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

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

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

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

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

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

A probability model lists the possible outcomes of a probability experiment and each outcome’s probability. A probability model must satisfy rules 1 and 2 of the rules of probabilities. 5-9 © 2010 Pearson Prentice Hall. All rights reserved

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

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

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

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

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. 5-17 © 2010 Pearson Prentice Hall. All rights reserved

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

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. EXAMPLEUsing Simulation 5-20 © 2010 Pearson Prentice Hall. All rights reserved

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. 5-22 © 2010 Pearson Prentice Hall. All rights reserved

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 people’s feelings about which horse will win the race. The probability is not based on a probability experiment or counting equally likely outcomes. 5-23 © 2010 Pearson Prentice Hall. All rights reserved

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

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

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 5-30 © 2010 Pearson Prentice Hall. All rights reserved

(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 5-31 © 2010 Pearson Prentice Hall. All rights reserved

5-34 © 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

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. 5-37 © 2010 Pearson Prentice Hall. All rights reserved

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 5-39 © 2010 Pearson Prentice Hall. All rights reserved

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 5-40 © 2010 Pearson Prentice Hall. All rights reserved

(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 5-41 © 2010 Pearson Prentice Hall. All rights reserved

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. 5-44 © 2010 Pearson Prentice Hall. All rights reserved

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. 5-45 © 2010 Pearson Prentice Hall. All rights reserved

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. 5-48 © 2010 Pearson Prentice Hall. All rights reserved

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

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 5-51 © 2010 Pearson Prentice Hall. All rights reserved

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

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. 5-57 © 2010 Pearson Prentice Hall. All rights reserved

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

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 don’t believe in God, but you do believe in a universal spirit or higher power, or you don’t 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 Don’t 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? 5-60 © 2010 Pearson Prentice Hall. All rights reserved

Believe in God Believe in universal spirit Don’t 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? 5-61 © 2010 Pearson Prentice Hall. All rights reserved

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 5-62 © 2010 Pearson Prentice Hall. All rights reserved

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. 5-65 © 2010 Pearson Prentice Hall. All rights reserved

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 5-72 © 2010 Pearson Prentice Hall. All rights reserved

●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 5-73 © 2010 Pearson Prentice Hall. All rights reserved

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 5-75 © 2010 Pearson Prentice Hall. All rights reserved

5-76 © 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

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 5-77 © 2010 Pearson Prentice Hall. All rights reserved

●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) 5-78 © 2010 Pearson Prentice Hall. All rights reserved

●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 ) 5-79 © 2010 Pearson Prentice Hall. All rights reserved

●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 ) 5-80 © 2010 Pearson Prentice Hall. All rights reserved

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 5-81 © 2010 Pearson Prentice Hall. All rights reserved

●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 5-83 © 2010 Pearson Prentice Hall. All rights reserved

●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) 5-85 © 2010 Pearson Prentice Hall. All rights reserved

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 5-88 © 2010 Pearson Prentice Hall. All rights reserved

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 Bayes’s Rule 5-89 © 2010 Pearson Prentice Hall. All rights reserved

●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 5-90 © 2010 Pearson Prentice Hall. All rights reserved

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 5-92 © 2010 Pearson Prentice Hall. All rights reserved

5-94 © 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 Bayes’s Rule

●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 Bayes’s Rule 5-95 © 2010 Pearson Prentice Hall. All rights reserved

●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 Bayes’s Rule (continued) 5-96 © 2010 Pearson Prentice Hall. All rights reserved

●Now do it using Bayes’s Rule – it’s the same calculation ●Bayes’s 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 Bayes’s Rule (continued) 5-97 © 2010 Pearson Prentice Hall. All rights reserved

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

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