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It's easier to figure out tough problems faster using Chegg Study. Freund's Mathematical Statistics with Applications solution manuals or printed answer keys, our experts show you how to solve each problem step-by-step.

No need to wait for office hours or assignments to be graded to find out where you took a wrong turn. You can check your reasoning as you tackle a problem using our interactive solutions viewer. Plus, we regularly update and improve textbook solutions based on student ratings and feedback, so you can be sure you're getting the latest information available. Our interactive player makes it easy to find solutions to John E.

Freund's Mathematical Statistics with Applications problems you're working on - just go to the chapter for your book. Hit a particularly tricky question? Bookmark it to easily review again before an exam. The best part? As a Chegg Study subscriber, you can view available interactive solutions manuals for each of your classes for one low monthly price. For only through such an understanding can appli- cations proceed without the serious mistakes that sometimes occur. The applications are illustrated by means of examples and a separate set of applied exercises, many of them involving the use of computers.

To this end, we have added at the end of the chapter a discussion of how the theory of the chapter can be applied in practice. We begin with a brief review of combinatorial methods and binomial coefficients. In connection with the latter, we often use the following theorem, sometimes called the basic principle of counting, the counting rule for compound events, or the rule for the multiplication of choices.

To justify this theorem, let us define the ordered pair xi , yj to be the outcome that arises when the first step results in possibility xi and the second step results in possibility yj. Find the number of different ways in which this can be done.

If an actual listing of all the possibilities is desirable, a tree diagram like that in Figure 1 provides a systematic approach. It is apparent that the 15 possible ways of taking the vacation are repre- sented by the 15 distinct paths along the branches of the tree. Oh plane a an di In. M ich ig an. Solution The red die can land in any one of six ways, and for each of these six ways the green die can also land in six ways.

Theorem 1 may be extended to cover situations where an operation consists of two or more steps. In this case, we have the following theorem.

EXAMPLE 3 A quality control inspector wishes to select a part for inspection from each of four different bins containing 4, 3, 5, and 4 parts, respectively.

In how many different ways can she choose the four parts? Frequently, we are interested in situations where the outcomes are the different ways in which a group of objects can be ordered or arranged.

For instance, we might want to know in how many different ways the 24 members of a club can elect a presi- dent, a vice president, a treasurer, and a secretary, or we might want to know in how many different ways six persons can be seated around a table. Different arrange- ments like these are called permutations.

A permutation is a distinct arrangement of n differ- ent elements of a set. Solution The possible arrangements are abc, acb, bac, bca, cab, and cba, so the number of distinct permutations is six. Using Theorem 2, we could have arrived at this answer without actually listing the different permutations. Since there are three choices to.

To simplify our notation. The number of permutations of n distinct objects is n!. Solution There are 5! The number of permutations of n distinct objects taken r at a time is n! Introduction select a letter for the first position. Generalizing the argument used in the preceding example. Solution We have two positions to fill. Generalizing the argument that we used in the preceding example. We denote this product by n Pr.

Many statistical software packages provide values of n Pr and other combinatorial quantities upon simple commands.

Permutations that occur when objects are arranged in a circle are called circular permutations. In problems concerning permutations. For example. Two circular permutations are not considered different and are counted only once if corresponding objects in the two arrangements have the same objects to their left and to their right. In how many different ways can this be done?

Solution The number of permutations of 24 distinct objects taken four at a time is 24! EXAMPLE 9 In how many ways can a local chapter of the American Chemical Society schedule three speakers for three different meetings if they are all available on any of five possible dates? Solution Since we must choose three of the five dates and the order in which they are chosen assigned to the three speakers matters. We have been assuming until now that the n objects from which we select r objects and form permutations are all distinct.

Generalizing the argument that we used in the two preceding examples. Solution If we denote the three copies of the first novel by a1. Solution If we arbitrarily consider the position of one of the four players as fixed.

In other words. There are many problems in which we are interested in determining the number of ways in which r objects can be selected from among n distinct objects without regard to the order in which they are selected.

Dividing n Pr by r! If we do not care about the order in which the households are selected. In general. The number of permutations of n objects of which n1 are of one kind. EXAMPLE 14 In how many different ways can a person gathering data for a market research orga- nization select three of the 20 households living in a certain apartment complex?

Solution If we care about the order in which the households are selected. A combination is a selection of r objects taken from n distinct objects without regard to the order of selection. EXAMPLE 16 How many different committees of two chemists and one physicist can be formed from the four chemists and three physicists on the faculty of a small college? A combination of r objects selected from a set of n distinct objects may be con- sidered a partition of the n objects into two subsets containing.

Theorem 1 shows that the 1! This result could also have been obtained by the rather tedious process of enumer- ating the various possibilities. Solution This question is the same as asking for the number of ways in which we can select the two tosses on which heads is to occur.

Had we not wanted to enumerate all the possibilities in the preceding example. Introduction Solution Denoting the four objects by a. The number of ways in which a set of n distinct objects can be partitioned into k subsets with n1 objects in the first subset.

Generalizing this argument. More generally. For instance. Their coefficients are 1. The calculation of binomial coefficients can often be simplified by making use of the three theorems that follow. To prove the theorem algebraically. Applying Theorem When no table is available. In this triangle To state the third theorem about binomial coefficients.. With reference to Exercise 1.

If the first step is made in the ith the constant n2. An operation consists of two steps.

Solution Substituting these values. Use the formula obtained in part the second step in the jth way. Using Theorem 8. Exercises 1. Feller cited among the references at the coefficients in terms of factorials and then simplifying end of this chapter. When n is large. Show that if n2i equals the constant n2 and n3ij equals loaves of bread to three customers. In a two-team basketball play-off.

We might argue the constant n3.

Expressing the binomial coefficients in terms of fac- 10! A derivation of this formula may be found in the book by W. Find an expression for the number the student can study at most 4 hours for the test on three of ways in which r indistinguishable objects can be dis- consecutive days. In some problems of occupancy theory we are con. In some problems of occupancy theory we are con- different ways. Prove Theorem 11 by expressing all the binomial logarithms.

Use Theorem 12 to show that In how many ways can she choose the three chips for assembly? Solution Using Theorem 6. Find the coefficient of x3 y2 z3 w in the expansion of Exercise 14 and part c of Exercise Use this generalized definition of binomial coefficients by expressing all these multinomial coefficients in terms to evaluate of factorials and simplifying algebraically.

With reference to the generalized definition of bino- mial coefficients in Exercise In how many ways can 4 of the 16 units be selected for inspection? If n is not a positive integer or zero. Repeatedly applying Theorem Find the coefficient of x2 y3 z3 in the expansion of with respect to y.

The following examples illustrate further applications of this theory. Rework Exercise 17 by making use of part a of In a primary election. Suppose that in a baseball World Series in which the these offices? This was the various possible play-off wild-card teams. In how many ways can the judges choose remaining game or games. Miss Belgium. A thermostat will call for heat 0. Draw a tree diagram to show the the third comes up with a different number of points.

Construct a tree diagram to show that if he dates for county attorney. Miss U. Introduction Solution According to Theorem 7. After the fourth flip of the coin. He assigned a virtue to each of these possibilities b Draw a tree diagram showing the various ways he and each sinner had to concentrate for some time on the can go to and from work without taking the same route virtue that corresponded to his cast of the dice.

On August 31 there are five wild-card terms in the American League that can make it to the play-offs.

There are four routes. Construct a tree diagram to show the Argentina. Draw a tree diagram which shows has been a popular pastime for many centuries. The five finalists in the Miss Universe contest are Miss three games to two. Counting the number of outcomes in games of chance only two will win spots.. If the NCAA has applications from six universities b the winner.

Miss Japan. In how many ways can each question be marked How many of these begin and end with the letter s? If they are all sold many different ways can three order chicken.

In how many ways can eight persons form a circle for home games of the Chicago Cubs. The price of a European tour includes four stopovers a with any two of the seven nominees. On a Friday morning. Introduction In how many ways can five persons line up to get on artists. If eight persons are having dinner together. How many permutations are there of the letters in the word A college team plays 10 football games during a sea.

In how many dif. A baseball fan has a pair of tickets for six different A shipment of 10 television sets includes three that b if the order of the stopovers does not matter?

In how many can choose 2 of 15 warehouses to ship a large order. In how many ways can she choose two of the b not get the one that is defective? How many distinct permutations are there of the let.

A carton of 15 light bulbs contains one that is defec. In how many differ. If he has five friends a folk dance? Rework Exercise 56 given that at least two of the cans true or false so that of tennis balls were sold on each of the three days.

Jones has four skirts. A multiple-choice test consists of 15 questions. An art collector. In how many different ways a bus? In how many ways can they line up if two of the can she leave these paintings to her three heirs? In Example 4 we showed that a true—false test consist. Determine the number of ways in which a distributor city council are three men and four women. In how many ways can the television director of Exer.

In how many ways can it end the season with five one pie? With reference to Exercise In how many ways can an inspector choose 3 of the sets? At the end of the day. Among the seven nominees for two vacancies on a How many different bridge hands are possible con- time slots allocated to commercials during a two-hour taining five spades. Find the number of ways in which one A.

In how many ways can a hotel download Upper Saddle History of Statistics and Probability. Princeton University Press. Studies in the and Whitworth. Prentice Hall. Choice and Chance. Applied Combinatorics Applied Combinatorial Mathe- and matics. Introduction References Among the books on the history of statistics there are Eisen. The History of Statistics. New York: Hafner Publishing Co.

Contributions to the History of Statis. Macmillan Publishing Co. Random and the more recent publications House. Harvard University Press. New Walker Elementary Combinatorial Analysis. New History of Statistics and Probability.

An Introduction to Combinatorial Analysis. More advanced treatments may be found in Beckenbach Combinatorial Mass. Eighth row: Studies in the Roberts. Studies in the History of Statistical York: Gordon and Breach. The Rise of Statistical Thinking.. Mathematics of Choice. Science Publishers. A wealth of material on combinatorial methods can be and found in Riordan. It is assumed. Marylees Miller. Eighth Edition. We can then say that if there are N equally likely possibili- ties. A major shortcoming of the classical probability concept is its limited applica- bility.

This would be the case. If we say that the probability is 0. Although equally likely possibilities are found mostly in games of chance. Irwin Miller. Among the various probability concepts.

The results one obtains from an experi- ment. Each outcome in a sample space is called an element of the sample space. We cannot guarantee what will happen on any particular occasion—the car may start and then it may not—but if we kept records over a long period of time.

If a sample space has a finite number of elements. The set of all possible outcomes of an experiment is called the sample space and it is usually denoted by the letter S. The approach to probability that we shall use in this chapter is the axiomatic approach. It is customary in statistics to refer to any process of observation or measure- ment as an experiment. Sample spaces with a large or infinite number of elements are best described by a statement or rule.

If an experiment consists of one roll of a die and we are interested in which face is turned up. Probability predicts that there is a 30 percent chance for rain that is. In this sense. The different colors are used to emphasize that the dice are distinct from one another. The outcomes of some experiments are neither finite nor countably infinite.

Such is the case. If a sample space contains a finite number of elements or an infinite though countable number of elements.

If we assume that distance is a variable that can be measured to any desired degree of accuracy. But even here the number of elements can be matched one-to-one with the whole numbers. A second sample space. Sample spaces are usually classified according to the number of elements that they contain. In the preceding example the sample spaces S1 and S2 contained a finite number of elements.

Probability not be discrete. Rolling a total of 7 with a pair of dice. Solution Among 1. If a sample space consists of a continuum. Continuous sample spaces arise in practice whenever the outcomes of experiments are measure- ments of physical properties. Solution Among the 36 possibilities. Solution If we let 0 and 1 represent a miss and a hit. Such a subset consists of all the elements of the sample space for which the event occurs.

EXAMPLE 6 Construct a sample space for the length of the useful life of a certain electronic component and indicate the subset that represents the event F that the component fails before the end of the sixth year. Sample space for Example 5.

An event is a subset of a sample space. Probability In the same way. This is discussed fur- ther in some of the more advanced texts listed among the references at the end of this chapter. Sample spaces and events. Figure 3. According to our definition.

In many problems of probability we are interested in events that are actually combinations of two or more events. For dis- crete sample spaces. Although the reader must surely be familiar with these terms.

When we are dealing with three events.

Some of the rules that control the formation of unions. Venn diagrams. Diagrams showing special relationships among events. Two events having no elements in com- mon are said to be mutually exclusive. The diagram on the right serves to indicate that A is contained in B. To indicate special relationships among events. Use Venn diagrams to verify the two De Morgan laws: Probability Figure 4.

Use Venn diagrams to verify that 3. When A and B are mutually exclusive.

Venn diagram. Figure 5. If one event occurs. Let us illustrate this in connection with the frequency interpretation. Before we study some of the immediate consequences of the postulates of prob- ability. Since proportions are always positive or zero. As far as the frequency interpretation is concerned. Explain why the following assignments of probabilities are not permissible: The following postulates of probability apply only to discrete sample spaces.

Probability 4 The Probability of an Event To formulate the postulates of probability. Taking the third postulate in the simplest case. The second postulate states indirectly that certainty is identified with a probability of 1. If A is an event in a discrete sample space S.

How this is done in some special situations is illustrated by the following examples. Proof Let O Since we assume that the coin is balanced. This is fortunate. To use this theorem. Instead of listing the probabilities of all possible subsets.

To assign a probability measure to a sample space. Probability Of course. Letting A denote the event that we will get at least one head. Solution Since the probabilities are all positive. Find P G. The probability measure of Example 10 would be appropriate. Since there are 13 ways of selecting the face value for the three of a kind and for each of these there are 12 ways of selecting the face value for the pair.

If an experiment is such that we can assume equal probabilities for all the sample points. Probability Here again we made use of the formula for the sum of the terms of an infinite geo- metric progression. EXAMPLE 11 A five-card poker hand dealt from a deck of 52 playing cards is said to be a full house if it consists of three of a kind and a pair.

Solution The number of ways in which we can be dealt a particular full house. If all the five-card hands are equally likely. If A is the union of n of these mutually exclusive N outcomes. If an experiment can result in any one of N different equally likely outcomes. ON represent the individual outcomes in S. Among them. In connection with the frequency interpretation.

In practice. Probability would not happen in a million years. If A and B are any two events in a sample space S. Proof Since A B. For any two events A and B. Probability A B b a c Figure 6. Venn diagram for proof of Theorem 7. What is the probability that a truck stopped at this roadblock will have faulty brakes as well as badly worn tires? Solution If B is the event that a truck stopped at the roadblock will have faulty brakes and T is the event that it will have badly worn tires.

What is the probability that a family owns either or both kinds of sets?

Solution If A is the event that a family in this metropolitan area owns a color television set and B is the event that it owns a HDTV set. What is the probability that a person visiting his dentist will have at least one of these things done to him?

Solution If C is the event that the person will have his teeth cleaned. F is the event that he will have a cavity filled. Prove by induction that Referring to Figure 6. Probability Exercises 5. Use the Venn diagram of Figure 7 with the prob. Use parts a and b of Exercise 3 to show that With reference to the Venn diagram of Figure 7.

Use the formula of Theorem 7 to show that a. C b Postulate 2. Use the Venn diagram of Figure 7 and the method by are A to B that an event will occur. The odds that an event will occur are given by the Venn diagram for Exercises Odds are usually quoted in terms of positive integers having no common factor. Figure 7. Subjective probabilities may be determined by expos- A B ing persons to risk-taking situations and finding the odds at which they would consider it fair to bet on the outcome.

Show that if the odds The odds are then converted into probabilities by means b g e of the formula of Exercise See also Exercise Show that if subjective probabilities are determined in this way. Since the choice of the sample space that is. It is also preferable when we want to refer to several sample spaces in the same example. One of them might apply to all those who are engaged in the private practice of law.

G is the event that a person is a law school graduate. Good service Poor service under warranty under warranty In business 10 years or more 16 4 In business less than 10 years 10 20 If a person randomly selects one of these new-car dealers. L is the event that a person is licensed to practice law.

Some ideas connected with conditional probabilities are illustrated in the fol- lowing example. If we let G denote the selection of a dealer who provides good service under warranty. EXAMPLE 15 A consumer research organization has studied the services under warranty provided by the 50 new-car dealers in a certain city. Probability 6 Conditional Probability Difficulties can easily arise when probabilities are quoted without specification of the sample space.

Generalizing from the preceding.

P G T is considerably higher than P G. Of these. Probability For the second question. This answers the second question and. What is the probability that such an order will be delivered on time given that it was ready for shipment on time? Solution If we let R stand for the event that an order is ready for shipment on time and D be the event that it is delivered on time.

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