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**A PowerPoint Presentation Package to Accompany**Applied Statistics in Business & Economics, 4th edition David P. Doane and Lori E. Seward Prepared by Lloyd R. Jaisingh**Chapter Contents**5.1 Random Experiments 5.2 Probability 5.3 Rules of Probability 5.4 Independent Events 5.5 Contingency Tables 5.6 Tree Diagrams 5.7 Bayes’ Theorem 5.8 Counting Rules Probability Chapter 5**Chapter Learning Objectives**LO5-1:Describe the sample space of a random variable. LO5-2: Distinguish among the three views of probability. LO5-3: Apply the definitions and rules of probability. LO5-4: Calculate odds from given probabilities. LO5-5: Determine when events are independent. Probability Chapter 5**Chapter Learning Objectives**LO5-6: Apply the concepts of probability to contingency tables. LO5-7: Interpret a tree diagram. LO5-8: Use Bayes’ Theorem to calculate revised probabilities. LO5-9: Apply counting rules to calculate possible event arrangements. Probability Chapter 5**5.1 Random Experiments**LO5-1 Chapter 5 LO5-1: Describe the sample space of a random experiment. Sample Space • A random experiment is an observational process whose results cannot be known in advance. • The set of all outcomes (S) is the sample space for the experiment. • A sample space with a countable number of outcomes is discrete.**5.1 Random Experiments**LO5-1 Chapter 5 Sample Space • For a single roll of a die, the sample space is: • When two dice are rolled, the sample space is the following pairs: 5A-6**5.1 Random Experiments**LO5-1 Chapter 5 Sample Space • If the outcome is a continuous measurement, the sample space cannot be listed but can be described by a rule. • For example, the sample space for the length of a randomly chosen cell phone call would be S = {all X such that X > 0}. • The sample space to describe a randomly chosen student’s GPA would be S = {all X such that 0.00 ≤ X ≤ 4.00}.**5.1 Random Experiments**LO5-1 Chapter 5 Events • An event is any subset of outcomes in the sample space. • A simple event or elementary event, is a single outcome. • A discrete sample space S consists of all the simple events (Ei): S = {E1, E2, …, En}. • For example, Amazon’s website for “Books & Music” • has seven categories that a shopper might choose: • S = {Books, DVD, VHS, Magazines, Newspapers, • Music, Textbooks}.**5.1 Random Experiments**LO5-1 Chapter 5 Events Within this sample space, we could define compound events “electronic media” as A = {Music, DVD, VHS} and “print periodicals” as B = (Newspapers, Magazines}. This can be shown in a Venn diagram.**If P(A) = 1, then the event is certain to occur.**If P(A) = 0, then the event cannot occur. 5.2 Probability LO5-2 Chapter 5 LO5-2:Distinguish among the three views of probability. Definitions • The probability of an event is a number that measures the relative likelihood that the event will occur. • The probability of event A [denoted P(A)] must lie within the interval from 0 to 1: 0 ≤ P(A) ≤ 1**Probability**5.2 Probability LO5-2 Chapter 5 Definitions • In a discrete sample space, the probabilities of all simple events must sum to one: P(S) = P(E1) + P(E2) + … + P(En) = 1 • For example, if on a shopping spree, the following percentages (on the left) were recorded for the four methods of payments. We can compute the equivalent probabilities which sum to 1.**number of defaults**number of loans = 5.2 Probability LO5-2 Chapter 5 Empirical Approach • Use the empirical or relative frequency approach to assign probabilities by counting the frequency (fi) of observed outcomes defined on the experimental sample space. • For example, to estimate the default rate on student loans: P(a student defaults) = f /n**5.2 Probability**LO5-2 Chapter 5 Law of Large Numbers The law of large numbers says that as the number of trials increases, any empirical probability approaches its theoretical limit. • Flip a coin 50 times. We would expect the proportion of heads to be near .50. • However, in a small finite sample, any ratio can be obtained (e.g., 1/3, 7/13, 10/22, 28/50, etc.). • A large n may be needed to get close to .50.**5.2 Probability**LO5-2 Chapter 5 Law of Large Numbers**5.2 Probability**LO5-2 Chapter 5 Classical Approach • A priori refers to the process of assigning probabilities before the event is observed or the experiment is conducted. • A priori probabilities are based on logic, not experience. • When flipping a coin or rolling a pair of dice, we do not actually have to perform an experiment because the nature of the process allows us to envision the entire sample space. • Instead of performing the experiment, we can use deduction to determine the probability of an event. • This is the classical approach to probability.**5.2 Probability**LO5-2 Chapter 5 Classical Approach • For example, the two-dice experiment has 36 equally likely simple events. The P(that the sum of the dots on the two faces equals 7) is • The probability is obtained a priori using the classical approach as shown in this Venn diagram for 2 dice:**5.2 Probability**LO5-2 Chapter 5 Subjective Approach • A subjective probability reflects someone’s informed judgment about the likelihood of an event. • Used when there is no repeatable random experiment. • For example,- What is the probability that a new truck product program will show a return on investment of at least 10 percent?- What is the probability that the price of Ford’s stock will rise within the next 30 days?**Which best exemplifies a subjective probability?**• a. The probability that a female aged 30 will have an accident in a week's car rental at Hertz • b. The probability that a pair of dice will come up 7 in a given throw • c. The probability that the summer Olympic games will be held in Chicago in 2020 • d. The probability that a checked bag on Flight 1872 will weigh more than 40 pounds • Which best exemplifies the classical definition of probability? • a. The probability that a male aged 50 will have an accident in a week's car rental at Alamo • b. The probability that a pair of dice will come up 7 when they are rolled • c. The probability that the winter Olympic games will be held in Europe in 2022 • d. The probability that a checked bag on Flight 1872 will weigh more than 30 pounds**5.3 Rules of Probability**LO5-3 Chapter 5 LO5-3: Apply the definitions and rules of probability. Complement of an Event • The complement of an event A is denoted by A′ and consists of everything in the sample space S except event A. • Since A and A′ together comprise the entire sample space, P(A) + P(A′ ) = 1 or P(A′ ) = 1 – P(A)**5.3 Rules of Probability**LO5-3 Chapter 5 Union of Two Events (Figure 5.5) • The union of two events consists of all outcomes in the sample space S that are contained either in event Aor in event Bor in both (denoted A B or “A or B”). may be read as “or” since one or the other or both events may occur.**5.3 Rules of Probability**LO5-3 Chapter 5 Intersection of Two Events • The intersection of two events A and B(denoted by A B or “A and B”) is the event consisting of all outcomes in the sample space S that are contained in both event A and event B. may be read as “and” since both events occur. This is a joint probability.**A and B**5.3 Rules of Probability LO5-3 Chapter 5 General Law of Addition • The general law of addition states that the probability of the union of two events A and B is: P(A B) = P(A) + P(B) – P(A B) So, you have to subtract P(A B) to avoid overstating the probability. When you add the P(A) and P(B) together, you count the P(A and B) twice. A B**Within a given population, 22 percent of the people are**smokers, 57 percent of the people are males, and 12 percent are males who smoke. If a person is chosen at random from the population, what is the probability that the selected person is either a male or a smoker? Information was collected on those who attended the opening of a new movie. The analysis found that 56 percent of the moviegoers were female, 26 percent were under age 25, and 17 percent were females under the age of 25. Find the probability that a moviegoer is either female or under age 25.**Q and R = 2/52**5.3 Rules of Probability LO5-3 Chapter 5 General Law of Addition • For a standard deck of cards: P(Q) = 4/52 (4 queens in a deck; Q = queen) P(R) = 26/52 (26 red cards in a deck; R = red) P(Q R) = 2/52 (2 red queens in a deck) P(Q R) = P(Q) + P(R) – P(Q R) = 4/52 + 26/52 – 2/52 Q4/52 R26/52 = 28/52 = .5385 or 53.85%**5.3 Rules of Probability**LO5-3 Chapter 5 Mutually Exclusive Events • Events A and B are mutually exclusive (or disjoint) if their intersection is the null set () which contains no elements. If A B = , then P(A B) = 0 Special Law of Addition • In the case of mutually exclusive events, the addition law reduces to: P(A B) = P(A) + P(B)**5.3 Rules of Probability**LO5-3 Chapter 5 Collectively Exhaustive Events • Events are collectively exhaustive if their union is the entire sample space S. • Two mutually exclusive, collectively exhaustive events are dichotomous (or binary) events. For example, a car repair is either covered by the warranty (A) or not (A’). Note: This concept can be extended to more than two events. See the next slide NoWarranty Warranty**5.3 Rules of Probability**LO5-3 Chapter 5 Collectively Exhaustive Events There can be more than two mutually exclusive, collectively exhaustive events, as illustrated below. For example, a Walmart customer can pay by credit card (A), debit card (B), cash (C), or check (D).**5.3 Rules of Probability**LO5-3 Chapter 5 Conditional Probability • The probability of event Agiven that event B has occurred. • Denoted P(A | B). The vertical line “ | ” is read as “given.”**5.3 Rules of Probability**LO5-3 Chapter 5 Conditional Probability • Consider the logic of this formula by looking at the Venn diagram. The sample space is restricted to B, an event that has occurred. A B is the part of B that is also in A. The ratio of the relative size of A B to B is P(A | B).**5.3 Rules of Probability**LO5-3 Chapter 5 Example: High School Dropouts • Of the population aged 16–21 and not in college: • What is the conditional probability that a member of this population is unemployed, given that the person is a high school dropout?**5.3 Rules of Probability**LO5-3 Chapter 5 Example: High School Dropouts • First define U = the event that the person is unemployed D = the event that the person is a high school dropout P(U) = .1350 P(D) = .2905 P(UD) = .0532 • P(U | D) = .1831 > P(U) = .1350 • Therefore, being a high school dropout is related to being unemployed.**5.3 Rules of Probability**LO5-4 Chapter 5 LO5-4: Calculate odds from given probabilities Odds of an Event • The oddsin favor of event A occurring are • The oddsagainst event A occurring are 5A-33**5.3 Rules of Probability**LO5-4 Chapter 5 Odds of an Event • If the odds against event A are quoted as b to a, then the implied probability of event A is: • For example, if a race horse has a 4 to 1 odds against winning, the P(win) is 5A-34**5.4 Independent Events**LO5-5 Chapter 5 LO5-5:Determine when events are independent • Event A is independentof event B if the conditional probability P(A | B) is the same as the marginal probability P(A). • P(U | D) = .1831 > P(U) = .1350, so U and D are not independent. That is, they are dependent. • Another way to check for independence: Multiplication Law If P(A B) = P(A)P(B) then event A is independent of event B since**5.4 Independent Events**LO5-5 Chapter 5 Application of the Multiplication Law (for Independent Events) • The probability of n independent events occurring simultaneously is: P(A1A2... An) = P(A1) P(A2) ... P(An) if the events are independent • To illustrate system reliability, suppose a website has 2 independent file servers. Each server has 99% reliability. What is the total system reliability? Let F1 be the event that server 1 fails F2 be the event that server 2 fails**5.4 Independent Events**LO5-5 Chapter 5 Application of the Multiplication Law (for Independent Events) • Applying the rule of independence: P(F1F2) = P(F1) P(F2)= (.01)(.01) = .0001 • So, the probability that both servers are down is .0001. • The probability that one or both servers is “up” is: 1 - .0001 = .9999 or 99.99%**5.5 Contingency Table**LO5-6 Chapter 5 LO5-6:Apply the concepts of probability to contingency tables. Example: Salary Gains and MBA Tuition • Consider the following cross-tabulation (contingency) table for n = 67 top-tier MBA programs:**5.5 Contingency Table**LO5-6 Chapter 5 Example: Salary Gains and MBA Tuition • Are large salary gains more likely to accrue to graduates of high-tuition MBA programs? • The frequencies indicate that MBA graduates of high-tuition • schools do tend to have large salary gains. • Also, most of the top-tier schools charge high tuition. • More precise interpretations of these data can be made using the concepts of probability.**5.5 Contingency Table**LO5-6 Chapter 5 Marginal Probabilities • The marginal probabilityof a single event is found by dividing a row or column total by the total sample size. • For example, find the marginal • probability of a medium salary gain (P(S2). P(S2) = 33/67 = .4925 • Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain).**5.5 Contingency Table**LO5-6 Chapter 5 Marginal Probabilities • Find the marginal probability of a low tuition P(T1). P(T1) = 16/67 = .2388 • There is a 24% chance that a top-tier school’s MBA tuition is under $40,000.**5.5 Contingency Table**LO5-6 Chapter 5 Joint Probabilities • A joint probability represents the intersection of two events in a cross-tabulation table. • Consider the joint event that the school has low tuition and large salary gains (denoted as P(T1S3)). P(T1S3) = 1/67 = .0149 • There is less than a 2% chance that a top-tier school has both low tuition and large salary gains.**5.5 Contingency Table**LO5-6 Chapter 5 Conditional Probabilities • Find the probability that the salary gains are small (S1) given that the MBA tuition is large (T3). P(T3|S1) = 5/32 = .1563 Independence • (S3) and (T1) are dependent.**Given the contingency table shown here:**a. find P(V). b. find P(V|W). c. find P(V ‘ ). d. find P(W ∩ S). e. find P(A or M).**Given the contingency table shown here**a. find P (A3 ∩ B2). b. find P(A2|B3). c. find P(A1 or B2). d. find P(A1 ∩ A2). e. find the probability that either event A2 or event B2 will occur.**5.5 Contingency Table**LO5-6 Chapter 5 Relative Frequencies • Here are the resulting probabilities (relative frequencies). Each value is divided by 67. For example, P(T1 and S1) = 5/67 P(T2 and S2) = 11/67 P(T3 and S3) = 15/67 P(S1) = 17/67 P(T2) = 19/67**5.6 Tree Diagrams**LO5-7 Chapter 5 LO5-7: Interpret a tree diagram. What is a Tree? • A tree diagramor decision treehelps you visualize all possible outcomes. • Start with a contingency table. • For example, this table gives expense ratios by fund type for 21 bond funds and 23 stock funds. • The tree diagram shows all events along with their marginal, conditional, and joint probabilities.**5.6 Tree Diagrams**LO5-7 Chapter 5 Tree Diagram for Fund Type and Expense Ratios**5.7 Bayes’ Theorem**LO5-8 Chapter 5 LO8:Use Bayes’ Theorem to compute revised probabilities • Thomas Bayes (1702-1761) provided a method (called Bayes Theorem) of revising probabilities to reflect new probabilities. • The prior (marginal) probability of an event B is revised after event A has been considered to yield a posterior (conditional) probability. • Bayes’s formula is: • In some situations P(A) is not given. Therefore, the most useful and common form of Bayes’ Theorem is:**5.7 Bayes’ Theorem**LO5-8 Chapter 5 General Form of Bayes’ Theorem • A generalization of Bayes’s Theorem allows event B to have as many • mutually exclusive and collectively exhaustive categories as we wish • (B1, B2, …, Bn) rather than just two dichotomous categories (B and B').

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