# Statistics Informed Decisions Using Data, 5e STATISTICS INFORMED DECISIONS USING DATA Fifth Edition Chapter 5 Probability Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules Learning Objectives 1. Apply the rules of probabilities 2. Compute and interpret probabilities using the empirical method 3. Compute and interpret probabilities using the classical method

4. Use simulation to obtain data based on probabilities 5. Recognize and interpret subjective probabilities Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules Introduction to Probability (1 of 5) Probability is a measure of the likelihood of a random phenomenon or chance behavior occurring. Probability describes the long-term proportion with which a certain outcome will occur in situations with short-term uncertainty. Use the probability applet to simulate flipping a coin 100 times. Plot the proportion of heads against the number of

flips. Repeat the simulation. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules Introduction to Probability (2 of 5) Probability deals with experiments that yield random shortterm results or outcomes, yet reveal long-term predictability. The long-term proportion in which a certain outcome is observed is the probability of that outcome. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules

Introduction to Probability (3 of 5) The Law of Large Numbers As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules Introduction to Probability (4 of 5)

In probability, an experiment is any process that can be repeated in which the results are uncertain. The sample space, S, of a probability experiment is the collection of all possible outcomes. An event is any collection of outcomes from a probability experiment. An event may consist of one outcome or more than one outcome. We will denote events with one outcome, sometimes called simple events, ei. In general, events are denoted using capital letters such as E. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules Introduction to Probability (5 of 5)

EXAMPLE Identifying Events and the Sample Space of a Probability Experiment Consider the probability experiment of having two children. (a) Identify the outcomes of the probability experiment. (b) Determine the sample space. (c) Define the event E = have one boy. (a) e1 = boy, boy, e2 = boy, girl, e3 = girl, boy, e4 = girl, girl (b) S = {(boy, boy), (boy, girl), (girl, boy), (girl, girl)} (c) E = {(boy, girl), (girl, boy)} Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.1 Apply the Rules of Probabilities (1 of 4)

Rules of probabilities 1. The probability of any event E, P(E),must be greater than or equal to 0 and less than or equal to 1. That is, 0 P(E) 1. 2. The sum of the probabilities of all outcomes must equal 1. That is, if the sample space S = {e1, e2, , en}, then P(e1) + P(e2) + + P(en) = 1 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.1 Apply the Rules of Probabilities (2 of 4)

A probability model lists the possible outcomes of a probability experiment and each outcomes probability. A probability model must satisfy rules 1 and 2 of the rules of probabilities. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.1 Apply the Rules of Probabilities EXAMPLE A Probability Model In a bag of peanut M&M milk chocolate candies, the colors of the candies can be brown, yellow, red, blue, orange, or green. Suppose that a candy is randomly selected from a bag. The table shows each color and

the probability of drawing that color. Verify this is a probability model. (3 of 4) Color Brown Yellow Red Blue Orange Green Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved Probability

0.12 0.15 0.12 0.23 0.23 0.15 5.1 Probability Rules 5.1.1 Apply the Rules of Probabilities (4 of 4) If an event is impossible, the probability of the event is 0. If an event is a certainty, the probability of the event is 1. An unusual event is an event that has a low probability of occurring.

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.2 Compute and Interpret Probabilities Using the Empirical Method (1 of 6) Approximating Probabilities Using the Empirical Approach The probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules

5.1.2 Compute and Interpret Probabilities Using the Empirical Method (2 of 6) EXAMPLE Building a Probability Model Pass the PigsTM is a MiltonBradley game in which pigs are used as dice. Points are earned based on the way the pig lands. There are six possible outcomes when one pig is tossed. A class of 52 students rolled pigs 3,939 times. The number of times each outcome occurred is recorded in the table at right. Outcome

Side with no dot Side with dot Razorback Trotter Snouter Leaning Jowler Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved Frequency 1344 1294 767 365 137 32

5.1 Probability Rules 5.1.2 Compute and Interpret Probabilities Using the Empirical Method (3 of 6) EXAMPLE Building a Probability Model (a) Use the results of the experiment to build a probability model for the way the pig lands. (b) Estimate the probability that a thrown pig lands on the side with dot. Outcome

Side with no dot Side with dot Razorback Trotter Snouter Leaning Jowler (c) Would it be unusual to throw a Leaning Jowler? Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved Frequency 1344 1294 767 365

137 32 5.1 Probability Rules 5.1.2 Compute and Interpret Probabilities Using the Empirical Method (4 of 6) Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.2 Compute and Interpret Probabilities Using the Empirical Method (5 of 6) Outcome Side with no dot Side with dot Razorback

Trotter Snouter Leaning Jowler Probability 0.341 0.329 0.195 0.093 0.035 0.008 (b) The probability a throw results in a side with dot is 0.329. In 1000 throws of the pig, we would expect about 329 to land on a side with dot. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

5.1 Probability Rules 5.1.2 Compute and Interpret Probabilities Using the Empirical Method (6 of 6) Outcome Side with no dot Side with dot Razorback Trotter Snouter Leaning Jowler Probability 0.341 0.329 0.195

0.093 0.035 0.008 (c) A Leaning Jowler would be unusual. We would expect in 1000 throws of the pig to obtain Leaning Jowler about 8 times. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.3 Compute and Interpret Probabilities Using the Classical Method (1 of 5) The classical method of computing probabilities requires equally likely outcomes. An experiment is said to have equally likely outcomes when each simple event has the same probability of

occurring. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.3 Compute and Interpret Probabilities Using the Classical Method (2 of 5) Computing Probability Using the Classical Method If an experiment has n equally likely outcomes and if the number of ways that an event E can occur is m, then the probability of E, P(E) is Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.3 Compute and Interpret Probabilities Using the Classical

Method (3 of 5) Computing Probability Using the Classical Method So, if S is the sample space of this experiment, Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.3 Compute and Interpret Probabilities Using the Classical Method (4 of 5) EXAMPLE Computing Probabilities Using the Classical Method Suppose a fun size bag of M&Ms contains 9 brown candies, 6 yellow candies, 7 red candies, 4 orange candies, 2 blue candies, and 2 green candies. Suppose that a candy is randomly selected. (a) What is the probability that it is yellow? (b) What is the probability that it is blue?

(c) Comment on the likelihood of the candy being yellow versus blue. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.3 Compute and Interpret Probabilities Using the Classical Method (5 of 5) EXAMPLE Computing Probabilities Using the Classical Method (a) There are a total of 9 + 6 + 7 + 4 + 2 + 2 = 30 candies, so N(S) = 30. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules

5.1.4 Use Simulation to Obtain Data Based on Probabilities EXAMPLE Using Simulation Use the probability applet to simulate throwing a 6-sided die 100 times. Approximate the probability of rolling a 4. How does this compare to the classical probability? Repeat the exercise for 1000 throws of the die. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.5 Recognize and Interpret Subjective Probabilities (1 of 2) The subjective probability of an outcome is a probability obtained on the basis of personal judgment. For example, an economist predicting there is a 20% chance of recession next year would be a subjective probability.

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.1 Probability Rules 5.1.5 Recognize and Interpret Subjective Probabilities (2 of 2) EXAMPLE Empirical, Classical, or Subjective Probability In his fall 1998 article in Chance Magazine, (A Statistician Reads the Sports Pages, pp. 17-21,) Hal Stern investigated the probabilities that a particular horse will win a race. He reports that these probabilities are based on the amount of money bet on each horse. When a probability is given that a particular horse will win a race, is this empirical, classical, or subjective probability? Subjective because it is based upon peoples feelings about which horse will win the race. The probability is not based on a probability experiment or counting equally likely outcomes. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

5.2 The Addition Rule and Complements Learning Objectives 1. Use the Addition Rule for Disjoint Events 2. Use the General Addition Rule 3. Compute the probability of an event using the Complement Rule Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.2 The Addition Rule and Complements 5.2.1 Use the Addition Rule for Disjoint Events (1 of 8) Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events.

(they are pairwise disjoint), then Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.2 The Addition Rule and Complements 5.2.1 Use the Addition Rule for Disjoint Events (6 of 8) EXAMPLE The Addition Rule for Disjoint Events The probability model to the right shows the distribution of the number of rooms in housing units in the United States. (a) Verify that this is a probability model. All probabilities are between 0

and 1, inclusive. Number of Rooms in Housing Unit One Two Three Four Five Six Seven Eight Nine or more 0.010 + 0.032 + + 0.080 = 1 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

Probability 0.010 0.032 0.093 0.176 0.219 0.189 0.122 0.079 0.080 5.2 The Addition Rule and Complements 5.2.1 Use the Addition Rule for Disjoint Events (7 of 8) EXAMPLE The Addition Rule for Disjoint Events

(b) What is the probability a randomly selected housing unit has two or three rooms? P(two or three) = P(two) + P(three) = 0.032 + 0.093 = 0.125 Number of Rooms in Housing Unit One Two Three Four Five Six

Seven Eight Nine or more Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved Probability 0.010 0.032 0.093 0.176 0.219 0.189 0.122 0.079 0.080

5.2 The Addition Rule and Complements 5.2.1 Use the Addition Rule for Disjoint Events (8 of 8) EXAMPLE The Addition Rule for Disjoint Events (c) What is the probability a randomly selected housing unit has one or two or three rooms? P(one or two or three) = P(one) + P(two) + P(three) = 0.010 + 0.032 + 0.093 = 0.135 Number of Rooms in Housing Unit

One Two Three Four Five Six Seven Eight Nine or more Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved Probability 0.010 0.032 0.093

0.176 0.219 0.189 0.122 0.079 0.080 5.2 The Addition Rule and Complements 5.2.2 Use the General Addition Rule (1 of 3) The General Addition Rule For any two events E and F, Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.2 The Addition Rule and Complements 5.2.2 Use the General Addition Rule (2 of 3)

Let S denote the sample space of a probability experiment and let E denote an event. The complement of E, denoted EC, is all outcomes in the sample space S that are not outcomes in the event E. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.2 The Addition Rule and Complements 5.2.3 Compute the Probability of an Event Using the Complement Rule (2 of 6) Complement Rule If E represents any event and EC represents the complement of E, then P(EC) = 1 P(E) Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

5.2 The Addition Rule and Complements 5.2.3 Compute the Probability of an Event Using the Complement Rule (3 of 6) EXAMPLE Illustrating the Complement Rule According to the American Veterinary Medical Association, 31.6% of American households own a dog. What is the probability that a randomly selected household does not own a dog? P(do not own a dog) = 1 P(own a dog) = 1 0.316 = 0.684 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.2 The Addition Rule and Complements 5.2.3 Compute the Probability of an Event Using the

Complement Rule (4 of 6) The data to the right represent the travel time to work for residents of Hartford County, CT. Travel Time (a) What is the probability a randomly selected resident has a travel time of 90 or more minutes? Frequency Less than 5 minutes

24,358 5 to 9 minutes 39,112 10 to 14 minutes 62,124 15 to 19 minutes 72,854 20 to 24 minutes

74,386 25 to 29 minutes 30,099 30 to 34 minutes 45,043 35 to 39 minutes 11,169 40 to 44 minutes

8,045 45 to 59 minutes 15,650 60 to 89 minutes 5,451 90 or more minutes 4,895 Source: United States Census Bureau Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

5.2 The Addition Rule and Complements 5.2.3 Compute the Probability of an Event Using the Complement Rule (5 of 6) There are a total of 24,358 + 39,112 + + 4,895 = 393,186 residents in Hartford County. Travel Time The probability a randomly selected resident will have a commute time of 90 or more minutes is Frequency

Less than 5 minutes 24,358 5 to 9 minutes 39,112 10 to 14 minutes 62,124 15 to 19 minutes 72,854

20 to 24 minutes 74,386 25 to 29 minutes 30,099 30 to 34 minutes 45,043 35 to 39 minutes 11,169

40 to 44 minutes 8,045 45 to 59 minutes 15,650 60 to 89 minutes 5,451 90 or more minutes 4,895

Source: United States Census Bureau Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.2 The Addition Rule and Complements 5.2.3 Compute the Probability of an Event Using the Complement Rule (6 of 6) (b) Compute the probability that a randomly selected resident of Hartford County, CT will have a commute time less than 90 minutes. P(less than 90 minutes) = 1 P(90 minutes or more) = 1 0.012 = 0.988 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule

Learning Objectives 1. Identify independent events 2. Use the Multiplication Rule for Independent Events 3. Compute at-least probabilities Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.1 Identify independent events (1 of 2) Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F. Two events are dependent if the occurrence of event E in a probability experiment affects the

probability of event F. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.1 Identify independent events (2 of 2) EXAMPLE Independent or Not? (a) Suppose you draw a card from a standard 52-card deck of cards and then roll a die. The events draw a heart and roll an even number are independent because the results of choosing a card do not impact the results of the die toss. (b) Suppose two 40-year old women who live in the United States are randomly selected. The events woman 1 survives the

year and woman 2 survives the year are independent. (c) Suppose two 40-year old women live in the same apartment complex. The events woman 1 survives the year and woman 2 survives the year are dependent. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.2 Use the Multiplication Rule for Independent Events (1 of 6) Multiplication Rule for Independent Events If E and F are independent events, then Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.2 Use the Multiplication Rule for Independent Events (2 of 6)

EXAMPLE Computing Probabilities of Independent Events The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that two randomly selected 60 year old females will survive the year? The survival of the first female is independent of the survival of the second female. We also have that P(survive) = 0.99186. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.2 Use the Multiplication Rule for Independent Events (3 of 6) EXAMPLE Computing Probabilities of Independent Events A manufacturer of exercise equipment knows that 10% of their products are defective. They also know that only 30% of their customers will actually use the equipment in the first year after it is

purchased. If there is a one-year warranty on the equipment, what proportion of the customers will actually make a valid warranty claim? We assume that the defectiveness of the equipment is independent of the use of the equipment. So, Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.2 Use the Multiplication Rule for Independent Events (4 of 6) Multiplication Rule for n Independent Events If E1, E2, E3, and En are independent events, then Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule

5.3.2 Use the Multiplication Rule for Independent Events (5 of 6) EXAMPLE Illustrating the Multiplication Principle for Independent Events The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that four randomly selected 60 year old females will survive the year? Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.2 Use the Multiplication Rule for Independent Events (6 of 6) EXAMPLE Illustrating the Multiplication Principle for Independent Events P(all 4 survive) = P(1st survives and 2nd survives and 3rd survives and 4th survives)

= P(1st survives) P(2nd survives) P(3rd survives) P(4th survives) = (0.99186) (0.99186) (0.99186) (0.99186) = 0.9678 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.3 Compute At-Least Probabilities (1 of 4) EXAMPLE Computing at least Probabilities The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that at least one of 500 randomly selected 60 year old females will die during the course of the year? P(at least one dies) = 1 P(none die) = 1 P(all survive)

= 1 (0.99186)500 = 0.9832 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.3 Compute At-Least Probabilities (2 of 4) Summary: Rules of Probability 1. The probability of any event must be between 0 and 1. If we let E denote any event, then 0 P(E) 1. 2. The sum of the probabilities of all outcomes in the sample space must equal 1. That is, if the sample space S = {e1, e2, , en}, then P(e1) + P(e2) + + P(en) = 1 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

5.3 Independence and the Multiplication Rule 5.3.3 Compute At-Least Probabilities (3 of 4) Summary: Rules of Probability 3. If E and F are disjoint events, then P(E or F) = P(E) + P(F). If E and F are not disjoint events, then P(E or F) = P(E) + P(F) P(E and F). 4. If E represents any event and EC represents the complement of E, then P(EC) = 1 P(E). Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.3 Independence and the Multiplication Rule 5.3.3 Compute At-Least Probabilities (4 of 4) Summary: Rules of Probability

5. If E and F are independent events, then Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule Learning Objectives 1. Compute conditional probabilities 2. Compute probabilities using the General Multiplication Rule Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule 5.4.1 Compute Conditional Probabilities

(1 of 8) Conditional Probability The notation P(F|E) is read the probability of event F given event E. It is the probability that the event F occurs given that event E has occurred. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule 5.4.1 Compute Conditional Probabilities (2 of 8) EXAMPLE An Introduction to Conditional Probability

Suppose that a single six-sided die is rolled. What is the probability that the die comes up 4? Now suppose that the die is rolled a second time, but we are told the outcome will be an even number. What is the probability that the die comes up 4? Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule 5.4.1 Compute Conditional Probabilities (3 of 8) The probability of event F occurring, given the occurrence of event E, is found by dividing the probability of E and F by the probability of E, or by dividing the number of outcomes in E and F by the

number of outcomes in E. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule 5.4.1 Compute Conditional Probabilities (4 of 8) EXAMPLE Conditional Probabilities on Belief about God and Region of the Country A survey was conducted by the Gallup Organization conducted May 8 11, 2008 in which 1,017 adult Americans were asked, Which of the following statements comes closest to your belief about God you believe in God, you dont believe in God, but you

do believe in a universal spirit or higher power, or you dont believe in either? The results of the survey, by region of the country, are given in the table on the next slide. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule 5.4.1 Compute Conditional Probabilities blank East Midwest South West

Believe in God 204 212 219 152 Believe in universal spirit 36 29 26 76 (5 of 8)

Dont believe in either 15 13 9 26 (a) What is the probability that a randomly selected adult American who lives in the East believes in God? (b) What is the probability that a randomly selected adult American who believes in God lives in the East?

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule 5.4.1 Compute Conditional Probabilities (6 of 8) EXAMPLE Conditional Probabilities on Belief about God and Region of the Country blank East Midwest South West

Believe in God 204 212 219 152 Believe in universal spirit 36 29 26 76 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

Dont believe in either 15 13 9 26 5.4 Conditional Probability and the General Multiplication Rule 5.4.1 Compute Conditional Probabilities (7 of 8) EXAMPLE Conditional Probabilities on Belief about God and Region of the Country

blank East Midwest South West Believe in God 204 212 219 152 Believe in universal spirit 36

29 26 76 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved Dont believe in either 15 13 9 26 5.4 Conditional Probability and the General Multiplication Rule 5.4.1 Compute Conditional Probabilities

(8 of 8) EXAMPLE Murder Victims In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 2005, 16.6% of all murder victims were 20 24 year old males. What is the probability that a randomly selected murder victim in 2005 was male given that the victim is 20 24 years old? Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule 5.4.2 Compute Probabilities Using the General Multiplication Rule (1 of 2)

In words, the probability of E and F is the probability of event E occurring times the probability of event F occurring, given the occurrence of event E. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.4 Conditional Probability and the General Multiplication Rule 5.4.2 Compute Probabilities Using the General Multiplication Rule (2 of 2) EXAMPLE Murder Victims In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 2005, 86.9% of murder victims were male given that the victim was 20 24 years old. What is the probability that a randomly selected murder victim in 2005 was a 20 24 year old male?

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques Learning Objectives 1. Solve counting problems using the Multiplication rule 2. Solve counting problems using permutations 3. Solve counting problems using combinations 4. Solve counting problems involving permutations with nondistinct items 5. Compute probabilities involving permutations and combinations Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques

5.5.1 Solve Counting Problems Using the Multiplication Rule (1 of 4) Multiplication Rule of Counting If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.1 Solve Counting Problems Using the Multiplication Rule (2 of 4) EXAMPLE Counting the Number of Possible Meals The fixed-price dinner at Mabenka Restaurant provides the following choices:

Appetizer: soup or salad Entre: baked chicken, broiled beef patty, baby beef liver, or roast beef au jus Dessert: ice cream or cheesecake How many different meals can be ordered? Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.1 Solve Counting Problems Using the Multiplication Rule (3 of 4) EXAMPLE Counting the Number of Possible Meals Ordering such a meal requires three separate decisions. Choose an appetizer. For each choice of appetizer, we have 4 choices of entre, and for each of these 2 4 = 8 parings, there are 2 choices for dessert. A total of

2 4 2 = 16 different meals can be ordered. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.1 Solve Counting Problems Using the Multiplication Rule (4 of 4) If n 0 is an integer, the factorial symbol, n!, is defined as follows: n! = n(n 1) 3 2 1 0! = 1 1! = 1 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

5.5 Counting Techniques 5.5.2 Solve Counting Problems using Permutations (1 of 3) A permutation is an ordered arrangement in which r objects are chosen from n distinct (different) objects so that r n and repetition is not allowed. The symbol nPr represents the number of permutations of r objects selected from n objects. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.2 Solve Counting Problems using Permutations (2 of 3) Number of Permutations of n Distinct Objects Taken r at a Time The number of arrangements of r objects chosen from n objects, in which 1. the n objects are distinct,

2. repetition of objects is not allowed, and 3. order is important, is given by the formula Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.2 Solve Counting Problems using Permutations (3 of 3) EXAMPLE Betting on the Trifecta In how many ways can horses in a 10-horse race finish first, second, and third? The 10 horses are distinct. Once a horse crosses the finish line, that horse will not cross the finish line again, and, in a race, order is important. We have a permutation of 10 objects taken 3 at a time. The top three horses can finish a 10-horse race in

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.3 Solve Counting Problems Using Combinations (1 of 3) A combination is a collection, without regard to order, in which r objects are chosen from n distinct objects with r n without repetition. The symbol nCr represents the number of combinations of n distinct objects taken r at a time. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.3 Solve Counting Problems Using Combinations (2 of 3) Number of Combinations of n Distinct Objects Taken r at a Time The number of different arrangements of r objects chosen from n

objects, in which 1. the n objects are distinct, 2. repetition of objects is not allowed, and 3. order is not important, is given by the formula Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.3 Solve Counting Problems Using Combinations (3 of 3) EXAMPLE Simple Random Samples How many different simple random samples of size 4 can be obtained from a population whose size is 20? The 20 individuals in the population are distinct. In addition, the order in which individuals are selected is unimportant. Thus, the number of simple random samples of size 4 from a population of size 20 is a combination of 20 objects taken 4 at a time.

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.4 Solve Counting Problems Involving Permutations with Nondistinct Items (1 of 2) Permutations with Nondistinct Items The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . . , and nk are of a kth kind is given by Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.4 Solve Counting Problems Involving Permutations with Nondistinct Items (2 of 2)

EXAMPLE Arranging Flags How many different vertical arrangements are there of 10 flags if 5 are white, 3 are blue, and 2 are red? We seek the number of permutations of 10 objects, of which 5 are of one kind (white), 3 are of a second kind (blue), and 2 are of a third kind (red). Using Formula (3), we find that there are Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.5 Compute Probabilities Involving Permutations and Combinations (1 of 3) EXAMPLE Winning the Lottery In the Illinois Lottery, an urn contains balls numbered 1 to 52. From this urn, six balls are randomly chosen without replacement. For a \$1 bet, a player chooses two sets of six numbers. To win, all

six numbers must match those chosen from the urn. The order in which the balls are selected does not matter. What is the probability of winning the lottery? Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.5 Compute Probabilities Involving Permutations and Combinations (2 of 3) EXAMPLE Winning the Lottery The probability of winning is given by the number of ways a ticket could win divided by the size of the sample space. Each ticket has two sets of six numbers, so there are two chances of winning for each ticket. The sample space S is the number of ways that 6 objects can be selected from 52 objects without replacement and without regard to order, so N(S) = 52C6.

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.5 Counting Techniques 5.5.5 Compute Probabilities Involving Permutations and Combinations (3 of 3) EXAMPLE Winning the Lottery Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? Learning Objectives 1. Determine the appropriate probability rule to use 2. Determine the appropriate counting technique to use

5.6.1 Determine the Appropriate Probability Rule to Use (3 of 8) Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? 5.6.1 Determine the Appropriate Probability Rule to Use (4 of 8) EXAMPLE Probability: Which Rule Do I Use? In the game show Deal or No Deal?, a contestant is presented with 26 suitcases that contain amounts ranging from \$0.01 to \$1,000,000. The contestant must pick an initial case that is set aside as the game progresses. The amounts are randomly distributed among the suitcases prior to the game. Consider the following breakdown: Prize

Number of Suitcases \$0.01\$100 8 \$200\$1000 6 \$5,000\$50,000 5 \$100,000\$1,000,000

7 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? 5.6.1 Determine the Appropriate Probability Rule to Use (5 of 8) EXAMPLE Probability: Which Rule Do I Use? The probability of this event is not compound. Decide among the empirical, classical, or subjective approaches. Each prize amount is randomly assigned to one of the 26 suitcases, so the outcomes are equally likely. From the table we see that 7 of the cases contain at least \$100,000. Letting E = worth at least \$100,000, we compute P(E) using the classical approach.

Prize Number of Suitcases \$0.01\$100 8 \$200\$1000 6 \$5,000\$50,000 5

\$100,000\$1,000,000 7 Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? 5.6.1 Determine the Appropriate Probability Rule to Use (6 of 8) The chance the contestant selects a suitcase worth at least \$100,000 is 26.9%. In 100 different games, we would expect about 27 games to result in a contestant choosing a suitcase worth at least \$100,000. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved

5.6 Putting It Together: Which Method Do I Use? 5.6.1 Determine the Appropriate Probability Rule to Use (7 of 8) EXAMPLE Probability: Which Rule Do I Use? According to a Harris poll in January 2008, 14% of adult Americans have one or more tattoos, 50% have pierced ears, and 65% of those with one or more tattoos also have pierced ears. What is the probability that a randomly selected adult American has one or more tattoos and pierced ears? Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use?

5.6.1 Determine the Appropriate Probability Rule to Use (8 of 8) EXAMPLE Probability: Which Rule Do I Use? The probability of a compound event involving AND. Letting E = one or more tattoos and F = ears pierced, we are asked to find P(E and F). The problem statement tells us that P(E) = 0.14, P(F) = 0.50 and P(F|E) = 0.65. Because P(F) P(F|E), the two events are not independent. We can find P(E and F) using the General Multiplication Rule. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? 5.6.2 Determine the Appropriate Counting Technique to Use (1 of 5)

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? 5.6.2 Determine the Appropriate Counting Technique to Use (2 of 5) EXAMPLE Counting: Which Technique Do I Use? The Hazelwood city council consists of 5 men and 4 women. How many different subcommittees can be formed that consist of 3 men and 2 women? Sequence of events to consider: select the men, then select the women. Since the number of choices at each stage is independent of previous choices, we use the Multiplication Rule of Counting to obtain N(subcommittees) = N(ways to pick 3 men) N(ways to pick 2 women)

Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? 5.6.2 Determine the Appropriate Counting Technique to Use (3 of 5) EXAMPLE Counting: Which Technique Do I Use? To select the men, we must consider the number of arrangements of 5 men taken 3 at a time. Since the order of selection does not matter, we use the combination formula. Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? 5.6.2 Determine the Appropriate Counting Technique to Use (4 of 5)

EXAMPLE Counting: Which Technique Do I Use? On February 17, 2008, the Daytona International Speedway hosted the 50th running of the Daytona 500. Touted by many to be the most anticipated event in racing history, the race carried a record purse of almost \$18.7 million. With 43 drivers in the race, in how many different ways could the top four finishers (1st, 2nd, 3rd, and 4th place) occur? Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved 5.6 Putting It Together: Which Method Do I Use? 5.6.2 Determine the Appropriate Counting Technique to Use (5 of 5) EXAMPLE Counting: Which Technique Do I Use?

The number of choices at each stage is independent of previous choices, so we can use the Multiplication Rule of Counting. The number of ways the top four finishers can occur is N(top four) = 43 42 41 40 = 2,961,840 We could also approach this problem as an arrangement of units. Since each race position is distinguishable, order matters in the arrangements. We are arranging the 43 drivers taken 4 at a time, so we are only considering a subset of r = 4 distinct drivers in each arrangement. Using our permutation formula, we get Copyright 2018, 2014, 2011 Pearson Education, Inc. All Rights Reserved