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3-1 Chapter 3 Probability © The McGraw-Hill Companies, Inc., 2000 3-2 Outline 3-1 3-2 3-3 3-4 3-5 Introduction Fundamentals Addition Rules for Probability Multiplication Rules: Basics Multiplication Rules: Beyond the Basics © The McGraw-Hill Companies, Inc., 2000 3-3 Objectives Determine Sample Spaces and find the probability of an event using classical probability. Find the probability of an event using empirical probability. Find the probability of compound events using the addition rules. © The McGraw-Hill Companies, Inc., 2000 3-4 Objectives Find the probability of compound events using the multiplication rules. Find the conditional probability of an event. © The McGraw-Hill Companies, Inc., 2000 3-5 3-2 Fundamentals A probability experiment is a process that leads to well-defined results called outcomes. An outcome is the result of a single trial of a probability experiment. NOTE: A tree diagram can be used as a systematic way to find all possible outcomes of a probability experiment. © The McGraw-Hill Companies, Inc., 2000 3-2 Tree Diagram for Tossing Two Coins 3-6 H H T Second Toss H T First Toss T © The McGraw-Hill Companies, Inc., 2000 3-7 3-2 Sample Spaces - Examples EXPERIM ENT SAM PLE SPACE Toss one coin H, T Roll a die 1, 2, 3, 4, 5, 6 Answer a truefalse question Toss two coins True, False H H , H T, TH , TT © The McGraw-Hill Companies, Inc., 2000 3-2 Formula for Classical Probability 3-8 Classical probability assumes that all outcomes in the sample space are equally likely to occur. That is, equally likely events are events that have the same probability of occurring. © The McGraw-Hill Companies, Inc., 2000 3-9 3-2 Formula for Classical Probability The probability of any event E is number of outcomes in E . total number of outcomes in the sample space This probability is denoted by n( E ) . n( S ) This probability is called classical probability , P( E ) = and it uses the sample space S . © The McGraw-Hill Companies, Inc., 2000 3-2 Classical Probability - Examples 3-10 For a card drawn from an ordinary deck, find the probability of getting (a) a queen (b) a 6 of clubs (c) a 3 or a diamond. Solution: (a) Since there are 4 queens and 52 cards, P(queen) = 4/52 = 1/13. (b) Since there is only one 6 of clubs, then P(6 of clubs) = 1/52. © The McGraw-Hill Companies, Inc., 2000 3-11 3-2 Classical Probability - Examples (c) There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in the listing. Hence there are only 16 possibilities of drawing a 3 or a diamond, thus P(3 or diamond) = 16/52 = 4/13. © The McGraw-Hill Companies, Inc., 2000 3-12 3-2 Classical Probability - Examples When a single die is rolled, find the probability of getting a 9. Solution: Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9. Hence, P(9) = 0/6 = 0. NOTE: The sum of the probabilities of all outcomes in a sample space is one. © The McGraw-Hill Companies, Inc., 2000 3-13 3-2 Complement of an Event The complement of an event E is the set of outcomes in the sample space that are not included in the outcomes of event E . The complement of E is denoted by E ( E bar ). E E © The McGraw-Hill Companies, Inc., 2000 3-2 Complement of an Event Example 3-14 Find the complement of each event. Rolling a die and getting a 4. Solution: Getting a 1, 2, 3, 5, or 6. Selecting a letter of the alphabet and getting a vowel. Solution: Getting a consonant (assume y is a consonant). © The McGraw-Hill Companies, Inc., 2000 3-2 Complement of an Event Example 3-15 Selecting a day of the week and getting a weekday. Solution: Getting Saturday or Sunday. Selecting a one-child family and getting a boy. Solution: Getting a girl. © The McGraw-Hill Companies, Inc., 2000 3-16 3-2 Rule for Complementary Event P(E ) 1 P(E ) or P(E ) = 1P(E ) or P ( E ) + P ( E ) = 1. © The McGraw-Hill Companies, Inc., 2000 3-17 3-2 Empirical Probability The difference between classical and empirical probability is that classical probability assumes that certain outcomes are equally likely while empirical probability relies on actual experience to determine the probability of an outcome. © The McGraw-Hill Companies, Inc., 2000 3-18 3-2 Formula for Empirical Probability Given a frequency distribution , the probability of an event being in a given class is frequency for the class P(E ) = total frequencies in the distribution f . n This probability is called the empirical probability and is based on observation . © The McGraw-Hill Companies, Inc., 2000 3-2 Empirical Probability Example 3-19 In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had AB blood. Set up a frequency distribution. © The McGraw-Hill Companies, Inc., 2000 3-20 3-2 Empirical Probability Example Type A B AB O Frequency 22 5 2 21 50 = n © The McGraw-Hill Companies, Inc., 2000 3-2 Empirical Probability Example 3-21 Find the following probabilities for the previous example. A person has type O blood. Solution: P(O) = f /n = 21/50. A person has type A or type B blood. Solution: P(A or B) = 22/50+ 5/50 = 27/50. © The McGraw-Hill Companies, Inc., 2000 3-22 3-3 Addition Rules for Probability Two events are mutually exclusive if they cannot occur at the same time (i.e., they have no outcomes in common). © The McGraw-Hill Companies, Inc., 2000 3-23 3-3 The Addition Rules for Probability A and B are mutually exclusive A B © The McGraw-Hill Companies, Inc., 2000 3-24 3-3 Addition Rule 1 When two events A and B are mutually exclusive, the probability that A or B will occur is P ( A or B ) P ( A ) + P ( B ) © The McGraw-Hill Companies, Inc., 2000 3-25 3-3 Addition Rule 1- Example At a political rally, there are 20 Liberals (L), 13 Conservatives (C), and 6 NDPs (N). If a person is selected, find the probability that he or she is either a Conservative or an NDP. Solution: P(C or N) = P(C) + P(N) = 13/39 + 6/39 = 19/39. © The McGraw-Hill Companies, Inc., 2000 3-26 3-3 Addition Rule 1- Example A day of the week is selected at random. Find the probability that it is a weekend. Solution: P(Saturday or Sunday) = P(Saturday) + P(Sunday) = 1/7 + 1/7 = 2/7. © The McGraw-Hill Companies, Inc., 2000 3-27 3-3 Addition Rule 2 When two events A and B are not mutually exclusive, the probabilityy that A or B will occur is P ( A or B ) P ( A) + P ( B) P ( A and B) © The McGraw-Hill Companies, Inc., 2000 3-28 3-3 Addition Rule 2 A and B A (common portion) B © The McGraw-Hill Companies, Inc., 2000 3-29 3-3 Addition Rule 2- Example In a hospital unit there are eight nurses and five physicians. Seven nurses and three physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male. The next slide has the data. © The McGraw-Hill Companies, Inc., 2000 3-30 3-3 Addition Rule 2 - Example STAFF STAFF FEMALES FEMALES MALES MALES TOTAL TOTAL NURSES NURSES 77 11 88 PHYSICIANS PHYSICIANS 33 22 55 TOTAL TOTAL 10 10 33 13 13 © The McGraw-Hill Companies, Inc., 2000 3-31 3-3 Addition Rule 2 - Example Solution: P(nurse or male) = P(nurse) + P(male) – P(male nurse) = 8/13 + 3/13 – 1/13 = 10/13. © The McGraw-Hill Companies, Inc., 2000 3-32 3-3 Addition Rule 2 - Example On New Year’s Eve, the probability that a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident? © The McGraw-Hill Companies, Inc., 2000 3-33 3-3 Addition Rule 2 - Example Solution: P(intoxicated or accident) = P(intoxicated) + P(accident) – P(intoxicated and accident) = 0.32 + 0.09 – 0.06 = 0.35. © The McGraw-Hill Companies, Inc., 2000 3-4 The Multiplication Rules and Conditional Probability 3-34 Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring. Example: Rolling a die and getting a 6, and then rolling another die and getting a 3 are independent events. © The McGraw-Hill Companies, Inc., 2000 3-35 3-4 Multiplication Rules W hen two events A and B are independent , the probability of both occurring is P ( A and B ) P ( A ) P ( B ). © The McGraw-Hill Companies, Inc., 2000 3-4 Multiplication Rule 1 Example 3-36 A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then an ace. Solution: Because these two events are independent (why?), P(queen and ace) = (4/52)(4/52) = 16/2704 = 1/169. © The McGraw-Hill Companies, Inc., 2000 3-4 Multiplication Rule 1 Example 3-37 A Decima pole found that 46% of Canadians say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer stress at least once a week. Solution: Let S denote stress. Then P(S and S and S) = (0.46)3 = 0.097. © The McGraw-Hill Companies, Inc., 2000 3-4 Multiplication Rule 1 Example 3-38 The probability that a specific medical test will show positive is 0.32. If four people are tested, find the probability that all four will show positive. Solution: Let T denote a positive test result. Then P(T and T and T and T) = (0.32)4 = 0.010. © The McGraw-Hill Companies, Inc., 2000 3-5 Multiplication Rules: Conditional Probability 3-39 When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent. Example: Having high grades and getting a scholarship are dependent events. © The McGraw-Hill Companies, Inc., 2000 3-5 Multiplication Rules: Conditional Probability 3-40 The conditional probability of an event B in relationship to an event A is the probability that an event B occurs after event A has already occurred. The notation for the conditional probability of B given A is P(B|A). NOTE: This does not mean B A. © The McGraw-Hill Companies, Inc., 2000 3-41 3-4 Multiplication Rule 2 When two events A and B are dependent , the probability of both occurring is P ( A and B ) P ( A) P ( B| A). © The McGraw-Hill Companies, Inc., 2000 3-5 The Multiplication Rules: Conditional Probability - Example 3-42 In a shipment of 25 microwave ovens, two are defective. If two ovens are randomly selected and tested, find the probability that both are defective if the first one is not replaced after it has been tested. Solution: See next slide. © The McGraw-Hill Companies, Inc., 2000 3-5 The Multiplication Rules and Conditional Probability - Example 3-43 Solution: Since the events are dependent, P(D1 and D2) = P(D1)P(D2| D1) = (2/25)(1/24) = 2/600 = 1/300. © The McGraw-Hill Companies, Inc., 2000 3-5 The Multiplication Rules and Conditional Probability - Example 3-44 The KW Insurance Company found that 53% of the residents of a city had homeowner’s insurance with its company. Of these clients, 27% also had automobile insurance with the company. If a resident is selected at random, find the probability that the resident has both homeowner’s and automobile insurance. © The McGraw-Hill Companies, Inc., 2000 3-5 The Multiplication Rules and Conditional Probability - Example 3-45 Solution: Since the events are dependent, P(H and A) = P(H)P(A|H) = (0.53)(0.27) = 0.1431. © The McGraw-Hill Companies, Inc., 2000 3-5 The Multiplication Rules and Conditional Probability - Example 3-46 Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls and one red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball. © The McGraw-Hill Companies, Inc., 2000 3-5 Tree Diagram for Example 3-47 P(R|B1) 2/3 P(B1) 1/2 P(B2) 1/2 Red (1/2)(2/3) Box 1 Blue (1/2)(1/3) P(B|B1) 1/3 P(R|B2) 1/4 Box 2 Red (1/2)(1/4) P(B|B2) 3/4 Blue (1/2)(3/4) © The McGraw-Hill Companies, Inc., 2000 3-5 The Multiplication Rules and Conditional Probability - Example 3-48 Solution: P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/24. © The McGraw-Hill Companies, Inc., 2000 3-49 3-5 Conditional Probability Formula The probability that the second event B occurs given that the first event A has occurred can be found by dividing the probability that both events occurred by the probability that the first event has occurred . The formula is P( A and B) . P( B| A) = P( A) © The McGraw-Hill Companies, Inc., 2000 3-5 Conditional Probability Example 3-50 The probability that Sam parks in a noparking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.2. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a ticket. © The McGraw-Hill Companies, Inc., 2000 3-5 Conditional Probability Example 3-51 Solution: Let N = parking in a noparking zone and T = getting a ticket. Then P(T|N) = [P(N and T) ]/P(N) = 0.06/0.2 = 0.30. © The McGraw-Hill Companies, Inc., 2000 3-5 Conditional Probability Example 3-52 A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results are shown in the table on the next slide. © The McGraw-Hill Companies, Inc., 2000 3-53 3-5 Conditional Probability Example Gender Gender Yes Yes No No Total Total Male Male 32 32 18 18 50 50 Female Female 88 42 42 50 50 Total Total 40 40 60 60 100 100 © The McGraw-Hill Companies, Inc., 2000 3-5 Conditional Probability Example 3-54 Find the probability that the respondent answered “yes” given that the respondent was a female. Solution: Let M = respondent was a male; F = respondent was a female; Y = respondent answered “yes”; N = respondent answered “no”. © The McGraw-Hill Companies, Inc., 2000 3-5 Conditional Probability Example 3-55 P(Y|F) = [P( F and Y) ]/P(F) = [8/100]/[50/100] = 4/25. Find the probability that the respondent was a male, given that the respondent answered “no”. Solution: P(M|N) = [P(N and M)]/P(N) = [18/100]/[60/100] = 3/10. © The McGraw-Hill Companies, Inc., 2000

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