Chapter 3. Probability

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1 Chapter 3 Probability

2 Every Day, each us makes decisions based on uncertainty. Should you buy an extended warranty for your new DVD player? It depends on the likelihood that it will fail during the warranty. Should you allow 45 min to get to 1 st period or is 35 min enough?

3 Lesson 3-1/3-2 Fundamentals

4 Rare Event Rule If, under a given assumption (such as a lottery being fair), the probability of a particular observed event (such as five consecutive lottery wins) is extremely small, we conclude that the assumption is probably not correct

5 Probability The probability of an event is the proportion of times the event is expected to occur in repeated experiments.

6 Sample Space (S) The set of all possible outcomes of an event is the sample space S of the event. Example: For the event roll a die and observe what number it lands on the sample space contains all possible numbers the die could land on. S 1,2,3,4,5,6

7 Event An event is an outcome (or a set of outcomes) from a sample space Example 1: When flipping three coins, an event may be getting the result HTH. In this case, the event is one outcome from the sample space. Example 2: When flipping three coins, an event may be getting two tails. In this case, the event is a set of outcomes (HTT, TTH, THT, ) from the sample space An event is usually denoted by a capital letter. Example: Call getting two tails event A. The probability of event A is denoted P(A).

8 Probability Properties P denotes probability The probability of an event, say event A, is denoted P(A). All probabilities are between 0 and 1. 0 PA ( ) 1 (i.e. ) The sum of the probabilities of all possible outcomes must be 1. The probability of an impossible event is 0. The probability of an event that is certain to occur is 1.

9 Assigning Probabilities to Events Rule 1: Relative frequency approach Rule 2: Classical approach Rule 3: Personal opinion approach (subjective)

10 Rule #1 Relative Frequency If an experiment is repeated n times under essentially identical conditions, and if the event A occurs m times, then as n grows large the ratio of m/n approaches a fixed limit, namely, the probability of A. PA ( ) number of times A occurred number of times the trial was repeated

11 Example Relative Frequency P(tack lands pointed down) we must repeat the procedure of tossing the tack many times and then find the ratio of the number of times the tack lands with the point down to the number of tosses.

12 Rule #2 Classical Approach Assume that a given procedure has n different simple events and that each those simple events has a equal chance of occurring. If event A can occur in s of these n ways then: PA ( ) number of ways A can occur number of different simple events s n

13 Example Classical When trying to determine P(2) with a balanced and fair dice, each of the six faces has an equal chance of occurring P(2) 1 6

14 Rule #3 Subjective P(A), the probability of event A, is found by simply guessing or estimating its value based on knowledge of the relevant circumstances.

15 Example Page 128, #6 In a study of 420,000 cell phones users in Denmark, it was found that 135 developed cancer of the brain or nervous system. Estimate the probability that a randomly selected cell phone user will develop such a cancer. Is the result very different from the probability that was found for the general population? What does this suggest about cell phones as a cause of such cancers, as has been claimed? 135 P( cancer ) ,000 No, this is not very different from the general population value of This suggests that cell phones are not the cause of cancer.

16 Law of Large Numbers As a procedure is repeated again and again, the relative frequency (from Rule 1) of an event tends to approach the actual probability.

17 Example Page 128, #4 Identifying Probability Values A. What is the probability of event that is certain to occur? P(A) = 1 B. What is the probability of an impossible event? P(A) = 0

18 Example Page 128, #4 C. A sample space consists of 10 separate events that are equally likely. What is the probability of each? PA ( ) 1 10 D. On a true/false test, what is the probability of answering the question correctly if you make a random guess? PA ( ) 1 2

19 Example Page 128, #4 E. On a multiple-choice test with five possible answers for each questions, what is the probability of answering the questions correctly, if you make a random guess? PA ( ) 1 5

20 Example Page 129, #10 Probability of a Wrong Result. Table 3-1 shows that among 14 women who were not pregnant, the test for pregnancy yielded the wrong conclusions 3 times. Table 3-1 Pregnancy Test Results Subject is Pregnant Subject Is not Pregnant Subject Test Result Pregnancy is indicated Subject Test Result Pregnancy not indicated

21 Example Page 129, #10 Probability of a Wrong Result. Table 3-1 shows that among 14 women who were not pregnant, the test for pregnancy yielded the wrong conclusions 3 times. A. Based on the available results, find the probability of wrong test conclusions for a women who is not pregnant. Let W = The test wrongly concludes a woman is not pregnant 3 PW ( )

22 Example Page 129, #10 Probability of a Wrong Result. Table 3-1 shows that among 14 women who were not pregnant, the test for pregnancy yielded the wrong conclusions 3 times. B. Is it unusual for the test conclusion to be wrong for women who are not pregnant? 3 PW ( ) No, since > 0.05, it is not unusual for the test to be wrong for women who are not pregnant.

23 Complement The complement of an event A, denoted by A c or Ā, is the set of outcomes that are not in A. A c means A does not occur P(A c ) = 1 P(A) Example: When flipping two coins, the probability of getting two heads is The probability of not getting two heads is P(A c ) = 1 P(A) = = 0.75

24 Odds Actual odds against event A occurring Is the ratio P(A c )/P(A), usually expressed in the form of a:b (or a to b ), where a and b are integers having no common factors. Actual odds in favor of a event A Is the reciprocal of the actual odds against that event P(A)/P(A c ), then odds in favor of A are b:a. Payoff odds against event A Represents the ratio of net profit (if you win) to the amount bet. Payoff odds against event A = (net profit) : (amount bet)

25 Example Page 131, #26 A roulette wheel has 38 slots. One slot is 0, another is 00, and the others are numbered 1 through 36 respectively. You are placing a bet that the outcome is an odd number. A. What is the probability of winning? 18 P( odd)

26 Example Page 131, #26 A roulette wheel has 38 slots. One slot is 0, another is 00, and the others are numbered 1 through 36 respectively. You are placing a bet that the outcome is an odd number. B. What are the actual odds against winning? actual odds against odd = : P(not odd) P(odd) =

27 Example Page 131, #26 C.When you bet the outcome is an odd number, the payoff odds are 1:1. How much profit do you make if you bet $18 and win? 1:1 means a profit of $1 for every $1 bet. A winning bet of $18 means a profit of $18.

28 Example Page 131, #26 D. How much profit would you make if on the $18 bet if you could somehow convince the casino to change its payoff odd so they are the same as the actual odds against winning? Payoff odds of 10:9 mean a profit of $10 for every $9 bet. A winning bet of $18 means a profit of $20.

29 Lesson 3-3 Addition Rule

30 Compound Event A compound event is any event combining two or more simple events. Compound probability are used to answer the following question: In rolling a single dice what is the probability of rolling even number or rolling a 1 or 2 When finding the probability that event A occurs or event B occurs, find the total number of ways A can occur and the number of ways B can occur, but find the total in such a way that no outcome is counted more than once.

31 Addition Rule The addition rule is used to find probabilities involving the word or. Rule For any two events A and B P(A 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 as an outcome in a trial or procedure.

32 Addition Rule Venn Diagram

33 Example Page 138, #6 Refer to figure 3-3. Find the probability of randomly selecting one of the peas and getting one with a yellow pod or a purple flower. P(Y or P) = P(Y) +P(P) P(Y and P)

34 Example Page 138, #10 If one of the Titanic passengers is randomly selected, find the probability of getting a man or someone who survived the sinking. Men Women Boys Girls Total Survived Died Total P(M or S) = P(M) + P(S) P(M and S)

35 Mutually Exclusive If events A and B have no simple events in common or cannot occur simultaneously, they are said to be disjoint or mutually exclusive. For two disjoint events A and B, the probability that one or the other occurs is the sum of the probabilities of the two events. P(A or B) = P ( A B ) P ( A) P ( B )

36 Example Page 137, #2 Determine whether events are disjoint. Are the two events disjoint for a single trial? a) Randomly selecting a head of household watching NBC on television at 8:15 tonight. Randomly selecting a head of household watching CBS on television at 8:15 tonight. Yes

37 Example Page 137, #2 Determine whether events are disjoint. Are the two events disjoint for a single trial? b) Receiving a phone call from a volunteer survey subject who opposes all government taxation. Receiving a phone call from a volunteer survey subject who approves all government taxation. Yes

38 Example Page 137, #2 Determine whether events are disjoint. Are the two events disjoint for a single trial? c) Randomly selecting a United States Senator currently holding office Randomly selecting a female elected official No

39 Mutually Exclusive Venn Diagram

40 Applying the Addition Rule

41 Rule of Complementary Events Complement Rule is used when you know the probability that some event will occur and you want to know the opposite: the chance it will not occur. Rules P(A) + P(A C ) = 1 P(A C ) = 1 P(A) P(A) = 1 P(A C ) P(A) and P(A C ) are mutually exclusive All simple events are either in A or A C.

42 Complement Venn Diagram

43 Example Page 138, #4 Finding the complements C A. Find PA ( ) given that PA ( ) C PA ( )

44 Example Page 138, #4 B. A Reuters/Zogby poll showed that 61% of Americans say they believe that life exists somewhere in the galaxy. What is the probability of randomly selecting someone not having that belief? Let B = selecting an American who believes that life exists elsewhere in the galaxy C PB ( )

45 Example Page 138, #8 If someone is randomly selected, find the probability that his or her birthday is not October. Ignore leap years. Let O = a person s birthday falls in October PO ( ) C P( O ) 1 P( O)

46 Example Page 138, #20 Refer to the accompanying figure, which describes the blood groups and Rh types of 100 people. In each case assume that 1 of the 100 is randomly selected and find the indicated probability. Group AB P(group A or O or type Rh + ) Group B 8 Rh + 2 Rh - 4 Rh + 1 Rh - Group A 35 Rh + 5 Rh - Group O 39 Rh + 6 Rh -

47 Group Example Page 138, #20 P(group A or O or type Rh + ) Group B 8 Rh + 2 Rh - Group AB 4 Rh + 1 Rh - Rh Factor + Total A B AB O Total Group A 35 Rh + 5 Rh - Group O 39 Rh + 6 Rh -

48 Group Example Page 138, #20 Rh Factor + Total A B AB O Total P(group A or O or type Rh + ) = P(A) + P(O) + P(Rh + ) P(A and Rh + ) P(O and Rh + ) =

49 Lesson 3-4 Multiplication Rule: Basics

50 Independent Events A and B are independent because the probability A does not change the probability of B. Example: Roll a yellow die and a red die. Event A is the yellow die landing on an even number, and event B is the red die landing on an odd number.

51 Multiplication Rule The multiplication rule is used to find probabilities involving the word and. Rule For any two events A and B P(A B) = P(A and B) = P(A occurs in a 1 st trial and event B occurs in a 2 nd trial For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events. P(A B) = P(A and B) = P(A) P(B)

52 Dependent If event A and event B are not independent, they are said to be dependent. The probability for the second event B should take into account the fact that the first event A has already occurred. The probability that event B occurs if we know for certain that event A will occur is called conditional probability. The conditional probability of B given A is: P(B A) which reads the probability of event B given event A has occurred. Rule For any two events A and B P( A B) = P(A and B) = P(A) P(B A)

53 Applying the Multiplication Rule

54 Example Page 146, #2 Identify events as independent or dependent. For each pair of events, classify the two events as independent or dependent. a) Finding that you calculator is not working. Finding that your refrigerator is not working. Independent

55 Example Page 146, #2 Identify events as independent or dependent. For each pair of events, classify the two events as independent or dependent. b) Finding that your kitchen light is not working. Finding that your refrigerator is not working. Dependent

56 Example Page 146, #2 Identify events as independent or dependent. For each pair of events, classify the two events as independent or dependent. c) Drinking until your driving ability is impaired. Being involved in a car crash. Dependent

57 Example Page 146, #4 A new computer owner creates a password consisting of two characters. She randomly selects a letter of the alphabet for the first character and a digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) for the second character. What is the probability that her password is K9? Would this password be effective as a deterrent against someone trying to gain access to her computer? Let A = Alphabet and N = Number P(A and N) P( A) P( N) P(K and 9)

58 Example - Page 147, #8 A study of hunting injuries and the wearing of hunter orange clothing showed that among 123 hunters injured when mistaken for game, 6 were wearing orange. If a follow-up study begins with the random selection of hunters from this sample of 123, find the probability that the first two selected hunters were both wearing orange. A. Assume that the first hunter is replaced before the next one is selected. P( O and O ) 1 2 P( O ) P( O O )

59 Example - Page 147, #8 B. Assume that the first hunter is not replaced before the next one is selected. P( O and O ) 1 2 P( O ) P( O O ) C. Given a choice between selecting with replacement and selecting without replacement, which choice makes more sense in this situation? Selecting with replacement could lead to re-interviewing

60 Example Page 148, #14 It is common for the public opinion polls to have a confidence level of 95%, meaning that there is a 0.95 probability that the poll results are accurate within the claimed margins of error. If five different organizations conduct independent polls, what is the probability that all five of them are accurate within the claimed margin of error? Does the result suggest that with a confidence level of 95%, we can expect that almost all polls will be within the claimed margin of error? A = a poll is accurate as claimed PA ( ) 0.95 for each poll 5 P( A, A... A ) (0.95) No, the more polls one considers, the less likely that they will be within the claimed margin of error.

61 Example Page 148, #16 Remote sensors are used to control each of two separate and independent values, denoted by p and q, that open to provide water for emergency cooling of a nuclear reactor. Each value has a probability of opening when triggered. For the given configuration, find the probability that when both sensors are triggered, water will get through the system so that cooling can occur. Is this result high enough to be considered safe? Let O = the value opens properly P(O) = , for each value

62 Example Page 148, #16 Let O = the value opens properly P(O) = , for each value P(water gets through) = P(O 1 and O 2 ) = P(O 1 ) P(O 2 ) = (0.9968) (0.9968) =

63 Example Page 148, #16 P(water gets through) = Is this result high enough to considered safe? Yes; since the systems is used in emergencies, and not on a routine basis but because the probability of failure is ( ), the system can be expected to fail about 64 times in every 10,000 uses. That means if it is used 28 times a day [ = 10,220 times a year], we expect about 64 failures annually and after one such failure in a nuclear reactor, there might not be a next time for anything.

64 Example Page 149, #22 Table 3-1 Pregnancy Test Results Positive Negative Yes 80 5 No If one of the subjects is randomly selected, find the probability of getting someone who test negative or someone who is not pregnant. P(Neg or No) P( Neg) P( No) P(Neg and No)

65 Example Page 149, #24 Table 3-1 Pregnancy Test Results Positive Negative Yes 80 5 NO If three people are randomly selected, find the probability that they all test negative P( N, N, N ) P( N ) P( N N ) P( N N, N )

66 Lesson 3-5 Multiplication Rule Complements and Conditional Probability

67 Complements: The 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 item of that type. To find the probability of at least one of something, calculate the probability of none, then subtract that result from 1. That is, P(at least one) = 1 P(none)

68 Example Page 153, #2 Provide a written description of the complement of the given event. Quality Control: When 50 HDTV units are shipped, all of them are free of defects. At least one unit is defective.

69 Example Page 153, #4 Provide a written description of the complement of the given event. A Hit with the Misses: When Mike ask five different women for a data, at least one of them accepts. None of the women accepts.

70 Example Page 154, #8 If a couple plans to have 12 children, what is the probability that there will be at least one girl? If the couple eventually has 12 children and they are all boys, what can the couple conclude? Let B = a child is a boy P(B) = 0.5, for each birth P(at least one girl) = 1 P(all boys) 1 P( B ) P( B )... P( B ) (.5) Either a very rare event has occurred or some genetic factor is causing the likelihood of boy to be much greater than 0.5

71 Example Page 154, #10 If you make random guesses for four multiple-choice test questions (each with five possible answers), what is the probability of getting at least one correct? If a very nondemanding instructor says that passing the test occurs if there is a least one correct answer, can you reasonably expect to pass by guessing? Let W = guessing a wrong answer PW ( ) P(at least one correct) = 1 P(all wrong) P( W ) P( W )... P( W )

72 Example Page 154, #10 Since > 0.50, you are more likely to pass than fail whether or not you can reasonably expect to pass depends on your perception of what is reasonable.

73 Conditional Probability A conditional probability of event occurs when the probability is affected by the knowledge of other circumstances. P(A and B) P(A) P( A) P( B A) P(A) P(B A)= P(A and B) P(A)

74 Example Page 154, #12 Refer to figure 3-3 in section 3-3 for the peas used in genetics experiment. If one of the peas is randomly selected and is found to have a green pod, what is the probability that is has a purple flower. P(Purple Green) = P(G and P) P(G)

75 Example Page 155, #20 If we randomly select some who died, what is the is the probability of getting a man? P(M D) Men Women Boys Girls Total Survived Died Total P( M and D) PD ( )

76 Example Page 155, #22 What is the probability of getting a man or women, given that the randomly selected person is someone who died.? Men Women Boys Girls Total Survived Died Total P([M or W] D) = P(D and [M or W]) P(D)

77 Example Page 155, #22 P([M or W] D) = Men Women Boys Girls Total Survived Died Total P(D and [M or W]) P(D) [ ] 2223 [ ]

78 Testing for Independence In Section 3-4 we stated that events A and B are independent if the occurrence of one does not affect the probability of occurrence of the other. This suggests the following test for independence: Two events A and B are Independent if: P(B A) = P(B) or P(A and B) = P(A) P(B) Two events A and B are dependent if: P(B A) P(B) or P(A and B) P(A) P(B)

79 Lesson 3-6 Probabilities Through Simulations

80 Simulation A simulation of a procedure is a process that behaves the same way as the procedure, so that similar results are produced. The numbers used in a simulation must be generated in such a way that they are equally likely

81 Obtaining Randomly Generated Numbers A table of random digits TI-83 graphing calculator

82 Example Page 160, #2 Assume that you want to use the digits in the accompanying list to simulate the rolling of singe die. If the digits 1, 2, 3, 4, 5, 6 are used while all other digits are ignored, list the outcomes from the first two rows , 6, 1, 6, 4, 3

83 Example Page 160, #8 When Mendel conducted his famous hybridization experiments, he used peas with green pods, and yellow pods. One experiment involved crossing peas in such a way that 25% of the offspring peas were expected to have yellow pods. Use the random digits in the margin to develop a simulation for finding the probability that when two offspring peas are produced, at least one of them has a yellow pods. How does the result compare to the correct probability of 7/16. (Hint: Because 25% of the offspring are expected to have yellow pods and the other 75% are expected to have green pods, let digit 1 represent yellow pods and let digits 2, 3, 4 represent green pods and ignore any other digits)

84 Example Page 160, # yellow pods 2, 3, 4 green pods 41, 43, 21, 13, 44, 44, 31, 14, 31, 14, 42, 12, 22, 14, 24, 42, 22, 32, 33, 34

85 Example Page 160, #8 41, 43, 21, 13, 44, 44, 31, 14, 31, 14, 42, 12, 22, 14, 24, 42, 32, 33, 34 1 yellow pods 2, 3, 4 green pods Find the number of yellow pods 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0 1, 0, 1, 0, 0, 0, 0, 0, 0 0 no yellow 1 yellow 2 yellow

86 Example Page 160, #8 Find the number of yellow pods 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0 1, 0, 1, 0, 0, 0, 0, 0, 0 0 no yellow 1 yellow 2 yellow x Frequency Relative Frequency /20 = /20 = /20 = 0.00 Total From the simulation we estimate P(x 1) 9/20 = 0.45 This compares very favorable with 7/16 =

87 Example Page 161, #10 Simulating Three Dice: In exercise 6 we used the digits in the margin to simulate the rolling of dice. Simulate the rolling of three dice 100 times. Describe the simulation, then use it to estimate the probability of getting a total of 10 when three dice are rolled Math PRB 5:randInt( STAT EDIT

88 Example Page 161, #10

89 Example Page 161, #10 P(Sum 10) = 11/100 = 0.11 This compares very favorable to the true value of 0.125, and in this instance the simulation produced a good estimate.

90 Example Page 161, #12 In Exercise 8 we used the digits in the margin as a basis for simulating the hybridization of peas. Again assume 25% of offspring peas are expected to have yellow pods, but develop you own simulation generating 100 pairs of offspring. Based on your results, estimate the probability of getting at least one pea with yellow pods when two offspring are obtain. 1 yellow pod 2, 3, 4 green pod

91 Example Page 161, #12 Math PRB 5:randInt( STAT EDIT P(at Least 1 yellow pod) = Compares favorable with the true probability of

92 Lesson 3-7 Counting

93 Fundamental Counting Rule For a sequence of two events in which the first event can occur m ways and the second event can occur n ways, the events together can occur a total of m n ways.

94 Example Counting the number of possible meals The fixed priced lunch at Sarasota High School provides the following chooses: Appetizer: soup or salad Entrée: baked chicken, pizza, sandwich, or hotdog Dessert: ice cream or cookie different meals can be ordered.

95 Example Airport Codes The International Airline Transportation Association uses 3 letter codes to assign airports. How many different airport codes ,576

96 Example Arranging Groups Arrange the sequence of 6 groups

97 Factorial Rule A collection of n different items can be arranged in order n! different ways. This factorial rule reflects the fact that the first item may be selected in n different ways, the second item may be selected in n 1 ways, and so on. Factorial symbol! Denotes the product of decreasing positive whole numbers. Example: 4! = = 24 By special definition: 0! = 1

98 Example Factorial 6! Math PRB 4:!

99 Number of Permutations of n Distinct Objects Taken r at a Time The number of arrangements of r objects chosen from n objects in which: The n objects are distinct once an object is used it cannot be repeated order is important Notation n P r = the number of permutations of r objects selected from n objects. Formula n P r n! n r!

100 Example - Permutations Evaluate the following 7! 7! P , ! 2! Math PRB 2:nPr

101 Example Page 170, #18 Singing legend Frank Sinatra recorded 381 songs. From a list of his top-10 songs, you must select three that will be sung in a medley as a tribute at the next MTV Music Awards ceremony. The order of the songs is important so that they fit together well. If you select three of Sinatra s top-10 songs, how many different sequences are possible. n r P3 720

102 Number of Combinations of n Distinct Objects Taken r at a Time The number of arrangements of n objects using r n of them in which: The n objects are distinct once an object is used it cannot be repeated order is not important Notation n C r = Represents the number of combinations n distinct objects taken r at a time Formula n C r n! r! n r!

103 Example - Combinations Evaluate the following C 6 4 6! 6! 6 5 4! ! 6 4! 4! 2! 4! Math PRB 2:nCr

104 Example Page 169, #12 Find the probability of winning the New York Take Five: the winning five numbers from 1, 2,..,39 n r C5 575,757 Since only one combination wins PW ( ) 1 575,757

105 Example Page 171, #26 Five Card Flush: A standard deck of cards contains 13 clubs, 13 diamonds, 13 hearts, and 13 spades. If five cards are randomly selected, find the probability of getting a flush. (A flush is obtained when all five cards are of the same suit. That is, they are all clubs, or all diamonds, or all hearts or all spades. Method #1 Counting Technique The total number of possible 5 card selections is 52! C 2,598,960 47!5! 52 5

106 Example Page 171, #26 The total number of possible 5 card selections from one particular suit is C ! 8!5! 1287 The total number of possible 5 card selections from any of the 4 suits is P(getting a flush)

107 Example Page 171, #26 Method #2 Probability Formulas Let F = selecting a card favorable for getting a flush. 52 PF ( ) 1 since the 1 52 st card could be anything PF ( ) since the 2 nd card must be from the same suit as the first P(getting a flush) = P(F 1 and F 2 and F 3 and F 4 and F 5 ) = P(F 1 ) P(F 2 ) P(F 3 ) P(F 4 ) P(F 5 )

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