Outline Lecture 8 5-1 Introduction 5-2 Sample Spaces and 5-3 The Addition Rules for 5-4 The Multiplication Rules and Conditional 5-11 Introduction 5-11 Introduction as a general concept can be defined as the chance of an event occurring. Games of chance (card games, slot machines, or lotteries). Used in the fields of insurance, investments, and weather forecasting and many others. Most important for us: probability is the basis of inferential statistics. Basic concepts of probability are explained in this lecture. 5-22 Sample Spaces and 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. 5-22 Tree Diagram for Tossing Two Coins H T First Toss H T Second Toss H T
5-22 Sample Spaces - s EXPERIM ENT Toss one coin SAM PLE SPACE H, T Roll a die 1, 2, 3, 4, 5, 6 Answer a truefalse question Toss two coins True, False HH, HT, TH, TT 5-22 Formula for Classical Classical probability assumes that all outcomes in a finite sample space are equally likely to occur. That is, equally likely events are events that have the same probability of occurring. 5-22 Formula for Classical 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 ) P( E ) =. n( S ) This probability is called classical probability, and it uses the sample space S. 5-22 Classical - s 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. 5-22 Classical - s (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. 5-22 Classical - s 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. 0
5-22 Classical There are four basic probability rules. These rules are helpful in solving probability problems, in understanding the nature of probability, and in deciding if your answers to the problems are correct. Rule 1 The probability of an event E is a number between 0 and 1: 0 P(E) 1 Rule 2 If an event E cannot occur (i.e., the event contains no members in the sample space), the probability of E is zero. Rule 3 If an event E is certain, then the probability of E equals 1. Rule 4 The sum of the probabilities of the outcomes in the sample space is 1. 5-22 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
5-22 Complement of an Event - 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). 5-22 Complement of an Event - 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. 5-22 Rule for Complementary Event P ( E ) = 1 P ( E ) or P ( E ) = 1 P ( E ) or P ( E ) + P ( E ) = 1. 5-22 Empirical 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. 5-22 Formula for Empirical 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. 5-22 Empirical - 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.
5-22 Empirical - 5-22 Empirical - Type A B AB O Frequency 22 5 2 21 50 = n 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. 5-33 The Addition Rules for Two events are mutually exclusive if they cannot occur at the same time (i.e. they have no outcomes in common). 5-33 The Addition Rules for A and B are mutually exclusive A B 5-33 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 ) 5-33 Addition Rule 1-1 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.
5-33 Addition Rule 1-1 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. 5-33 Addition Rule 2 When two events A and B are not mutually exclusive, the probability that A or B will occur is P( A or B) = P( A) + P( B) P( A and B) 5-33 Addition Rule 2 5-33 Addition Rule 2-2 A A and B (common portion) B 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. 5-33 Addition Rule 2-5-33 Addition Rule 2 - STAFF FEMALES MALES TOTAL NURSES 7 1 8 PHYSICIANS 3 2 5 Solution: P(nurse or male) = P(nurse) + P(male) P(male nurse) = 8/13 + 3/13 1/13 = 10/13. TOTAL 10 10 3 13 13
5-33 Addition Rule 2 - 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? 5-33 Addition Rule 2 - Solution: P(intoxicated or accident) = P(intoxicated) + P(accident) P(intoxicated and accident) = 0.32 + 0.09 0.06 = 0.35. Conditional Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring. : Rolling a die and getting a 6, and then rolling another die and getting a 3 are independent events. 5-44 Multiplication Rule 1 When two events A and B are independent, the probability of both occurring is P( A and B) = P( A) P( B). 5-44 Multiplication Rule 1 - 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. 5-44 Multiplication Rule 1 - 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.
5-44 Multiplication Rule 1 - 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. Conditional 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. : Having high grades and getting a scholarship are dependent events. Conditional 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). 5-44 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). Conditional - 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. Conditional - Solution: Since the events are dependent, P(D 1 and D 2 ) = P(D 1 ) P(D 2 D 1 ) = (2/25)(1/24) = 2/600 = 1/300.
Conditional - 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. Conditional - Solution: Since the events are dependent, P(H and A) = P(H) P(A H) = (0.53)(0.27) = 0.1431. 5 Conditional - 5-44 Tree Diagram for P(R B 1 ) 2/3 Red (1/2)(2/3) 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. P(B 1 ) 1/2 P(B 2 ) 1/2 Box 1 Blue (1/2)(1/3) P(B B 1 ) 1/3 P(R B 2 ) 1/4 Box 2 Red (1/2)(1/4) P(B B 2 ) 3/4 Blue (1/2)(3/4) Conditional - Solution: P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/24. 5-44 Conditional - 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)
5-44 Conditional - 5-44 Conditional - 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. 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. 5-44 Conditional - 5-44 Conditional - 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. Gender Yes No Total Male 32 18 50 Female 8 42 50 Total 40 60 100 5-44 Conditional - 5-4 Conditional 4 Conditional - 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. 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.