Math 227 Elementary Statistics Bluman 5 th edition
CHAPTER 4 Probability and Counting Rules 2
Objectives Determine sample spaces and find the probability of an event using classical probability or empirical probability. Find the probability of compound events using the addition rules. Find the probability of compound events using the multiplication rules. Find the conditional probability of an event. 3
Objectives (cont.) Determine the number of outcomes of a sequence of events using a tree diagram. Find the total number of outcomes in a sequence of events using the fundamental counting rule. Find the number of ways r objects can be selected from n objects using the permutation rule. Find the number of ways r objects can be selected from n objects without regard to order using the combination rule. Find the probability of an event using the counting rules. 4
Introduction Probability as a general concept can be defined as the chance of an event occurring. In addition to being used in games of chance, probability is used in the fields of insurance, investments, and weather forecasting, and in various areas. Rules such as the fundamental counting rule, combination rule and permutation rules allow us to count the number of ways in which events can occur. Counting rules and probability rules can be used together to solve a wide variety of problems. 5
Section 4.1 Sample Spaces and Probability I. Basic Concepts A probability experiment is a chance process that leads to well-defined results called outcomes. An outcome is the result of a single trial of a probability experiment. A sample space is the set of all possible outcomes of a probability experiment. 6
Sample Spaces and Events Examples of some sample space: Experiment Sample Space Toss one coin H(head), T(tail) Roll a die 1, 2, 3, 4, 5, 6 Answer a true/false question True, false Toss two coins H-H, T-T, H-T, T-H 7
Sample Spaces and Events An event consists of a set of outcomes of a probability experiment. Simple event - an event with one outcome Compound event - an event with two or more outcomes. 8
Basic Concepts (cont.) Equally likely events are events that have the same probability of occurring. A tree diagram is a device consisting of line segments emanating from a starting point and also from the outcome point. It is used to determine all possible outcomes of a probability experiment. 9
Example: A tree diagram to find the sample space for tossing two coins. H T H T H T Outcomes H H H T T H T T 10
Example 1 : A die is tossed one time. (a) List the elements of the sample space S. S = {1, 2, 3, 4, 5, 6} (b) List the elements of the event consisting of a number that is greater than 4. Event A: x > 4 A = {5, 6} 11
Example 2 : A coin is tossed twice. List the elements of the sample space S, and list the elements of the event consisting of at least one head. H T H T H T S = {HH, HT, TH, TT} Event A : At least one head (H) A = {HT, TH, HH} 12
Example 3 : A randomly selected citizen is interviewed and the following information is recorded: employment status and level of education. The symbols for employment status are Y = employed and N = unemployed, and the symbols for level of education are 1 = did not complete high school, 2 = completed high school but did not complete college, and 3 = completed college. List the elements of sample space, and list the elements of the following events. Y N 1 2 3 1 2 3 S = {Y1, Y2, Y3, N1, N2, N3} (a) Did not complete high school (b) Is unemployed {Y1, N1} {N1, N2, N3} 13
II. Calculating Probabilities A. Empirical Probability (Relative Frequency Approximation of Probability) Empirical probability relies on actual experience to determine the likelihood of outcomes. 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 14
Example 1 : The age distribution of employees for this college is shown below: Age # of employees Under 20 25 20 29 48 30 39 32 40 49 15 50 and over 10 n = 130 If an employee is selected at random, find the probability that he or she is in the following age groups. (a) Between 30 and 39 years of age (b) Under 20 or over 49 years of age 15
Example 2 : During a sale at men s store, 16 white sweaters, 3 red sweaters, 9 blue sweaters, and 7 yellow sweaters were purchased. If a customer is selected at random, find the probability that he bought a sweater that was not white. 16
B. Classical Probability Classical probability uses sample spaces to determine the numerical probability that an event will happen. Classical probability assumes that all outcomes in the sample space are equally likely to occur. PE ( ) ne ( ) Number of outcomes in E ns ( ) total number of outcomes in the sample space 17
Example 1 : A statistics class contains 14 males and 20 females. A student is to be selected by chance and the gender of the student recorded. (Ref: General Statistics by Chase/Bown, 4 th Ed.) (a) Give a sample space S for the experiment. S = {M, F} (b) Is each outcome equally likely? Explain. P(M) P(F). It is not equally likely because we have more females than males. (c) Assign probabilities to each outcome. 18
Example 2 : Two dice are tossed. Find the probability that the sum of two dice is greater than 8? S = { (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) } 19
Example set of 52 playing cards; 13 of each suit clubs, diamonds, hearts, and spades 20
Example 3 : If one card is drawn from a deck, find the probability of getting (a) a club 52 cards in a deck, 13 of those are clubs. (b) a 4 and a club 21
Example 4 : Three equally qualified runners, Mark (M), Bill (B) and Alan (A), run a 100-meter, sprint, and the order of finish is recorded. (a) Give a sample space S. B A M A B B M A A M S = {MBA, MAB, BMA, BAM, AMB, ABM} M B A B M (b) What is the probability that Mark will finish last? {BAM, ABM} 22
III. Rounding Rule for Probabilities Probabilities should be expressed as reduced fractions or rounded to two or three decimal places. 23
Probability Rules 1. The probability of an event E is a number (either a fraction or decimal) between and including 0 and 1. This is denoted by: 0 P( E) 1 *Rule 1 states that probabilities cannot be negative or greater than one. 24
Probability Rules (cont.) 2. If an event E cannot occur (i.e., the event contains no members in the sample space), the probability is zero. 3. If an event E is certain, then the probability of E is 1. 4. The sum of the probabilities of the outcomes in the sample space is 1. 25
Example 1: A probability experiment is conducted. Which of these can be considered a probability of an outcome? a) 2/5 Yes b) -0.28 No c) 1.09 No 26
Example 2 : Given : S = {E 1, E 2, E 3, E 4 } Find : P(E 4 ) P(E 1 ) = P(E 2 ) = 0.2 and P(E 3 ) = 0.5 P(E 1 ) + P(E 2 ) + P(E 3 ) +P(E 4 ) = 1 0.2 + 0.2 + 0.5 +P(E 4 ) = 1 P(E 4 ) = 1 0.9 P(E 4 ) = 0.1 27
IV. Complementary Events 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. Rule for Complementary Events P( E ) 1 P( E), OR P( E) P( E) 1 E P( E) 1 P( E) 28
Example 1 : The chance of raining tomorrow is 70%. What is the probability that it will not rain tomorrow? P (No Rain) = 1 0.70 = 0.3 30% chance it will not rain tomorrow. 29
Section 4.2 The Addition Rules for Probability Mutually Exclusive Events Two events are mutually exclusive if they cannot occur at the same time (i.e., they have no outcomes in common). The probability of two or more events can be determined by the addition rules. 30
I. Addition Rules 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) 31
Ex1) A single card is drawn from a deck. Find the probability of selecting a club or a diamond. P (club or diamond) mutually exclusive P (club or diamond) = P (club) + P (diamond) 13 13 52 52 26 1 52 2 32
Ex2) In a large department store, there are 2 managers, 4 department heads, 16 clerks, and 4 stock persons. If a person is selected at random, find the probability that the person is either a clerk or a manager. P (clerk or manager) mutually exclusive P (clerk or manager) = P (clerk) + P (manager) 16 2 26 26 18 9 26 13 33
I. Addition Rules (cont.) Addition Rule 2 If A and B are not mutually exclusive, then: P( A or B) P( A) P( B) P( A and B) 34
Example1 : A single card is drawn from a deck. Find the probability of selecting a jack or a black card. P (Jack or Black) = P (Jack) + P (Black) P (Jack and Black) 35
Example 2 : In a certain geographic region, newspapers are classified as being published daily morning, daily evening, and weekly. Some have a comics section and other do not. The distribution is shown here. Having comics Section Morning Evening Weekly Yes 2 3 1 No 3 4 2 If a newspaper is selected at random, find these probabilities. (a) The newspaper is weekly publication. S = 2 + 3 + 1 + 3 + 4 +2 = 15 (b) The newspaper is a daily morning publication or has comics. P (Daily morning or has comics) = P (Daily morning) + P (Has comics) P (Daily morning and has comics) 36
(c) The newspaper is published weekly or does not have comics. P (Weekly or No comics) = P (Weekly) + P (No comics) P (Weekly and no comics) 37
Section 4.3 The Multiplication Rules and Conditional Probability Independent Events Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring. 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. 38
I. Multiplication Rules The multiplication rules can be used to find the probability of two or more events that occur in sequence. Multiplication Rule 1 When two events are independent, the probability of both occurring is: P( A and B) P( A) P( B) 39
Example 1 : If 36% of college students are overweight, find the probability that if three college students are selected at random, all will be overweight. P (1st overweight and 2nd overweight and 3rd overweight) = 0.36 0.36 0.36 = 0.046656 0.047 40
Example 2 : If 25% of U.S. federal prison inmates are not U.S. citizens, find the probability that two randomly selected federal prison inmates will be U.S. citizens. P (1 st U.S. citizen and 2 nd U.S. citizen) = 0.75 0.75 Example 3 : = 0.5625.563 Suppose the probability of remaining with a particular company 10 years or longer is 1/6. A man and a woman start work at the company on the same day. (a) What is the probability that the man will work there less than 10 years? (b) What is the probability that both the man and woman will work there less than 10 years? 41
Example 4 : A smoke-detector system uses two devices, A and B. If smoke is present, the probability that it will be detected by device A is 0.95; by device B, 0.98; and by both devices, 0.94. (a) If smoke is present, find the probability that the smoke will be detected by device A or device B or by both devices. P ( A or B) = P (A ) + P (B) P (A and B) = 0.95 + 0.98 0.94 = 0.99 (b) Find the probability that the smoke will not be detected. P (Neither ) = 1 0.99 = 0.01 42
Example 5 : 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? P (All wrong) = P (1 st wrong & 2 nd wrong & 3 rd wrong & 4 th wrong ) P (At least 1 correct) = 1 P (All wrong) 43
Example 6 : There are 2000 voters in a town. Consider the experiment of randomly selecting a voter to be interviewed. The event A consists of being in favor of more stringent building codes; the event B consists of having lived in the town less than 10 years. The following table gives the number of voters in various categories. (Ref: General statistics by Chase/Bown, 4 th Ed.) Favor more Stringent codes Do not favor more stringent codes Less than 10 years At least 10 years 100 700 1000 200 Find each of the following: (a) (b) (c) 44
Multiplication Rules (cont.) Multiplication Rule 2 When two events are dependent (e.g. the outcome of event A affects the outcomes of event B), the probability of both occurring is: P( A and B) P( A) P( B A) 45
Example 1 : A box has 5 red balls and 2 white balls. If two balls are randomly selected (one after the other), what is the probability that they both are red? (a) With replacement independent case (b) Without replacement dependent case 46
Example 2 : Three cards are drawn from a deck without replacement. Find the probability that all are Jacks. 47
Multiplication Rules (cont.) The conditional probability of an event B in relationship to an event A is the probability that event B occurs after event A has already occurred. The notation for conditional probability is P(B A). 48
Formula for Conditional Probability The probability that the second event B occurs given that the first event A has occurred can be found dividing the probability that both events occurred by the probability that the first event has occurred. The formula is: P( B A) P( A and B) P( A) 49
Example 1: Two fair dice are tossed. Consider the following events. A = sum is 7 or more, B = sum is even, and C = a match (both numbers are the same). A = { (1,6) (2,5) (2,6) (3,4) (3,5) (3, 6) (4, 3) (4,4) (4,5) (4,6) (5,2) (5,3) (5, 4) (5, 5) (5,6) (6,1) (6,2) (6,3) (6,4) (6, 5) (6, 6) } B = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6) } C = { (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) } 50
(a) P (A and B) = A = { (1,6) (2,5) (2,6) (3,4) (3,5) (3, 6) (4, 3) (4,4) (4,5) (4,6) (5,2) (5,3) (5, 4) (5, 5) (5,6) (6,1) (6,2) (6,3) (6,4) (6, 5) (6, 6) } B = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5) (4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6) } (b) P (A or B) = P (A) + P (B) P (A and B) 51
(c) P (A B) = (d) P (B C) = Example 2 : At a local Country Club, 65% of the members play bridge and swim, and 72% play bridge. If a member is selected at random, find the probability that the member swims, given that the member plays bridge. 52
Example 3 : Eighty students in a school cafeteria were asked if they favored a ban on smoking in the cafeteria. The results of the survey are shown in the table. Class Favor Oppose No opinion Freshman 15 27 8 Sophomore 23 5 2 If a student is selected at random, find these probabilities. (a) The student is a freshman or favors the ban. P (Freshman or Favor ban) = P (Freshman) + P (Favor ban) P (Freshman and Favor ban) (b) Given that the student favors the ban, the student is a sophomore. 53
Section 4.4 Counting Rules I. The Fundamental Counting Rule In a sequence of n events in which the first one has k 1 possibilities and the second event has k 2 and the third has k 3, and so forth, the total number of possibilities of the sequence will be: k 1 k 2 k 3 k n Note: And in this case means to multiply. 54
Example 1 : Two dice are tossed. How many outcomes are in S. How many outcomes for Task 1? How many outcome for Task 2? 6 6 = 36 outcomes Example 2 : A password consists of two letters followed by one digit. How many different passwords can be created? (Note: Repetitions are allowed) 1st letter 26 outcomes (A Z) 2 nd letter 26 outcomes (A Z) One digit 10 outcomes (0-9) 26 26 10 = 6,760 ways 55
Example 3 : Suppose four digits are to be randomly selected (with repetitions allowed). (Note: the set of digits is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.) If the first digit must be a 2 and repetitions are permitted, how many different 4-digit can be made? 1 st digit has to be a 2 1 outcome 2 nd digit any number (0 9) 10 outcomes 3 rd digit any number (0 9) 10 outcomes 4 th digit any number (0 9) 10 outcomes 1 10 10 10 = 1,000 ways 56
II. Permutations and Combinations Factorial Notation 5! = 5 4 3 2 1 7! = 7 6 5 4 3 2 1 In general, n! = n (n-1) (n-2) (n-3) 1 0!=1 57
Permutation A permutation is an arrangement of n objects in a specific order. The arrangement of n objects in a specific order using r objects at a time is called a permutation of n objects taking r objects at a time. Note: Order does matter. It is written as n P r, and the formula is: n Pr n! ( n r)! 58
Combinations A selection of distinct objects without regard to order is called a combination. (Order does NOT matter!) The number of combinations of r objects selected from n objects is denoted n C r and is given by the formula: n C r n! ( n r)! r! Note: Combinations are always less than permutations for the same n and r. 59
Example 1 : (a) 5! 5 4 3 2 1 = 120 (b) 25P8 (c) 12P4 (d) 25C8 60
Example 2 : A television news director wishes to use three news stories on an evening show. One story will be the lead story, one will be the second story, and the last will be a closing story. If the director has a total of eight stories to choose from, how many possible ways can the program be set up? Example 3 : If a person can select 3 presents from 10 presents, how many different combinations are there? 61
Example 4 : Your family vacation involves a cross-country air flight, a rental car, and a hotel stay in Boston. If you can choose from four major air carriers, five car rental agencies, and three major hotel chains, how many options are available for your vacation accommodations? Multiplication Rule 4 5 3 = 60 ways 62
Section 4.5 Probability and Counting Rules I. Basic Concepts By using the fundamental counting rule, the permutation rules, and the combination rule, the probability of outcomes of many experiments can be computed. 63
Example 1 : A combination lock consists of 26 letters of the alphabet. If a three-letter combination is needed, find the probability that the combination will consist of the first two letters AB in that order. The same letter can be used more than once. 64
Example 2 : Five cards are selected from a 52-card deck for a poker hand. (A poker hand consists of 5 cards dealt in any order.) (a) How many outcomes are in the sample space? (b) A royal flush is a hand that contains that A, K, Q, J, 10, all in the same suit. How many ways are there to get a royal flush? A, K, Q, J, 10 - A, K, Q, J, 10 - A, K, Q, J, 10 - A, K, Q, J, 10 - n(a) = 4 combinations (c) What is the probability of being dealt a royal flush? 65
Example 4-52: Committee Selection A store has 6 TV Graphic magazines and 8 Newstime magazines on the counter. If two customers purchased a magazine, find the probability that one of each magazine was purchased. TV Graphic: One magazine of the 6 magazines Newstime: One magazine of the 8 magazines Total: Two magazines of the 14 magazines C C 6 8 48 C 91 91 6 1 8 1 14 2 Bluman, Chapter 4 66
Summary The three types of probability are classical, empirical, and subjective. Classical probability uses sample spaces. Empirical probability uses frequency distributions and is based on observations. In subjective probability, the researcher makes an educated guess about the chance of an event occurring. 67
Summary (cont) An event consists of one or more outcomes of a probability experiment. Two events are said to be mutually exclusive if they cannot occur at the same time. Events can also be classified as independent or dependent. If events are independent, whether or not the first event occurs does not affect the probability of the next event occurring. 68
Summary (cont.) If the probability of the second event occurring is changed by the occurrence of the first event, then the events are dependent. The complement of an event is the set of outcomes in the sample space that are not included in the outcomes of the event itself. Complementary events are mutually exclusive. 69
Summary The number of ways a sequence of n events can occur; if the first event can occur in k 1 ways, the second event can occur in k 2 ways, etc. (Multiplication rule) k 1 k 2 k 3 k n The arrangement of n objects in a specific order using r objects at a time (Permutation rule) n! n Pr ( n r)! The number of combinations of r objects selected from n objects (order is not important) (Combination rule) n C r n! ( n r)! r! 70
Conclusions Probability can be defined as the chance of an event occurring. It can be used to quantify what the odds are that a specific event will occur. Some examples of how probability is used everyday would be weather forecasting, 75% chance of snow or for setting insurance rates. 71
Conclusions A tree diagram can be used when a list of all possible outcomes is necessary. When only the total number of outcomes is needed, the multiplication rule, the permutation rule, and the combination rule can be used. 72