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Chapter Probability © 2010 Pearson Prentice Hall. All rights reserved 3 5

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Section 5.1 Probability Rules 5-2 © 2010 Pearson Prentice Hall. All rights reserved

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Probability is a measure of the likelihood of a random phenomenon or chance behavior. 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. 5-3 © 2010 Pearson Prentice Hall. All rights reserved

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Probability deals with experiments that yield random short-term results or outcomes, yet reveal long-term predictability. The long-term proportion with which a certain outcome is observed is the probability of that outcome. 5-4 © 2010 Pearson Prentice Hall. All rights reserved

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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. 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. 5-5 © 2010 Pearson Prentice Hall. All rights reserved

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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 even 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, e i. In general, events are denoted using capital letters such as E. 5-6 © 2010 Pearson Prentice Hall. All rights reserved

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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. EXAMPLEIdentifying Events and the Sample Space of a Probability Experiment (a) e 1 = boy, boy, e 2 = boy, girl, e 3 = girl, boy, e 4 = girl, girl (b) {(boy, boy), (boy, girl), (girl, boy), (girl, girl)} (c) {(boy, girl), (girl, boy)} 5-7 © 2010 Pearson Prentice Hall. All rights reserved

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5-8 © 2010 Pearson Prentice Hall. All rights reserved

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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. 5-9 © 2010 Pearson Prentice Hall. All rights reserved

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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. ColorProbability Brown0.12 Yellow0.15 Red0.12 Blue0.23 Orange0.23 Green0.15 All probabilities are between 0 and 1, inclusive. Because 0.12 + 0.15 + 0.12 + 0.23 + 0.23 + 0.15 = 1, rule 2 (the sum of all probabilities must equal 1) is satisfied. 5-10 © 2010 Pearson Prentice Hall. All rights reserved

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If an event is a certainty, the probability of the event is 1. If an event is impossible, the probability of the event is 0. An unusual event is an event that has a low probability of occurring. 5-11 © 2010 Pearson Prentice Hall. All rights reserved

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Copyright © 2010 Pearson Education, Inc. True or false. The following represents a probability model. A. True B. False Cell Phone Provider Probability AT&T0.271 Sprint0.236 T–Mobile0.111 Verizon0.263 Slide 5- 12

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Copyright © 2010 Pearson Education, Inc. True or false. The following represents a probability model. A. True B. False Cell Phone Provider Probability AT&T0.271 Sprint0.236 T–Mobile0.111 Verizon0.263 Slide 5- 13

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5-14 © 2010 Pearson Prentice Hall. All rights reserved

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5-15 © 2010 Pearson Prentice Hall. All rights reserved

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Pass the Pigs TM is a Milton-Bradley 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. (Source: http://www.members.tripod.com/~passpigs/prob.html) EXAMPLE Building a Probability Model OutcomeFrequency Side with no dot1344 Side with dot1294 Razorback767 Trotter365 Snouter137 Leaning Jowler32 (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. (c)Would it be unusual to throw a Leaning Jowler? 5-16 © 2010 Pearson Prentice Hall. All rights reserved

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(a) OutcomeProbability Side with no dot Side with dot0.329 Razorback0.195 Trotter0.093 Snouter0.035 Leaning Jowler0.008 (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. (c) A Leaning Jowler would be unusual. We would expect in 1000 throws of the pig to obtain Leaning Jowler about 8 times. 5-17 © 2010 Pearson Prentice Hall. All rights reserved

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5-18 © 2010 Pearson Prentice Hall. All rights reserved

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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. 5-19 © 2010 Pearson Prentice Hall. All rights reserved

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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. (a)There are a total of 9 + 6 + 7 + 4 + 2 + 2 = 30 candies, so N(S) = 30. (b) P(blue) = 2/30 = 0.067. (c) Since P(yellow) = 6/30 and P(blue) = 2/30, selecting a yellow is three times as likely as selecting a blue. 5-20 © 2010 Pearson Prentice Hall. All rights reserved

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Copyright © 2010 Pearson Education, Inc. A box contains 6 twenty-five watt light bulbs, 9 sixty-watt light bulbs, and 5 hundred-watt light bulbs. What is the probability a randomly selected light bulb is sixty-watts? A. 0.45 B. 0.3 C. 0.05 D. 0.25 Slide 5- 21

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Copyright © 2010 Pearson Education, Inc. A box contains 6 twenty-five watt light bulbs, 9 sixty-watt light bulbs, and 5 hundred-watt light bulbs. What is the probability a randomly selected light bulb is sixty-watts? A. 0.45 B. 0.3 C. 0.05 D. 0.25 Slide 5- 22

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Use the probability applet on your calculator (instructor will show you how) 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. EXAMPLEUsing Simulation 5-24 © 2010 Pearson Prentice Hall. All rights reserved

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5-25 © 2010 Pearson Prentice Hall. All rights reserved

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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. 5-26 © 2010 Pearson Prentice Hall. All rights reserved

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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? EXAMPLEEmpirical, 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. 5-27 © 2010 Pearson Prentice Hall. All rights reserved

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Section 5.2 Probability Rules 5-28 © 2010 Pearson Prentice Hall. All rights reserved

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5-29 © 2010 Pearson Prentice Hall. All rights reserved

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Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events. 5-30 © 2010 Pearson Prentice Hall. All rights reserved

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We often draw pictures of events using Venn diagrams. These pictures represent events as circles enclosed in a rectangle. The rectangle represents the sample space, and each circle represents an event. For example, suppose we randomly select a chip from a bag where each chip in the bag is labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Let E represent the event choose a number less than or equal to 2, and let F represent the event choose a number greater than or equal to 8. These events are disjoint as shown in the figure. 5-31 © 2010 Pearson Prentice Hall. All rights reserved

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5-32 © 2010 Pearson Prentice Hall. All rights reserved

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The probability model to the right shows the distribution of the number of rooms in housing units in the United States. Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 Source: American Community Survey, U.S. Census Bureau EXAMPLE The Addition Rule for Disjoint Events (a) Verify that this is a probability model. All probabilities are between 0 and 1, inclusive. 0.010 + 0.032 + … + 0.080 = 1 5-33 © 2010 Pearson Prentice Hall. All rights reserved

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Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 (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 5-34 © 2010 Pearson Prentice Hall. All rights reserved

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(c) What is the probability a randomly selected housing unit has one or two or three rooms? Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 P(one or two or three) = P(one) + P(two) + P(three) = 0.010 + 0.032 + 0.093 = 0.135 5-35 © 2010 Pearson Prentice Hall. All rights reserved

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Copyright © 2010 Pearson Education, Inc. The data shows the distance employees of a company travel to work. One of these employees is randomly selected. Determine the probability the employee travels between 10 and 29 miles to work. A. 0.401 B. 0.566 C. 0.334 D. 0.735 Slide 5- 36 Distance (miles)Number of employees 0 – 9124 10 – 19309 20 – 29257 30 – 3978 40 – 492

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Copyright © 2010 Pearson Education, Inc. The data shows the distance employees of a company travel to work. One of these employees is randomly selected. Determine the probability the employee travels between 10 and 29 miles to work. A. 0.401 B. 0.566 C. 0.334 D. 0.735 Slide 5- 37 Distance (miles)Number of employees 0 – 9124 10 – 19309 20 – 29257 30 – 3978 40 – 492

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Copyright © 2010 Pearson Education, Inc. The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefers sausage. A. 0.458 B. 0.583 C. 0.125 D. 0.556 Slide 5- 38 CheesePepperoniSausageTotal Male85215 Female2439 Total109524

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Copyright © 2010 Pearson Education, Inc. The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefers sausage. A. 0.458 B. 0.583 C. 0.125 D. 0.556 CheesePepperoniSausageTotal Male85215 Female2439 Total109524 Slide 5- 39

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5-42 © 2010 Pearson Prentice Hall. All rights reserved Suppose that a pair of dice are thrown. Let E = the first die is a two and let F = the sum of the dice is less than or equal to 5. Find P(E or F) using the General Addition Rule. EXAMPLEIllustrating the General Addition Rule

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Complement of an Event Let S denote the sample space of a probability experiment and let E denote an event. The complement of E, denoted E C, is all outcomes in the sample space S that are not outcomes in the event E. 5-45 © 2010 Pearson Prentice Hall. All rights reserved

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Complement Rule If E represents any event and E C represents the complement of E, then P(E C ) = 1 – P(E) 5-46 © 2010 Pearson Prentice Hall. All rights reserved

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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 EXAMPLE Illustrating the Complement Rule 5-47 © 2010 Pearson Prentice Hall. All rights reserved

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The data to the right represent the travel time to work for residents of Hartford County, CT. (a) What is the probability a randomly selected resident has a travel time of 90 or more minutes? EXAMPLE Computing Probabilities Using Complements Source: United States Census Bureau There are a total of 24,358 + 39,112 + … + 4,895 = 393,186 residents in Hartford County, CT. The probability a randomly selected resident will have a commute time of 90 or more minutes is 5-48 © 2010 Pearson Prentice Hall. All rights reserved

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(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 5-49 © 2010 Pearson Prentice Hall. All rights reserved

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Section 5.3 Independence and Multiplication Rule 5-50 © 2010 Pearson Prentice Hall. All rights reserved

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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. 5-52 © 2010 Pearson Prentice Hall. All rights reserved

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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. 5-53 © 2010 Pearson Prentice Hall. All rights reserved

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5-54 © 2010 Pearson Prentice Hall. All rights reserved

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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? EXAMPLE Computing Probabilities of Independent Events The survival of the first female is independent of the survival of the second female. We also have that P(survive) = 0.99186. 5-56 © 2010 Pearson Prentice Hall. All rights reserved

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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? EXAMPLE Computing Probabilities of Independent Events We assume that the defectiveness of the equipment is independent of the use of the equipment. So, 5-57 © 2010 Pearson Prentice Hall. All rights reserved

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5-58 © 2010 Pearson Prentice Hall. All rights reserved

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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? P(all four survive) = P (1 st survives and 2 nd survives and 3 rd survives and 4 th survives) = P(1 st survives). P(2 nd survives). P(3 rd survives). P(4 th survives) = (0.99186) (0.99186) (0.99186) (0.99186) = 0.9678 EXAMPLE Illustrating the Multiplication Principle for Independent Events 5-59 © 2010 Pearson Prentice Hall. All rights reserved

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5-60 © 2010 Pearson Prentice Hall. All rights reserved

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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 EXAMPLE Computing at least Probabilities 5-61 © 2010 Pearson Prentice Hall. All rights reserved

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Copyright © 2010 Pearson Education, Inc. Forty-four percent of college students have engaged in binge drinking. If five college students are randomly selected, what is the probability that at least one of the five has engaged in binge drinking? A. 0.055 B. 0.216 C. 0.945 D. 0.016 Slide 5- 62

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Copyright © 2010 Pearson Education, Inc. Forty-four percent of college students have engaged in binge drinking. If five college students are randomly selected, what is the probability that at least one of the five has engaged in binge drinking? A. 0.055 B. 0.216 C. 0.945 D. 0.016 Slide 5- 63

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Section 5.4 Conditional Probability and the General Multiplication Rule 5-65 © 2010 Pearson Prentice Hall. All rights reserved

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5-66 © 2010 Pearson Prentice Hall. All rights reserved

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Conditional Probability The notation P(F | E) is read the probability of event F given event E. It is the probability of an event F given the occurrence of the event E. 5-67 © 2010 Pearson Prentice Hall. All rights reserved

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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? First roll: S = {1, 2, 3, 4, 5, 6} Second roll: S = {2, 4, 6} 5-68 © 2010 Pearson Prentice Hall. All rights reserved

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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 below. Believe in God Believe in universal spirit Dont believe in either East2043615 Midwest2122913 South219269 West1527626 (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? 5-70 © 2010 Pearson Prentice Hall. All rights reserved

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Believe in God Believe in universal spirit Dont believe in either East2043615 Midwest2122913 South219269 West1527626 (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? 5-71 © 2010 Pearson Prentice Hall. All rights reserved

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In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 1998, 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? EXAMPLE Murder Victims = 0.869 5-72 © 2010 Pearson Prentice Hall. All rights reserved

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Copyright © 2010 Pearson Education, Inc. The table shows the favorite pizza topping for a sample of students. What is the probability that a randomly selected student who was male preferred pepperoni? A. 0.333 B. 0.375 C. 0.6 D. 0.556 Slide 5- 73 CheesePepperoniSausageTotal Male85215 Female2439 Total109524

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Copyright © 2010 Pearson Education, Inc. The table shows the favorite pizza topping for a sample of students. What is the probability that a randomly selected student who was male preferred pepperoni? A. 0.333 B. 0.375 C. 0.6 D. 0.556 Slide 5- 74 CheesePepperoniSausageTotal Male85215 Female2439 Total109524

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Section 5.5 Counting Techniques 5-75 © 2010 Pearson Prentice Hall. All rights reserved

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76 One, two, three, were… Counting

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77 Counting problems are of the following kind: How many different 8-letter passwords are there? How many possible ways are there to pick 11 soccer players out of a 20-player team? Most importantly, counting is the basis for computing probabilities of discrete events. What is the probability of winning the lottery? BASIC COUNTING PRINCIPLES

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78 The sum rule: If a task can be done in n 1 ways and a second task in n 2 ways, and if these two tasks cannot be done at the same time, then there are n 1 + n 2 ways to do either task. BASIC COUNTING PRINCIPLES Example: The department will award a free computer to either a student or a teacher. How many different choices are there, if there are 530 students and 45 teachers? There are 530 + 45 = 575 choices.

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79 Generalized sum rule: If we have tasks T 1, T 2, …, T m that can be done in n 1, n 2, …, n m ways, respectively, and no two of these tasks can be done at the same time, then there are n 1 + n 2 + … + n m ways to do one of these tasks. BASIC COUNTING PRINCIPLES

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80 The product rule: Suppose that a procedure can be broken down into two successive tasks. If there are n 1 ways to do the first task and n 2 ways to do the second task after the first task has been done, then there are n 1 n 2 ways to do the procedure. BASIC COUNTING PRINCIPLES

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81 Example: How many different license plates are there that contain exactly three English letters ? Solution: There are 26 possibilities to pick the first letter, then 26 possibilities for the second one, and 26 for the last one. So there are 26 26 26 = 17576 different license plates. BASIC COUNTING PRINCIPLES

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82 Generalized product rule: If we have a procedure consisting of sequential tasks T 1, T 2, …, T m that can be done in n 1, n 2, …, n m ways, respectively, then there are n 1 n 2 … n m ways to carry out the procedure. BASIC COUNTING PRINCIPLES

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0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 83 Tree Diagrams How many bit strings of length four do NOT have two consecutive 1s? There are 8 strings. 0 1 1 0 0 0 1 1 1 0 0 1 0 1

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84 The pigeonhole principle: If (k + 1) or more objects are placed into k boxes, then there is at least one box containing two or more of the objects. Example 1: If there are 11 players in a soccer team that wins 12-0, there must be at least one player in the team who scored at least twice. Example 2: If you have 6 classes from Monday to Friday, there must be at least one day on which you have at least two classes. THE PIGEONHOLE PRINCIPLE

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85 The generalized pigeonhole principle: If N objects are placed into k boxes, then there is at least one box containing at least N/k of the objects. Example 1: In a 60-student class, at least 12 students will get the same letter grade (A, B, C, D, or F). THE PIGEONHOLE PRINCIPLE

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86 Example 2: Assume you have a drawer containing a random distribution of a dozen brown socks and a dozen black socks. It is dark, so how many socks do you have to pick to be sure that among them there is a matching pair? Solution: There are two types of socks, so if you pick at least 3 socks, there must be either at least two brown socks or at least two black socks. Generalized pigeonhole principle: 3/2 = 2. THE PIGEONHOLE PRINCIPLE

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Copyright © 2010 Pearson Education, Inc. How many 4-letter television call signs are possible, if each sign must start with either a K or a W? A. 35,152 B. 456,976 C. 16 D. 104 Slide 5- 87

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Copyright © 2010 Pearson Education, Inc. How many 4-letter television call signs are possible, if each sign must start with either a K or a W? A. 35,152 B. 456,976 C. 16 D. 104 Slide 5- 88

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Copyright © 2010 Pearson Education, Inc. There are 15 dogs entered in a show. How many ways can first, second, and third place be awarded? A. 45 B. 455 C. 2,730 D. 3,375 Slide 5- 89

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Copyright © 2010 Pearson Education, Inc. There are 15 dogs entered in a show. How many ways can first, second, and third place be awarded? A. 45 B. 455 C. 2,730 D. 3,375 Slide 5- 90

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92 Permutations and Combinations How many ways are there to pick a set of 3 people from a group of 6? The answer to this depends on whether we want the order in which they are picked to matter or not. For example, picking person C, then person A, and then person E leads to the same group as first picking E, then C, and then A. There are 6 choices for the first person, 5 for the second one, and 4 for the third one, so there are 6 5 4 = 120 ways to do this. Since in the original statement, it does not seem that order is important. This is not the correct result! However, these cases are counted separately in the above equation.

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93 So how can we compute how many different subsets of people can be picked (that is, we want to disregard the order of picking) ? To find out about this, we need to first look at permutations. A permutation of a set of distinct objects is an ordered arrangement of these objects. An ordered arrangement of r elements of a set is called an r-permutation. Permutations and Combinations

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94 Example: Let S = {1, 2, 3}. The arrangement 3, 1, 2 is a permutation of S. The arrangement 3, 2 is a 2-permutation of S. The number of r-permutations of a set with n distinct elements is denoted by P(n, r) or nPr. We can calculate P(n, n) with the product rule: P(n, n) = n (n – 1) (n – 2) … 3 2 1. (n choices for the first element, (n – 1) for the second one, (n – 2) for the third one…) Permutations and Combinations

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95 Example: 8 P 3 = (8 7 6 5 4 3 2 1)/(5 4 3 2 1) = 8 7 6 = 336 General formula: P(n, r) = n!/(n – r)! = n P r Knowing this, we can return to our initial question: How many ways are there to pick a set of 3 people from a group of 6 (disregarding the order of picking)? Permutations and Combinations

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A combination is an arrangement, without regard to order, of n distinct objects without repetitions. The symbol n C r represents the number of combinations of n distinct objects taken r at a time, where r < n. 5-96 © 2010 Pearson Prentice Hall. All rights reserved

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97 An r-combination of elements of a set is an unordered selection of r elements from the set. Thus, an r-combination is simply a subset of the set with r elements. Example: Let S = {1, 2, 3, 4}. Then {1, 3, 4} is a 3-combination from S. The number of r-combinations of a set with n distinct elements is denoted by C(n, r) or nCr. Example: C(4, 2) = 6, since, for example, the 2- combinations of a set {1, 2, 3, 4} are {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}. PERMUTATIONS AND COMBINATIONS

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98 How can we calculate C(n, r)? Consider that we can obtain the r-permutation of a set in the following way: First, we form all the r-combinations of the set (there are C(n, r) such r-combinations). Then, we generate all possible orderings in each of these r-combinations (there are P(r, r) such orderings in each case). Therefore, we have: P(n, r) = C(n, r) P(r, r) PERMUTATIONS AND COMBINATIONS

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99 C(n, r) = nCr = P(n, r)/P(r, r) = n!/(n – r)!/(r!/(r – r)!) = n!/(r!(n – r)!) Now we can answer our initial question: How many ways are there to choose a set of 3 people from a group of 6 (disregarding the order of picking)? C(6, 3) = 6!/(3! 3!) = 720/(6 6) = 720/36 = 20 There are 20 different ways, that is, 20 different groups to be picked. PERMUTATIONS AND COMBINATIONS

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100 Corollary: Let n and r be nonnegative integers with r n. Then C(n, r) = C(n, n – r). Note that choosing a group of r people from a group of n people is the same as splitting a group of n people into a group of r people and another group of (n – r) people. PERMUTATIONS AND COMBINATIONS

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101 Example: A soccer club has 8 female and 7 male members. For todays match, the coach wants to have 6 female and 5 male players on the grass. How many possible configurations are there? 8 C 6 7 C 5 = 28 21 = 588 PERMUTATIONS AND COMBINATIONS

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Determine the value of 9 C 3. 5-103 © 2010 Pearson Prentice Hall. All rights reserved

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The United States Senate consists of 100 members. In how many ways can 4 members be randomly selected to attend a luncheon at the White House? 5-104 © 2010 Pearson Prentice Hall. All rights reserved

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Copyright © 2010 Pearson Education, Inc. There are 13 students in a club. How many ways can four students be selected to attend a conference? A. 17,160 B. 52 C. 28,561 D. 715 Slide 5- 105

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Copyright © 2010 Pearson Education, Inc. There are 13 students in a club. How many ways can four students be selected to attend a conference? A. 17,160 B. 52 C. 28,561 D. 715 Slide 5- 106

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Section 5.6 Bayess Rule The material in this section is available on the CD that accompanies the text. 5-107 © 2010 Pearson Prentice Hall. All rights reserved

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EXAMPLE Introduction to the Rule of Total Probability At a university 55% of the students are female and 45% are male 15% of the female students are business majors 20% of the male students are business majors What percent of students, overall, are business majors? The percent of the business majors in the university contributed by females 55% of the students are female 15% of those students are business majors Thus 15% of 55%, or 0.55 0.15 = 0.0825 or 8.25% of the total student body are female business majors Contributed by males In the same way, 20% of 45%, or 0.45 0.20 =.09 or 9% are male business majors 5-110 © 2010 Pearson Prentice Hall. All rights reserved

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Altogether 8.25% of the total student body are female business majors 9% of the total student body are male business majors So … 17.25% of the total student body are business majors EXAMPLE Introduction to the Rule of Total Probability 5-111 © 2010 Pearson Prentice Hall. All rights reserved

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Another way to analyze this problem is to use a tree diagram Female Male 0.55 0.45 EXAMPLE Introduction to the Rule of Total Probability 5-112 © 2010 Pearson Prentice Hall. All rights reserved

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EXAMPLE Introduction to the Rule of Total Probability Female Male 0.55 0.45 Business Not Business Business Not Business 0.550.15 0.550.85 0.450.20 0.450.80 5-113 © 2010 Pearson Prentice Hall. All rights reserved

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5-114 © 2010 Pearson Prentice Hall. All rights reserved Multiply out, and add the two business branches EXAMPLE Introduction to the Rule of Total Probability Business Not Business Business Not Business Female Male 0.55 0.45 0.550.15 0.550.85 0.450.20 0.450.80 0.0825 0.0900 0.4675 0.3600 0.0825 0.0900 Total = 0.1725 0.4675 0.3600

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This is an example of the Rule of Total Probability P(Bus)= 55% 15% + 45% 20% = P(Female) P(Bus | Female) + P(Male) P(Bus | Male) This rule is useful when the sample space can be divided into two (or more) disjoint parts 5-115 © 2010 Pearson Prentice Hall. All rights reserved

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A partition of the sample space S are two non-empty sets A 1 and A 2 that divide up S In other words A 1 Ø A 2 Ø A 1 A 2 = Ø (there is no overlap) A 1 U A 2 = S (they cover all of S) 5-116 © 2010 Pearson Prentice Hall. All rights reserved

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Let E be any event in the sample space S Because A 1 and A 2 are disjoint, E A 1 and E A 2 are also disjoint Because A 1 and A 2 cover all of S, E A 1 and E A 2 cover all of E This means that we have divided E into two disjoint pieces E = (E A 1 ) U (E A 2 ) 5-117 © 2010 Pearson Prentice Hall. All rights reserved

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Because E A 1 and E A 2 are disjoint, we can use the Addition Rule P(E) = P(E A 1 ) + P(E A 2 ) We now use the General Multiplication Rule on each of the P(E A 1 ) and P(E A 2 ) terms P(E) = P(A 1 ) P(E | A 1 ) + P(A 2 ) P(E | A 2 ) 5-118 © 2010 Pearson Prentice Hall. All rights reserved

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P(E) = P(A 1 ) P(E | A 1 ) + P(A 2 ) P(E | A 2 ) This is the Rule of Total Probability (for a partition into two sets A 1 and A 2 ) It is useful when we want to compute a probability (P(E)) but we know only pieces of it (such as P(E | A 1 )) The Rule of Total Probability tells us how to put the probabilities together 5-119 © 2010 Pearson Prentice Hall. All rights reserved

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The general Rule of Total Probability assumes that we have a partition (the general definition) of S into n different subsets A 1, A 2, …, A n Each subset is non-empty None of the subsets overlap S is covered completely by the union of the subsets This is like the partition before, just that S is broken up into many pieces, instead of just two pieces 5-121 © 2010 Pearson Prentice Hall. All rights reserved

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In a particular town 30% of the voters are Republican 30% of the voters are Democrats 40% of the voters are independents This is a partition of the voters into three sets There are no voters that are in two sets (disjoint) All voters are in one of the sets (covers all of S) EXAMPLE The Rule of Total Probability For a particular issue 90% of the Republicans favor it 60% of the Democrats favor it 70% of the independents favor it These are the conditional probabilities E = {favor the issue} The above probabilities are P(E | political party) 5-123 © 2010 Pearson Prentice Hall. All rights reserved

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The total proportion of votes who favor the issue 0.3 0.9 + 0.3 0.6 + 0.4 0.7 = 0.73 So 73% of the voters favor this issue 5-124 © 2010 Pearson Prentice Hall. All rights reserved

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In our male / female and business / non- business majors examples before, we used the rule of total probability to answer the question What percent of students are business majors? We solved this problem by analyzing male students and female students separately 5-126 © 2010 Pearson Prentice Hall. All rights reserved

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We could turn this problem around We were told the percent of female students who are business majors We could also ask What percent of business majors are female? This is the situation for Bayess Rule 5-127 © 2010 Pearson Prentice Hall. All rights reserved

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For this example We first choose a random business student (event E) What is the probability that this student is female? (partition element A 1 ) This question is asking for the value of P(A 1 | E) Before, we were working with P(E | A 1 ) instead The probability (15%) that a female student is a business major 5-128 © 2010 Pearson Prentice Hall. All rights reserved

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The Rule of Total Probability – Know P(A i ) and P(E | A i ) – Solve for P(E) Bayess Rule – Know P(E) and P(E | A i ) – Solve for P(A i | E) 5-129 © 2010 Pearson Prentice Hall. All rights reserved

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Bayes Rule, for a partition into two sets U 1 and U 2, is This rule is very useful when P(U 1 |B) is difficult to compute, but P(B|U 1 ) is easier 5-130 © 2010 Pearson Prentice Hall. All rights reserved

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5-132 © 2010 Pearson Prentice Hall. All rights reserved The business majors example from before Business Not Business Business Not Business Female Male 0.55 0.45 0.550.15 0.550.85 0.450.20 0.450.80 0.0825 0.0900 Total = 0.1725 0.4675 0.3600 EXAMPLE Bayess Rule

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If we chose a random business major, what is the probability that this student is female? A 1 = Female student A 2 = Male student E = business major We want to find P(A 1 | E), the probability that the student is female (A 1 ) given that this is a business major (E) EXAMPLE Bayess Rule 5-133 © 2010 Pearson Prentice Hall. All rights reserved

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Do it in a straight way first We know that 8.25% of the students are female business majors We know that 9% of the students are male business majors Choosing a business major at random is choosing one of the 17.25% The probability that this student is female is 8.25% / 17.25% = 47.83% EXAMPLE Bayess Rule (continued) 5-134 © 2010 Pearson Prentice Hall. All rights reserved

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Now do it using Bayess Rule – its the same calculation Bayess Rule for a partition into two sets (n = 2) P(A 1 ) =.55, P(A 2 ) =.45 P(E | A 1 ) =.15, P(E | A 2 ) =.20 We know all of the numbers we need EXAMPLE Bayess Rule (continued) 5-135 © 2010 Pearson Prentice Hall. All rights reserved

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EXAMPLE Bayess Rule (continued) 5-136 © 2010 Pearson Prentice Hall. All rights reserved

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