5-1 Chapter 5 Probability 1

Outline 5-2 5-1 Introduction 5-2 Sample Spaces and Probability 5-3 The Addition Rules for Probability 5-4 The Multiplication Rules and Conditional Probability 2 2

Objectives 5-3 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. 4

Objectives 5-4 Find the probability of compound events using the multiplication rules. Find the conditional probability of an event. 5

5-2 Sample Spaces and Probability 5-5 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. 7

5-2 Tree Diagram for Tossing Two Coins 5-6 H T Second Toss First Toss 8

5-2 Sample Spaces - Examples 5-7 9

5-2 Formula for Classical Probability 5-8 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. 10

5-2 Formula for Classical Probability 5-9 11

5-2 Classical Probability - Examples 5-10 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. 12

5-2 Classical Probability - Examples 5-11 (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. 13

5-2 Classical Probability - Examples 5-12 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. 14

5-2 Complement of an Event 5-13 E 15

5-2 Complement of an Event - Example 5-14 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). 16

5-2 Complement of an Event - Example 5-15 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. 17

5-2 Rule for Complementary Event 5-16 P ( E )  1  P ( E ) or P ( E ) = 1  P ( E ) or P ( E ) + P ( E ) = 1 . 18

5-2 Empirical Probability 5-17 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. 19

5-2 Formula for Empirical Probability 5-18 20

5-2 Empirical Probability - Example 5-19 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. 21

5-2 Empirical Probability - Example 5-20 Type Frequency A B AB O 22 5 2 21 50 = n 22

5-2 Empirical Probability - Example 5-21 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. 23

5-3 The Addition Rules for Probability 5-22 Two events are mutually exclusive if they cannot occur at the same time (i.e. they have no outcomes in common). 24

5-3 The Addition Rules for Probability 5-23 A and B are mutually exclusive A B 25

5-3 Addition Rule 1 5-24 When two events A and B are mutually exclusive, the probability that A or B will occur is P A or B ( ) = + 26

5-3 Addition Rule 1- Example 5-25 At a political rally, there are 20 Republicans (R), 13 Democrats (D), and 6 Independents (I). If a person is selected, find the probability that he or she is either a Democrat or an Independent. Solution: P(D or I) = P(D) + P(I) = 13/39 + 6/39 = 19/39. 27

5-3 Addition Rule 1- Example 5-26 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. 28

5-3 Addition Rule 2 5-27 When two events A and B are not mutually exclusive, the probability y that A or B will occur is P ( A or B )  P ( A )  P ( B )  P ( A and B ) 29

5-3 Addition Rule 2 5-28 A and B (common portion) A B 30

5-3 Addition Rule 2- Example 5-29 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. 31

5-3 Addition Rule 2 - Example 5-30 32

5-3 Addition Rule 2 - Example 5-31 Solution: P(nurse or male) = P(nurse) + P(male) – P(male nurse) = 8/13 + 3/13 – 1/13 = 10/13. 33

5-3 Addition Rule 2 - Example 5-32 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? 34

5-3 Addition Rule 2 - Example 5-33 Solution: P(intoxicated or accident) = P(intoxicated) + P(accident) – P(intoxicated and accident) = 0.32 + 0.09 – 0.06 = 0.35. 35

5-4 The Multiplication Rules and Conditional Probability 5-34 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. 36

5-4 Multiplication Rule 1 5-35 37

5-4 Multiplication Rule 1 - Example 5-36 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. 38

5-4 Multiplication Rule 1 - Example 5-37 A Harris pole found that 46% of Americans 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. 39

5-4 Multiplication Rule 1 - Example 5-38 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. 40

5-4 The Multiplication Rules and Conditional Probability 5-39 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. 41

5-4 The Multiplication Rules and Conditional Probability 5-40 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. 42

5-4 Multiplication Rule 2 5-41 43

5-4 The Multiplication Rules and Conditional Probability - Example 5-42 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. 44

5-4 The Multiplication Rules and Conditional Probability - Example 5-43 Solution: Since the events are dependent, P(D1 and D2) = P(D1)P(D2| D1) = (2/25)(1/24) = 2/600 = 1/300. 45

5-4 The Multiplication Rules and Conditional Probability - Example 5-44 The WW 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. 46

5-4 The Multiplication Rules and Conditional Probability - Example 5-45 Solution: Since the events are dependent, P(H and A) = P(H)P(A|H) = (0.53)(0.27) = 0.1431. 47

5-4 The Multiplication Rules and Conditional Probability - Example 5-46 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. 48

5-4 Tree Diagram for Example 5-47 P(B1) 1/2 Red Blue Box 1 P(B2) 1/2 Box 2 P(R|B1) 2/3 P(B|B1) 1/3 P(R|B2) 1/4 P(B|B2) 3/4 (1/2)(2/3) (1/2)(1/3) (1/2)(1/4) (1/2)(3/4) 50

5-4 The Multiplication Rules and Conditional Probability - Example 5-48 Solution: P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/24. 51

5-4 Conditional Probability - Formula 5-49 52

5-4 Conditional Probability - Example 5-50 The probability that Sam parks in a no-parking 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. 53

5-4 Conditional Probability - Example 5-51 Solution: Let N = parking in a no-parking zone and T = getting a ticket. Then P(T |N) = [P(N and T) ]/P(N) = 0.06/0.2 = 0.30. 54

5-4 Conditional Probability - Example 5-52 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. 55

5-4 Conditional Probability - Example 5-53 56

5-4 Conditional Probability - Example 5-54 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”. 57

5-4 Conditional Probability - Example 5-55 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. 58