Chapter 6 Probability The Study of Randomness
Lesson 6-1 The Idea of Probability
Randomness Behavior is random if, while individual outcomes are uncertain, for a large number of repetitions, outcomes are regularly distributed. Example: If I roll a die once, I can t predict with certainty what number it will land on, but if I roll sixty times, I can expect it to land on 1 ten times, 2 ten times, 3 ten times, etc.
Probability Models The probability of an outcome is the proportion of times the outcome would occur for a large number of repetitions. Example: The probability of a die landing on 4 is the proportion of times a die lands on 4 for a large number of repetitions.
Chance Behavior is unpredictable in the short run but has a regular and predictable pattern in the long run. Tossing a Coin
Example Page 335, #6.10 A recent opinion poll showed that about 73% of married women agree that their husbands do at least their fair share of household chores. Suppose that this is exactly true. Choosing a married woman at random that has probability 0.73 of getting one who agrees that her husband does his share. A). Simulate drawing 20 women, then 80 women, then 320. What proportion agree in each case? We expect (but because of chance variation we can t be sure) that proportion will be closer to 0.73 in longer runs of trials.
Example Page 335, #6.10 A). Use Randint and let 0 Agree and 1 disagree. 12 women disagreed and 8 women agreed 8 20 0.40
Example Page 335, #6.10 A). Use Randint and let 0 Agree and 1 disagree. Sample 80 Sample 320 36 women disagreed and 44 women agreed 44 0.55 80 167 women disagreed and 153 women agreed 153 320 0.49
Example Page 335, #6.10 B). Simulate drawing 20 women 10 times and record the percents in each trial who agreed. Then simulate drawing 320 women 10 times and again record the 10 percents. Which set of 10 results is less variable? We expect the results of 320 trials to be more predictable (less variable) than the results of 20 trials. That is long-run regularity showing itself.
Example Page 335, #6.10 20 Women 320 Women Trail 1 0.50.51 Trial 2.70.54 Trial 3.54.55 Trial 4.55.53 Trial 5.45.54 Trial 6.55.52 Trial 7.45.48 Trial 8.50.50 Trail 9.35.46 Trial 10.65.45
Example Page 334, #6.4 Probability is a measure of how likely an event is to occur. Match one of the probabilities that follow with each statement about an event. (The probability is usually a much more exact measure of likelihood than is the verbal statement.)
Example Page 334, #6.4 0, 0.01, 0.03, 0.6, 0.99, 1 0 1 0.01 0.6 A). This event is impossible. It can never occur. B). This event is certain. It will occur on every trial of the random phenomenon. C). This event is very unlikely, but it will occur once in a while in a long sequence of trials. D). This event will occur more often than not.
Lesson 6-2, Part 1 Probability Models Addition/Complements
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
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).
Identifying Outcomes Tree Diagrams Multiplication Rule One task a number of ways and a second task b number of ways, then both tasks can be done a x b number of ways.
Tree Diagram Toss a coin and rolling a dice Possible outcomes 2 6 12
With or Without Replacement Assume you are drawing index cards with the digits 1 10. You are picking 3 cards. With Replacement 10 10 10 1000 Without Replacement different ways to arrange a 3-digit number 10 9 8 720
Example Page 341, #6.14 For each of the following, use a tree diagram or the multiplication principal to determine the number of outcomes in the sample space. Then write the sample space using set notation. A). Toss 2 coins. 2 2 4 S HH, HT, TT, TH
Example Page 341, #6.14 B). Toss 3 coins. 2 2 2 8 S HHH, HHT, HTT, TTH, HTH, TTT, THT, THH
Example Page 341, #6.14 C). Toss 4 coins. 2 2 2 2 16 S HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THTH, TTHH, THHT, HTTT, THTT, TTHT, TTTH, TTTT
Example Page 342, #6.18 Suppose you select a card from a standard deck of 52 playing cards. In how many ways can the selected card be: A). A red card? 26 B). A heart? 13 C). A queen and a heart? 1 D). A queen or a heart? 16 E). A queen that is not a heart? 3
Probability Rules The probability of any event is between 0 and 1. A probability of 0 indicates the event will never occur. A probability of 1 indicates the event will always occur 0 P(A) 1 If S is the sample space, then P(S) = 1. The sum of the probabilities of all possible outcomes must be 1
Example Page 348, #6.22 The New York Times (August 21, 1989) reported a poll that interviewed a random sample of 1025 women. The married women in the sample were asked whether their husbands did their fair share of household chores. Here are the results: Outcome Does more than his fair share 0.12 Does his fair share 0.61 Does less than his fair share? Probability These proportions are probabilities for the random phenomenon of choosing a married women at random and asking her opinion.
Example Page 348, #6.22 Outcome Does more than his fair share 0.12 Does his fair share 0.61 Does less than his fair share? Probability A). What must be the probability that women chosen says that her husband does less than his fair share? Why? All three probabilities must add up to 1 1 (0.12 0.61) 0.27
Example Page 348, #6.22 Outcome Does more than his fair share 0.12 Does his fair share 0.61 Does less than his fair share? Probability B). The event I think my husband does at least his fair share contains the first two outcomes. What is the probability. (0.12 0.61) 0.73
Complement The complement of an event A, denoted by A c, is the set of outcomes that are not in A. P(A c ) = 1 P(A) Example: When flipping two coins, the probability of getting two heads is 0.25. The probability of not getting two heads is 1 0.25 = 0.75
Complement S A C A The sample space S P( S) 1 The set A and its complement P ( A) 1 P ( A C )
Addition Rule If events A and B are disjoint if they have no outcomes in common. Events like this, that can t occur together, are called 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 )
Addition Rule Example: Let event A be rolling a die and landing on an even number, and event B be rolling a die and landing on an odd number. The outcomes for A are {2, 4, 6} and the outcomes for B are {1, 3, 5}. These events are disjoint because they have no outcomes in common. The probability of A or B (landing on either even or an odd number) P(A or B) = P(A) + P(B)
Disjoint S B Two disjoint sets, A and B We write P(A or B) as P(A B). The symbol means union, representing the outcomes in event A or event B (or both). A
Lesson 6-2, Part 2 Probability Models Independence
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.
Multiplication Rule 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 and B) = P ( A B ) P ( A) P ( B )
Multiplication Rule Example: The probability that a yellow die lands on an even number and the red die lands on an odd number is: P(A and B) = P(A) P(B) = ½ x ½ = ¼ If events A and B are independent, then their complements, A c and B c are also independent, and A c independent of B
Intersection B A and B A Two sets A and B are not disjoint. We write P(A and B) as P(A B). The symbol means intersection, representing the outcomes that are in both event A and event B.
Example Page 349, #6.26 Example 6.10 (page 345) states that the first digits of numbers in legitimate records often follow a distribution know as Benford s Law. Here is the distribution: 1 st Digit 1 2 3 4 5 6 7 8 9 Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046 P(A) = P(first digit is 1) = 0.301 P(B) = P(first digit is 6 or greater) = 0.222 P(C) = P(first digit is odd) = 0.609 P(D) = P(first digit is less than 4) = 0.602
Example Page 349, #6.26 1 st Digit 1 2 3 4 5 6 7 8 9 Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046 P(A) = P(first digit is 1) = 0.301 P(B) = P(first digit is 6 or greater) = 0.222 P(C) = P(first digit is odd) = 0.609 P(D) = P(first digit is less than 4) = 0.602 A). P(D) = P(1, 2, or 3) = 0.301 + 0.176 + 0.125 = 0.602 B). P ( B D) P ( B ) P ( D) 0.222 0.602 0.824 C C). P( D ) 1 PD ( ) 1 0.602 0.398
Example Page 349, #6.26 1 st Digit 1 2 3 4 5 6 7 8 9 Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046 P(A) = P(first digit is 1) = 0.301 P(B) = P(first digit is 6 or greater) = 0.222 P(C) = P(first digit is odd) = 0.609 P(D) = P(first digit is less than 4) = 0.602 D). E). P ( C D) P(1 and 3) P(1) P(3) 0.301 0.125 0.426 P ( B C ) P(7 or 9) P(7) P(9) 0.058 0.046 0.104
Example Page 354, #28 An automobile manufacturer buys computer chips from a supplier. The supplier sends a shipment containing 5% defective chips. Each chip chosen from this shipment has probability 0.05 of being defective, and each automobile uses 12 chips selected independently. What is the probability that all 12 chips in a car will work properly? P(all chips work properly) = 1 P(defective chips) 12 12 1.05 0.95 0.5404
Example Page 354, #6.30 Choose at random a U.S. resident at least 25 years of age. We are interested in the events A = {The person chosen completed 4 years of college} B = {The person chosen is 55 years old or older}
Example Page 354, #6.30 Government data recorded in table 4.6 on page 241 allows us to assign probabilities to these events.
Example Page 354, #6.30 A). Explain why P(A) = 0.256 44845 P( A) 0.256 175230
Example Page 354, #6.30 B). Find P(B). 56008 P( B) 0.320 175230
Example Page 354, #6.30 C). Find the probability that the person is at least 55 years old and had 4 years of college education, P(A and B). Are events A and B independent?
Example Page 354, #6.30 P(A and B) 10596 175230 0.0604 A and B are not independent because P(A and B) P ( A) P ( B ) 0.0604 0.256(0.320) 0.0604 0.08192
Example Page 355, #6.32 An athlete suspected of having used steroids is given two tests that operate independently of each other. Test A has a probability 0.9 of being positive if steroids have been used. Test B has probability 0.8 of being positive if steroids have been used. What is the probability that neither test is positive if steroids have used? P(neither test is positive) P A C C P B (1 0.9)(1 0.8) 0.02
Lesson 6-3, Part 1 General Probability Rules
Basic Properties of Probability 0 P(A) 1 for any event A P(S) = 1 Complement Rule For any event A, P(A c ) = 1 P(A) Addition Rule Events A and B are disjoint P(A or B) = P(A B) = P(A) + P(B) Multiplication Rule Events A and B are independent P(A and B) = P(A B) = P(A) P(B)
Addition Rule Not Disjoint The union of two or more events that at least one of those occurs. The addition Rule for the Union to Two Events: P(A or B) = P(A) + P(B) P(A and B) P(A B) = P(A) +P(B) P(A B) Mutually Exclusive Two events are mutually exclusive: If the P(A and B) =0
Addition Rule Not Disjoint P(B) P(A or B) = P(A) +P(B) P(A and B) P P(A B) = P(A) +P(B) P(A B) Two events are mutually exclusive if the P(A and B) = 0 P(A)
Example 1 Addition Rule Not Disjoint The accompanying table represents the blood groups and RH types of 100 people. RH Types of 100 People RH Group A B AB O Total Positive 35 8 4 39 86 Negative 5 2 1 6 14 Total 40 10 5 45 100 Based on the table what is the probability a person will have group B blood or be RH Positive?
Example 1 Addition Rule Not Disjoint RH Types of Group 100 People A B AB O Total Positive 35 8 4 39 86 RH Negative 5 2 1 6 14 Total 40 10 5 45 100 P(B Blood or RH +) = P(B Blood) + P(RH +) P(B Blood and RH +) 10 86 8 0.88 100 100 100
Example 2 Addition Rule Mutually Exclusive A study of credit card fraud was conducted by Master Card International, and the table below is based on the results Method of Fraud Number Stolen Card 243 Counterfeit Card 85 Mail/Phone Order 52 Other 46 Total 426 Based on the table, what is the probability that fraud resulted from a stolen card or a mail/phone order?
Example 2 Addition Rule Mutually Exclusive Method of Fraud Number Stolen Card 243 Counterfeit Card 85 Mail/Phone Order 52 Other 46 Total 426 P(stolen or Mail/Phone)= P(stolen card) + P(Mail/Phone) P(Stolen and Mail/Phone) 243 52 0 0.69 426 426
Example Page 365, #6.46 Call a household prosperous if its income exceeds $100,000. Call the household educated if the household completed college. Select an American household at random, and let A be the event that the selected household is prosperous and B the event that it is educated. According the Census Bureau, P(A) = 0.134, P(B) = 0.254, and the joint probability that a household is both prosperous and educated is P(A and B) = 0.080. What is the probability P(A or B) that the household selected is either prosperous or educated? P(A or B) = P(A) + P(B) P(A and B) = 0.134 + 0.254 0.080 = 0.308
Example Page 364, #6.48 Consolidated builders has bid on two large construction projects. The company president believes that the probability of winning the first contract (event A) is 0.6, that the probability of winning the second (event B) is 0.5, and that the joint probability of winning both jobs (event {A and B}) is 0.3. What is the probability of event {A or B} that Consolidated will win at least one of the jobs? P(at least one contract) = P(A or B) = P(A) + P(B) P(A and B) = 0.6 + 0.5 0.3 = 0.80
Example Page 365, #6.52 Musical styles other than rock and pop are becoming more popular. A survey of college students finds that 40% like country music, 30% like gospel music, and 10% like both. A) Make a Venn Diagram with these results. Neither 0.4 Gospel Only 0.2 Both 0.1 Country Only 0.30
Example Page 365, #6.52 Musical styles other than rock and pop are becoming more popular. A survey of college students finds that 40% like country music, 30% like gospel music, and 10% like both. B) What percent of college students like country but not gospel? P(country but not gospel) = P(C) P(C and G) 0.40.10 0.30
Example Page 365, #6.52 Musical styles other than rock and pop are becoming more popular. A survey of college students finds that 40% like country music, 30% like gospel music, and 10% like both. C) What percent like neither country nor gospel. P(neither country or gospel) = 1 P(C or G) = 1 [P(C) + P(G) P(C and G)] 1 [.40.30.10] 0.40
Basic Properties of Probability 0 P(A) 1 for any event A P(S) = 1 Complement Rule For any event A, P(A c ) = 1 P(A) Addition Rule Events A and B are not disjoint P(A or B) = P(A B) = P(A) + P(B) P(A and B) Events A and B are disjoint P(A or B) = P(A B) = P(A) + P(B) Multiplication Rule Events A and B are independent P(A and B) = P(A B) = P(A) P(B)
Conditional Probability Notation 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. If events A and B are independent, then knowing that event A will occur does not change the probability of B. P(A B) = P(A) or P(B A) = P(B)
Conditional Probability Independent Example: When flipping a coin twice, what is the probability of getting heads on the second flip if the first flip was head? Event A: getting head on first flip Event B: getting head on second flip Events A and B are independent since the outcome of the first flip does not change the probability of the second flip. P ( B A) P ( B ) 1 2
Multiplication Rule Not Independent The intersection of two or more events is that all of those events occur. The probability that two events, A and B both occur is the probability that event A occurs multiplied by the probability that event B also occurs that is by the probability that event B occurs given that event A occurs. P(A and B) = P(A) P(B A) P(A B) = P(A) P(B A)
Conditional Probability To find the probability of event B given the event A, we restrict our attention to the outcomes in A. We then find in what fraction of those outcomes B also occurred. P(B A) = P(B A) = P(A and B) P(A) P(A B) P(A)
Example Conditional Probability RH Types of 100 People RH Group A B AB O Total Positive 35 8 4 39 86 Negative 5 2 1 6 14 Total 40 10 5 45 100 What is the probability that a person will be RH positive given they have blood type O? 39 P(O and RH +) 39 P(RH + O Blood) = 100 0.87 P(O blood) 45 45 100
Example Are the Events Disjoint? Independent? Police report that 78% of drivers are given a breath test, 36% a blood test, and 22% both tests 1. Are giving a DWI suspect a blood test and a breath test mutually exclusive? 2. Are giving the two tests independent? State the events we re interest in. Let A = {suspect is given a breath test}. Let B = {suspect is given a blood test}.
Example Are the Events Disjoint? Independent? Police report that 78% of drivers are given a breath test, 36% a blood test, and 22% both tests State the events we re interest in. Let A = {suspect is given a breath test}. Let B = {suspect is given a blood test}. State the given probabilities. P(A) = 0.78 P(B) = 0.36 P(A and B) = P(A B) = 0.22
Example Are the Events Disjoint? Independent? Let A = {suspect is given a breath test}. Let B = {suspect is given a blood test}. P(A) = 0.78 P(B) = 0.36 P(A and B) = P(A B) = 0.22 1. Are giving a DWI suspect a blood test and a breath test mutually exclusive? Disjoint events cannot both happen at the same time, so check to see if P(A and B) = 0. P(A and B) = 0.22, the events cannot be mutually exclusive. 22% of all suspects get both tests, so a breath test and blood test are not disjoint events.
Lesson 6-3, Part 2 General Probability Rules Conditional Probability
Example Are the Events Disjoint? Independent? Let A = {suspect is given a breath test}. Let B = {suspect is given a blood test}. P(A) = 0.78 P(B) = 0.36 P(A and B) = P(A B) = 0.22 2. Are the two tests independent? Does getting a breath test change the probability of getting a blood test? That is, does P(B A) = P(B)? P ( A B ) 0.22 P ( B A) 0.28 PB ( ) 0.36 P( A) 0.78 Because the two probabilities are not the same, the two events are not independent.
Example Are the Events Disjoint? Independent? Overall, 36% of the drivers get blood tests, but only 22% of those get a breath test too. Since suspects who get a breath test are less likely to have a blood test, the two events are not independent.
Example Page 369, #6.54 Choose an adult American woman at random. Table 6.1 describes the population from which we draw. Use the information in that table to answer the following questions. A). What is the probability that the women chosen is 65 years old or older? B). What is the conditional probability that the women chosen is married, given that she is 65 or over? C). How many women are both married and in the over 65 age group? What is the probability that the women we choose is a married women at least 65 years old?
Example Page 369, #6.54 A). P( 65 years old or older) = 18669 0.18 103870
Example Page 369, #6.54 B). P(M 65) = P(65 and M) P(65) 8270 103870 8270 0.443 18669 18669 103870
Example Page 369, #6.54 C). P(M and 65) = 8270 103870 0.08
Example Page 370, #6.56 Choose an employed person at random. Let A be the event that the person chosen is a women and B the event that the person holds a managerial or professional job. Government data tell us that P(A) = 0.46 and the probability of managerial and professional jobs among women is P(B A) = 0.32. Find the probability that a randomly chosen employed person is a women holding a managerial or professional position. P(Women and Managerial/Professional Position) P(A and B) = P(A) P(B A) = 0.46 0.32 0.1472
Example Page 370, #6.58 A poker player holds a flush when all 5 cards in the hand belong to the same suit. We will find the probability of a flush when 5 cards are dealt. Remember that a deck contains 52 cards, 13 of each suit, and that when the deck is well shuffled, each card dealt is equally likely to be any of those that remain in the deck.
Example Page 370, #6.58 A). We will concentrate on spades. What is the probability that the first card dealt is a spade? What is the conditional probability that the second card is a spade, given that the first is a spade? P(1 st card Spade) = 13 0.25 52 P(2 nd card spade 1 st card spade) = 12 0.2353 51
Example Page 370, #6.58 B). Continue to count the remaining cards to find the conditional probability of a spade on the third, fourth, and fifth card, given that all previous cards are spades. P(3 rd card spade 2 nd card spade) = P(4 th card spade 3 rd card spade) = P(5 th card spade 4 th card spade) = 11 50 0.22 10 49 0.2041 9 48 0.1875
Example Page 370, #6.58 C). The probability of being dealt 5 spades is the product of the five probabilities you have found? Why? What is this probability? P(5 spades) = 0.25 0.2353 0.22 0.2041 0.1875 0.0004952 The product of these conditional probabilities gives the probability of a flush in spades by the extended multiplication rule: We must draw a spade then another, and then a third, fourth, and fifth.
Example Page 370, #6.58 D). The probability of being dealt 5 hearts, or 5 diamonds or 5 clubs is the same as the probability of being dealt 5 spades. What is the probability of being dealt a flush? P(flush) = (0.0004952)4 0.001981
Example Page 378, #6.64 Enzyme immunoassay (EIA) tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. Antibodies indicate the presence of the virus. The test is quite accurate but is not always correct. Here are approximate probabilities of positive and negative EIA outcomes when the blood tested does and does not actually contain antibodies to HIV. Positive Test result Negative Antibodies Present 0.9985 0.0015 Antibodies Absent 0.006 0.994 Suppose that 1% of a large population carries antibodies to HIV in their blood.
Example Page 378, #6.64 Test result Positive Negative Antibodies Present 0.9985 0.0015 Antibodies Absent 0.006 0.994 A). Draw a tree diagram for selecting a person from this population (outcomes: antibodies present or absent) and for testing his or her blood (outcomes: EIA positive or negative).
Example Page 378, #6.64 Test result Positive Negative Antibodies Present 0.9985 0.0015 Antibodies Absent 0.006 0.994 Subject 0.01 Antibody Present 0.9985 0.0015 EIA + EIA 0.99 Antibody Absent 0.006 0.994 EIA + EIA
Example Page 378, #6.64 B). What is the probability that the EIA is positive for an randomly chosen person from this population? P(Test +) = P(antibody and Test +) + P(no antibody and Test +) = (0.01)(0.9985) (0.99)(0.006) 0.016 Subject 0.01 Antibody Present 0.9985 0.0015 EIA + EIA 0.99 Antibody Absent 0.006 0.994 EIA + EIA
Example Page 378, #6.64 C). What is the probability that a person has the antibody given that the EIA test is positive? P(Antibody Test +) = P(Test + and Antibody) P(Test +) (0.9985)(0.01) 0.016 0.624 Subject 0.01 Antibody Present 0.9985 0.0015 EIA + EIA 0.99 Antibody Absent 0.006 0.994 EIA + EIA