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

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Section 5.1 Probability Rules 4-2 © 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.

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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.

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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.

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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.

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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)}

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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.

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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.

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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.

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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”?

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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.

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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.

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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.

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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 dice. EXAMPLEUsing Simulation

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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.

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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 people’s feelings about which horse will win the race. The probability is not based on a probability experiment or counting equally likely outcomes.

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Section 5.2 Probability Rules

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Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events.

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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.

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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

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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

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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

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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.

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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)

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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

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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

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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

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Section 5.3 Independence and Multiplication Rule

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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.

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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.

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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.

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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,

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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

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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

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Section 5.5 Counting Techniques

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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.

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Determine the value of 9 C 3.

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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?

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Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 5 Section 1 – Slide 1 of 33 Chapter 5 Section 1 Probability Rules.

Sullivan – Fundamentals of Statistics – 2 nd Edition – Chapter 5 Section 1 – Slide 1 of 33 Chapter 5 Section 1 Probability Rules.

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