont’d Solution: The odds in favor of E are 3:7. The odds against E are 5:13.

109. Example 10 Find the odds in favor of event E , given the following probabilities: P( E ) = 1/4. P( E ) = 3/5.

110. Example 10, cont’d Solution: The odds in favor of E are 1:3. The odds in favor of E are 3:2.

111. 10.3 Initial Problem Solution A roulette wheel has 38 slots numbered 00, 0 and 1 through 36. You place a bet on a number or combination of numbers. If you bet on the winning number, you win your bet plus 35 times your bet. If you lose, you lose the money you bet. If you bet $100 on one number, what is your expected gain or loss?

112. Initial Problem Solution Solution: The probability of winning is 1/38 and the amount won would be $3500. The probability of losing is 37/38 and the amount lost would be $100.

113. Initial Problem Solution The expected value is approximately -$5.26. You should expect to lose this amount, on average, for every $100 you bet.

114. Chapter 10 Assignments Section 10.1 pg 645 (7,11,15,17,31,35) Section 10.2 pg 664 (1,3,11,13,19,25) Section 10.3 pg 686 (5a,6a,15,27,3339,41)