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Suppose someone asked us which is more likely, that the next birthday of a person chosen at random will fall on a Monday or that the birthdays of two persons chosen at random will fall on the same day of the week. Not giving the matter much thought, we may answer quickly that the second alternative is less likely to happen because—there are so many more possibilities. Well then, let us see how many possibilities there are. In the first case, there are seven possibilities, depending on whether the person's next birthday falls on a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday, and of these only Monday represents a "success." Thus, in the first case one-seventh of the outcomes are favorable. In the second case there are, indeed, more possibilities: both persons might have their birthday on a Monday, the first person might have his birthday on a Wednesday and the second person on a Tuesday, the first person might have his birthday on a Friday and the second person on a Sunday, and so on. Carefully preparing a list, we would arrive at the result that there are altogether 49 possibilities, of which seven (Monday and Monday, Tuesday and Tuesday, ..., and Sunday and Sunday) provide a "success." Thus, in the second case seven of the 49 possibilities are favorable, which is again one-seventh, and we may well conclude from this that randomly choosing a person whose next birthday falls on a Monday is as likely as choosing two persons whose birthdays fall on the same day of the week. If the reader feels that this is a foregone conclusion, or that it was obvious from the start, suppose we substitute days of the year for days of the week. Is it as likely to choose a person whose birthday is on a given day of the year, say, March 12th, as it is to choose two persons whose birthdays are on the same day of the year? The answer is "Yes," and the argument is the same as before, but the result may not be quite so obvious.
Our example has shown how counting possibilities can go a long way toward evaluating chances. We must be careful, though, because we tacitly assumed in each case that the possibilities were all equally likely, and if this is not the case, we can run into all sorts of difficulties. Suppose, for instance, that in a baseball World Series (in which the winner is the first team to win four games) the National League champion leads the American League champion 3 games to 2. What are the chances that the American League team will nevertheless win the series? Clearly, there are the following possibilities:
(1)The National League team will win the sixth game and the series.
(2)The American League team will win the sixth game and then the National League team will win the seventh game and the series.
(3)The American League team will win the sixth game and also the seventh game to win the series.
Thus, one of the three possibilities is favorable to the American League team, which is one-third, but can we conclude from this that the chances of the American League team winning the series equal, say, those of rolling a 1 or a 2 with a balanced die (so that two of the six faces are regarded as a "success")? The answer is "No"—even if the two teams are evenly matched—and this should be apparent from the fact that in the third case the American League team has to win two games to win the series, while in the first case the National League team has to win only one. All this goes to show that although counting possibilities may help, it does not necessarily provide the answers; very often, counting possibilities constitutes but the first step of a more detailed analysis.
In connection with the counting of possibilities there are essentially two kinds of problems, the first of which will be discussed in the section that follows, and the second in the section beginning on page 8. The first kind of problem is that of listing everything that can happen in a given situation, and even though this may give the impression of being straightforward, it is often easier said than done. It would be quite a job, for example, to list the millions of five-man committees which can be selected from among a company's 124 employees, and even when the number of possibilities is relatively small, it can be difficult to make sure that none have been left out. The second kind of problem is that of determining how many different things can happen without actually compiling a list, and it is here that we shall learn, for example, how to arrive at the result that 225,150,024 different five-man committees can be selected from among a company's 124 employees.
On page 3 we said that there are 49 ways in which the birthdays of two persons can fall on the various days of the week, pointing out at the time that we could arrive at this result by carefully preparing a list. Perhaps "systematically" would have been a better word than "carefully," for proceeding systematically we find that when the first person's birthday falls on a Monday there are seven possibilities corresponding to the second person's birthday falling on a Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, or Sunday. Similarly, when the first person's birthday falls on a Tuesday there are the same seven possibilities corresponding to the second person's birthday falling on the different days of the week, and this argument applies also when the first person's birthday falls on a Wednesday, Thursday, Friday, Saturday, or Sunday. Thus, there are altogether seven times seven, or 49 possibilities.
In this kind of systematic listing of possibilities it often helps to refer to a diagram like that of Figure 1.1, which is called a tree diagram. This diagram shows that for the first person there are seven possibilities (branches) corresponding to his birthday falling on Monday, Tuesday, ..., or Sunday. Then, for the second person there are seven branches emanating from each of the original branches again corresponding to the seven days of the week, and it can be seen that altogether there are 49 possibilities corresponding to the different paths along the "branches" of the tree. Starting at the top, the first path along the branches represents the case where the two birthdays fall on Monday and Monday, the second path represents the case where they fall on Monday and Tuesday, ..., the eighteenth path represents the case where they fall on Wednesday and Thursday, ..., and the forty-ninth path represents the case where they fall on Sunday and Sunday.
For the other example of the preceding section, the one dealing with the World Series, we obtain the tree diagram of Figure 1.2. First, there are two possibilities corresponding to which team wins the sixth game. The first of these does not figure in the next step since the National League team has already won, but emanating from the second branch there are again two branches corresponding to which team wins the seventh game. Clearly, there are three paths along the branches of this tree, and they represent the three possibilities listed on page 4. Of course, the original result was obtained very easily without the tree diagram, but this is not the case, for example, in Exercise 4 on page 11, where the reader will be asked to verify that there are ten possibilities (sequences of wins and losses for either team) after the National League team leads 2 games to 1 (instead of 3 games to 2).
To give another example in which the listing of possibilities is not straightforward and a tree diagram will help, suppose that a boy slips three pieces of candy in his pocket before going on a hike with his brother and a friend. What we are interested in is the number of different ways in which he can distribute all this "wealth." Clearly, there are many possibilities—he could keep all of the candy for himself, he could give two to his brother and one to the friend, he could keep one for himself and give one each to his brother and the friend, and so forth. Continuing this way, we may be able to complete the list, but unless we are very careful, the chances are that we will omit at least one of the possibilities.
Referring to the tree diagram of Figure 1.3, we find that to begin with there are four possibilities (branches) corresponding to the boy's eating zero, one, two, or three pieces himself. Considering the brother next, we find that there are four branches emanating from the top branch, three from the second branch, two from the third branch, and one from the bottom branch. Evidently, there are again four possibilities (zero, one, two, or three) for the brother when the boy, himself, does not eat any of the candy, but only three possibilities (zero, one, or two) when the boy eats one piece of candy, two possibilities (zero or one) when the boy eats two pieces, and one possibility (zero) when the boy eats all of the candy. So far as the third step is concerned, there is only one possibility in each case, since the friend will get whatever candy, if any, remains. Thus, it can be seen that there are altogether 10 possibilities corresponding to the different paths along the branches of the tree diagram of Figure 1.3—starting at the top, the first path corresponds to the boy's giving all of the candy to the friend, the second path corresponds to his giving one piece to his brother and two to the friend, ..., and the tenth path corresponds to his eating all of the candy himself.
The tree diagram of Figure 1.1 shows that there are seven times seven ways in which the birthdays of two persons can fall on the various days of the week. There are seven possibilities for the birthday of the first person, and for each of these there are seven possibilities for the birthday of the second person. Similarly, Figure 1.4 shows that if a restaurant offers pie, ice cream, cake, pudding, or fruit for dessert, which it serves with coffee, tea, or milk, there are five times three ways in which a person can order a dessert and a drink. These two examples illustrate the following rule, which is basic to most of the work of this chapter:
If a choice consists of two steps, of which one can be made inmways and the other innways, then the whole choice can be made inm·nways.
This expresses what is sometimes called the "multiplication of choices," and to prove it, we have only to draw a tree diagram like those of Figures 1.1 and 1.4. First there are m branches corresponding to the possibilities in the first step, and then there are n branches emanating from each of these branches to represent the possibilities in the other step. This leads to m·n paths along the branches of the tree diagram, and hence m·n possibilities.
Thus, if there are 14 boys in a class and 11 girls, there are 14·11 = 154 ways in which an award can be given to one of the boys and one of the girls; if in a city election there are three candidates for mayor and two candidates for treasurer, there are 3·2 = 6 ways in which a person can vote for one candidate for each office; and if somebody wants to vacation in one of the eight Mountain states and travel by car, bus, train, or plane, he can plan his vacation in 8·4 = 32 different ways.
By using suitable tree diagrams, we can easily generalize the above rule for the "multiplication of choices" so that it applies to choices involving more than two steps. For k steps, where k is a positive integer, we have
If a choice consists ofksteps, of which the first can be made inn1ways, the second inn2ways, ..., and thekth innkways, then the whole choice can be made inn1 ·n2· ... ·nkways.
Thus, we simplify the number of ways in which each step can be made, and if a new-car buyer has the choice of four body styles, three different engines, and twelve colors, we find that he can make his choice in 4·3·12 = 144 different ways. Furthermore, if he has the option of choosing a car with or without airconditioning, with or without an automatic transmission, and with or without power brakes, the total number of ways in which the new-car buyer can make his choice goes up to 4·3·12·2·2·2 = 1,152.
To consider another example, suppose that a test consists of 12 multiple-choice questions, with each permitting a choice of four alternatives. Applying the above rule with n1 = 4, n2 = 4, ..., and nl2 = 4, we find that there are
4·4·4·4·4·4·4·4·4·4·4·4 = 16,777,216
ways in which a student can check off his answers, and in only one of these cases will all his answers be correct. To make matters worse, in
3·3·3·3·3·3·3·3·3·3·3·3· = 531,441
of the 16,777,216 possibilities all of his answers will be wrong.
1. In a union election, Mr. Brown, Mr. Morris, and Mr. Mathews are running for president, while Mr. Adams, Mr. Mason, and Mr. Perkins are running for vice-president. Construct a tree diagram showing the nine possible outcomes of this election, and use it to determine the number of ways in which these union officials can be elected so that
(a) both of their names begin with the letter M;
(b) neither of their names begins with the letter M.
Can we conclude that it is as likely that both of their names will begin with the letter M as it is that neither of their names will begin with the letter M? Explain.
2. A businessman can gp to work along four different routes, Routes A, B, C, or D, of which Route C is one-way, so that he cannot take it on the way home. Draw a tree diagram showing the various ways in which he can go to and from work, and use it to determine in what percentage of the possibilities he takes
(a)Route A to work;
(b)the same route both ways.
Can we conclude that he is just as likely to take Route A to work as he is to take the same route both ways? Explain.
3. In a traffic court, violators are classified according to whether or not they are properly licensed, whether their violations are major or minor, and whether or not they have committed any other violations in the proceeding 12 months.
(a) Construct a tree diagram showing the various ways in which this traffic court classifies violators.
(b) If there are 10 violators in each of the eight categories of part (a) and the judge gives each violator who is not properly licensed a stern lecture, how many of the violators will receive a stern lecture?
(c) If, furthermore, the judge gives a $50 fine to everybody who has committed a major violation and/or another violation in the preceding 12 months, how many of the violators will receive a $50 fine?
(d) How many of the violators will receive a stern lecture from the judge as well as a $50 fine?
4. Draw a tree diagram to show that there would have been ten possibilities in the example on page 4 if the National League team had led the American League team only by 2 games to 1. Since more than half of these possibilities require a seventh game, can we conclude that there is a better than fifty-fifty chance that there will be a seventh game? Explain.
5. With reference to the example on page 7, draw a tree diagram to show that there are altogether 20 ways in which the boy can distribute all, some, or none of the candy among himself, his brother, and the friend. Can we conclude from this that he is more likely to distribute two of the three pieces of candy as he is to distribute one or none? Explain.
6. To cut down on his television viewing, a child is allowed to watch at most three cartoon shows per day. Construct a tree diagram to determine the number of ways in which he can watch at least four cartoon shows on two consecutive days.
7. The pro at a golf course stocks two identical sets of women's clubs, reordering at the end of each day (for delivery early the next morning) if and only if he has sold them both. Construct a tree diagram to show that if he starts on a Monday with two sets of these clubs, there are altogether eight different ways in which he can make sales on the first two days of that week. In how many ways can he make sales on the first three days of that week?
8. A student can study either zero, one, or two hours for a history examination on any given night. Construct a tree diagram to show that there are ten different ways in which he can study altogether six hours for the examination on four consecutive nights.
9. If a college schedules four lecture sections and eight laboratory sections for Freshman Chemistry, in how many different ways can a student sign up for one of each ? How many of these possibilities are left, if a student finds that two of the lecture sections and three of the laboratory sections are already filled?
10. A chain of furniture stores in California has four warehouses, of which two are in the Los Angeles area, and 15 retail outlets, of which six are in the Los Angeles area.
(a) In how many different ways can they ship an item from one of the warehouses to one of the stores?
(b) What percentage of the possibilities of part (a) involve only shipments from warehouses in the Los Angeles area to retail outlets in the Los Angeles area?
(c) What percentage of the possibilities of part (a) involve only shipments from warehouses in the Los Angeles area to retail outlets that are not in the Los Angeles area?
Excerpted from Introduction to Probability by John E. Freund. Copyright © 1973 John E. Freund. Excerpted by permission of Dover Publications, Inc..
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Posted July 7, 2007
This clearly and carefully written text introduces probability at a level suitable for first year university students. It begins with a discussion of enumerative combinatorics 'In how many ways ...?' because probability is classically defined as the ratio of the number of ways an event can occur divided by the total number of possible events. This leads to a discussion of the different approaches to probability, including the classical, frequentist, and subjective approaches. From there, Freund covers expectation and decision making, events, rules of probability, conditional probability, probability distributions (including the binomial, multinomial, hypergeometric, and geometric), Chebyshev's Theorem, and the Law of Large Numbers. Each topic is explained clearly and is copiously illustrated with examples that make the relevance and utility of probability clear. The exercises are tractable. Answers to the odd-numbered ones are provided, making the book suitable for self-study. It is not as good a reference. The book is meant to be read in the order in which it is written. Exercises often refer to preceding exercises or examples from earlier in the text. The text does what it is designed to do. It gives the reader who is willing to work through it an understanding of basic probability.
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Posted June 7, 2013
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