Probability - Walpole Probability & Statistics for Engineers & Scientists 9th Edition PDF

Chapter 2 Probability 2.1 Sample Space In the study of statistics, we are concerned basically with the presentation and interpretation of chance outcomes that occur in a planned study or scientiﬁc investigation. For example, we may record the number of accidents that occur monthly at the intersection of Driftwood Lane and Royal Oak Drive, hoping to justify the installation of a traﬃc light; we might classify items coming oﬀ an as- sembly line as “defective” or “nondefective”; or we may be interested in the volume of gas released in a chemical reaction when the concentration of an acid is varied. Hence, the statistician is often dealing with either numerical data, representing counts or measurements, or categorical data, which can be classiﬁed according to some criterion. We shall refer to any recording of information, whether it be numerical or categorical, as an observation. Thus, the numbers 2, 0, 1, and 2, representing the number of accidents that occurred for each month from January through April during the past year at the intersection of Driftwood Lane and Royal Oak Drive, constitute a set of observations. Similarly, the categorical data N, D, N, N, and D, representing the items found to be defective or nondefective when ﬁve items are inspected, are recorded as observations. Statisticians use the word experiment to describe any process that generates a set of data. A simple example of a statistical experiment is the tossing of a coin. In this experiment, there are only two possible outcomes, heads or tails. Another experiment might be the launching of a missile and observing of its velocity at speciﬁed times. The opinions of voters concerning a new sales tax can also be considered as observations of an experiment. We are particularly interested in the observations obtained by repeating the experiment several times. In most cases, the outcomes will depend on chance and, therefore, cannot be predicted with certainty. If a chemist runs an analysis several times under the same conditions, he or she will obtain diﬀerent measurements, indicating an element of chance in the experimental procedure. Even when a coin is tossed repeatedly, we cannot be certain that a given toss will result in a head. However, we know the entire set of possibilities for each toss. Given the discussion in Section 1.7, we should deal with the breadth of the term experiment. Three types of statistical studies were reviewed, and several examples were given of each. In each of the three cases, designed experiments, observational studies, and retrospective studies, the end result was a set of data that of course is 35 36 Chapter 2 Probability subject to uncertainty. Though only one of these has the word experiment in its description, the process of generating the data or the process of observing the data is part of an experiment. The corrosion study discussed in Section 1.2 certainly involves an experiment, with measures of corrosion representing the data. The ex- ample given in Section 1.7 in which blood cholesterol and sodium were observed on a group of individuals represented an observational study (as opposed to a designed experiment), and yet the process generated data and the outcome is subject to un- certainty. Thus, it is an experiment. A third example in Section 1.7 represented a retrospective study in which historical data on monthly electric power consump- tion and average monthly ambient temperature were observed. Even though the data may have been in the ﬁles for decades, the process is still referred to as an experiment. Deﬁnition 2.1: The set of all possible outcomes of a statistical experiment is called the sample space and is represented by the symbol S. Each outcome in a sample space is called an element or a member of the sample space, or simply a sample point. If the sample space has a ﬁnite number of elements, we may list the members separated by commas and enclosed in braces. Thus, the sample space S, of possible outcomes when a coin is ﬂipped, may be written S = {H, T }, where H and T correspond to heads and tails, respectively. Example 2.1: Consider the experiment of tossing a die. If we are interested in the number that shows on the top face, the sample space is S1 = {1, 2, 3, 4, 5, 6}. If we are interested only in whether the number is even or odd, the sample space is simply S2 = {even, odd}. Example 2.1 illustrates the fact that more than one sample space can be used to describe the outcomes of an experiment. In this case, S1 provides more information than S2. If we know which element in S1 occurs, we can tell which outcome in S2 occurs; however, a knowledge of what happens in S2 is of little help in determining which element in S1 occurs. In general, it is desirable to use the sample space that gives the most information concerning the outcomes of the experiment. In some experiments, it is helpful to list the elements of the sample space systematically by means of a tree diagram. Example 2.2: An experiment consists of ﬂipping a coin and then ﬂipping it a second time if a head occurs. If a tail occurs on the ﬁrst ﬂip, then a die is tossed once. To list the elements of the sample space providing the most information, we construct the tree diagram of Figure 2.1. The various paths along the branches of the tree give the distinct sample points. Starting with the top left branch and moving to the right along the ﬁrst path, we get the sample point HH, indicating the possibility that heads occurs on two successive ﬂips of the coin. Likewise, the sample point T 3 indicates the possibility that the coin will show a tail followed by a 3 on the toss of the die. By proceeding along all paths, we see that the sample space is S = {HH, HT, T 1, T 2, T 3, T 4, T 5, T 6}. 2.1 Sample Space 37 First Second Sample Outcome Outcome Point H HH H T HT 1 T1 2 T2 3 T3 T 4 T4 5 T5 6 T6 Figure 2.1: Tree diagram for Example 2.2. Many of the concepts in this chapter are best illustrated with examples involving the use of dice and cards. These are particularly important applications to use early in the learning process, to facilitate the ﬂow of these new concepts into scientiﬁc and engineering examples such as the following. Example 2.3: Suppose that three items are selected at random from a manufacturing process. Each item is inspected and classiﬁed defective, D, or nondefective, N. To list the elements of the sample space providing the most information, we construct the tree diagram of Figure 2.2. Now, the various paths along the branches of the tree give the distinct sample points. Starting with the ﬁrst path, we get the sample point DDD, indicating the possibility that all three items inspected are defective. As we proceed along the other paths, we see that the sample space is S = {DDD, DDN, DN D, DN N, N DD, N DN, N N D, N N N }. Sample spaces with a large or inﬁnite number of sample points are best de- scribed by a statement or rule method. For example, if the possible outcomes of an experiment are the set of cities in the world with a population over 1 million, our sample space is written S = {x | x is a city with a population over 1 million}, which reads “S is the set of all x such that x is a city with a population over 1 million.” The vertical bar is read “such that.” Similarly, if S is the set of all points (x, y) on the boundary or the interior of a circle of radius 2 with center at the origin, we write the rule S = {(x, y) | x2 + y 2 ≤ 4}. 38 Chapter 2 Probability First Second Third Sample Item Item Item Point D DDD D N DDN D D DND N N DNN D NDD D N NDN N D NND N N NNN Figure 2.2: Tree diagram for Example 2.3. Whether we describe the sample space by the rule method or by listing the elements will depend on the speciﬁc problem at hand. The rule method has practi- cal advantages, particularly for many experiments where listing becomes a tedious chore. Consider the situation of Example 2.3 in which items from a manufacturing process are either D, defective, or N , nondefective. There are many important statistical procedures called sampling plans that determine whether or not a “lot” of items is considered satisfactory. One such plan involves sampling until k defec- tives are observed. Suppose the experiment is to sample items randomly until one defective item is observed. The sample space for this case is S = {D, N D, N N D, N N N D,... }. 2.2 Events For any given experiment, we may be interested in the occurrence of certain events rather than in the occurrence of a speciﬁc element in the sample space. For in- stance, we may be interested in the event A that the outcome when a die is tossed is divisible by 3. This will occur if the outcome is an element of the subset A = {3, 6} of the sample space S1 in Example 2.1. As a further illustration, we may be inter- ested in the event B that the number of defectives is greater than 1 in Example 2.3. This will occur if the outcome is an element of the subset B = {DDN, DN D, N DD, DDD} of the sample space S. To each event we assign a collection of sample points, which constitute a subset of the sample space. That subset represents all of the elements for which the event is true. 2.2 Events 39 Deﬁnition 2.2: An event is a subset of a sample space. Example 2.4: Given the sample space S = {t | t ≥ 0}, where t is the life in years of a certain electronic component, then the event A that the component fails before the end of the ﬁfth year is the subset A = {t | 0 ≤ t < 5}. It is conceivable that an event may be a subset that includes the entire sample space S or a subset of S called the null set and denoted by the symbol φ, which contains no elements at all. For instance, if we let A be the event of detecting a microscopic organism by the naked eye in a biological experiment, then A = φ. Also, if B = {x | x is an even factor of 7}, then B must be the null set, since the only possible factors of 7 are the odd numbers 1 and 7. Consider an experiment where the smoking habits of the employees of a man- ufacturing ﬁrm are recorded. A possible sample space might classify an individual as a nonsmoker, a light smoker, a moderate smoker, or a heavy smoker. Let the subset of smokers be some event. Then all the nonsmokers correspond to a diﬀerent event, also a subset of S, which is called the complement of the set of smokers. Deﬁnition 2.3: The complement of an event A with respect to S is the subset of all elements of S that are not in A. We denote the complement of A by the symbol A. Example 2.5: Let R be the event that a red card is selected from an ordinary deck of 52 playing cards, and let S be the entire deck. Then R is the event that the card selected from the deck is not a red card but a black card. Example 2.6: Consider the sample space S = {book, cell phone, mp3, paper, stationery, laptop}. Let A = {book, stationery, laptop, paper}. Then the complement of A is A = {cell phone, mp3}. We now consider certain operations with events that will result in the formation of new events. These new events will be subsets of the same sample space as the given events. Suppose that A and B are two events associated with an experiment. In other words, A and B are subsets of the same sample space S. For example, in the tossing of a die we might let A be the event that an even number occurs and B the event that a number greater than 3 shows. Then the subsets A = {2, 4, 6} and B = {4, 5, 6} are subsets of the same sample space S = {1, 2, 3, 4, 5, 6}. Note that both A and B will occur on a given toss if the outcome is an element of the subset {4, 6}, which is just the intersection of A and B. Deﬁnition 2.4: The intersection of two events A and B, denoted by the symbol A ∩ B, is the event containing all elements that are common to A and B. Example 2.7: Let E be the event that a person selected at random in a classroom is majoring in engineering, and let F be the event that the person is female. Then E ∩ F is the event of all female engineering students in the classroom. 40 Chapter 2 Probability Example 2.8: Let V = {a, e, i, o, u} and C = {l, r, s, t}; then it follows that V ∩ C = φ. That is, V and C have no elements in common and, therefore, cannot both simultaneously occur. For certain statistical experiments it is by no means unusual to deﬁne two events, A and B, that cannot both occur simultaneously. The events A and B are then said to be mutually exclusive. Stated more formally, we have the following deﬁnition: Deﬁnition 2.5: Two events A and B are mutually exclusive, or disjoint, if A ∩ B = φ, that is, if A and B have no elements in common. Example 2.9: A cable television company oﬀers programs on eight diﬀerent channels, three of which are aﬃliated with ABC, two with NBC, and one with CBS. The other two are an educational channel and the ESPN sports channel. Suppose that a person subscribing to this service turns on a television set without ﬁrst selecting the channel. Let A be the event that the program belongs to the NBC network and B the event that it belongs to the CBS network. Since a television program cannot belong to more than one network, the events A and B have no programs in common. Therefore, the intersection A ∩ B contains no programs, and consequently the events A and B are mutually exclusive. Often one is interested in the occurrence of at least one of two events associated with an experiment. Thus, in the die-tossing experiment, if A = {2, 4, 6} and B = {4, 5, 6}, we might be interested in either A or B occurring or both A and B occurring. Such an event, called the union of A and B, will occur if the outcome is an element of the subset {2, 4, 5, 6}. Deﬁnition 2.6: The union of the two events A and B, denoted by the symbol A ∪ B, is the event containing all the elements that belong to A or B or both. Example 2.10: Let A = {a, b, c} and B = {b, c, d, e}; then A ∪ B = {a, b, c, d, e}. Example 2.11: Let P be the event that an employee selected at random from an oil drilling com- pany smokes cigarettes. Let Q be the event that the employee selected drinks alcoholic beverages. Then the event P ∪ Q is the set of all employees who either drink or smoke or do both. Example 2.12: If M = {x | 3 < x < 9} and N = {y | 5 < y < 12}, then M ∪ N = {z | 3 < z < 12}. The relationship between events and the corresponding sample space can be illustrated graphically by means of Venn diagrams. In a Venn diagram we let the sample space be a rectangle and represent events by circles drawn inside the rectangle. Thus, in Figure 2.3, we see that A ∩ B = regions 1 and 2, B ∩ C = regions 1 and 3, 2.2 Events 41 S A B 2 7 6 1 4 3 5 C Figure 2.3: Events represented by various regions. A ∪ C = regions 1, 2, 3, 4, 5, and 7, B ∩ A = regions 4 and 7, A ∩ B ∩ C = region 1, (A ∪ B) ∩ C = regions 2, 6, and 7, and so forth. S A B C Figure 2.4: Events of the sample space S. In Figure 2.4, we see that events A, B, and C are all subsets of the sample space S. It is also clear that event B is a subset of event A; event B ∩ C has no elements and hence B and C are mutually exclusive; event A ∩ C has at least one element; and event A ∪ B = A. Figure 2.4 might, therefore, depict a situation where we select a card at random from an ordinary deck of 52 playing cards and observe whether the following events occur: A: the card is red, / / 42 Chapter 2 Probability B: the card is the jack, queen, or king of diamonds, C: the card is an ace. Clearly, the event A ∩ C consists of only the two red aces. Several results that follow from the foregoing deﬁnitions, which may easily be veriﬁed by means of Venn diagrams, are as follows: 1. A ∩ φ = φ. 6. φ = S. 2. A ∪ φ = A. 7. (A ) = A. 3. A ∩ A = φ. 4. A ∪ A = S. 8. (A ∩ B) = A ∪ B . 5. S = φ. 9. (A ∪ B) = A ∩ B . Exercises 2.1 List the elements of each of the following sample comes up 3 followed by a head and then a tail on the spaces: coin, construct a tree diagram to show the 18 elements (a) the set of integers between 1 and 50 divisible by 8; of the sample space S. (b) the set S = {x | x2 + 4x − 5 = 0}; 2.6 Two jurors are selected from 4 alternates to serve (c) the set of outcomes when a coin is tossed until a at a murder trial. Using the notation A1 A3 , for exam- tail or three heads appear; ple, to denote the simple event that alternates 1 and 3 (d) the set S = {x | x is a continent}; are selected, list the 6 elements of the sample space S. (e) the set S = {x | 2x − 4 ≥ 0 and x < 1}. 2.7 Four students are selected at random from a chemistry class and classiﬁed as male or female. List 2.2 Use the rule method to describe the sample space the elements of the sample space S1 , using the letter S consisting of all points in the ﬁrst quadrant inside a M for male and F for female. Deﬁne a second sample circle of radius 3 with center at the origin. space S2 where the elements represent the number of females selected. 2.3 Which of the following events are equal? (a) A = {1, 3}; 2.8 For the sample space of Exercise 2.4, (b) B = {x | x is a number on a die}; (a) list the elements corresponding to the event A that (c) C = {x | x2 − 4x + 3 = 0}; the sum is greater than 8; (d) D = {x | x is the number of heads when six coins (b) list the elements corresponding to the event B that are tossed}. a 2 occurs on either die; (c) list the elements corresponding to the event C that a number greater than 4 comes up on the green die; 2.4 An experiment involves tossing a pair of dice, one green and one red, and recording the numbers that (d) list the elements corresponding to the event A ∩ C; come up. If x equals the outcome on the green die (e) list the elements corresponding to the event A ∩ B; and y the outcome on the red die, describe the sample (f) list the elements corresponding to the event B ∩ C; space S (g) construct a Venn diagram to illustrate the intersec- (a) by listing the elements (x, y); tions and unions of the events A, B, and C. (b) by using the rule method. 2.9 For the sample space of Exercise 2.5, 2.5 An experiment consists of tossing a die and then (a) list the elements corresponding to the event A that ﬂipping a coin once if the number on the die is even. If a number less than 3 occurs on the die; the number on the die is odd, the coin is ﬂipped twice. (b) list the elements corresponding to the event B that Using the notation 4H, for example, to denote the out- two tails occur; come that the die comes up 4 and then the coin comes up heads, and 3HT to denote the outcome that the die (c) list the elements corresponding to the event A ; / / Exercises 43 (d) list the elements corresponding to the event A ∩ B; (b) Given that A is the set of nonmedicated subjects (e) list the elements corresponding to the event A ∪ B. and B is the set of walkers, list the elements of A ∪ B. 2.10 An engineering ﬁrm is hired to determine if cer- (c) List the elements of A ∩ B. tain waterways in Virginia are safe for ﬁshing. Samples are taken from three rivers. 2.13 Construct a Venn diagram to illustrate the pos- (a) List the elements of a sample space S, using the sible intersections and unions for the following events letters F for safe to ﬁsh and N for not safe to ﬁsh. relative to the sample space consisting of all automo- biles made in the United States. (b) List the elements of S corresponding to event E that at least two of the rivers are safe for ﬁshing. F : Four door, S : Sun roof, P : Power steering. (c) Deﬁne an event that has as its elements the points {F F F, N F F, F F N, N F N }. 2.14 If S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and A = {0, 2, 4, 6, 8}, B = {1, 3, 5, 7, 9}, C = {2, 3, 4, 5}, and D = {1, 6, 7}, list the elements of the sets correspond- 2.11 The resumés of two male applicants for a college ing to the following events: teaching position in chemistry are placed in the same (a) A ∪ C; ﬁle as the resumés of two female applicants. Two po- sitions become available, and the ﬁrst, at the rank of (b) A ∩ B; assistant professor, is ﬁlled by selecting one of the four (c) C ; applicants at random. The second position, at the rank (d) (C ∩ D) ∪ B; of instructor, is then ﬁlled by selecting at random one (e) (S ∩ C) ; of the remaining three applicants. Using the notation M2 F1 , for example, to denote the simple event that (f) A ∩ C ∩ D. the ﬁrst position is ﬁlled by the second male applicant and the second position is then ﬁlled by the ﬁrst female 2.15 Consider the sample space S = {copper, sodium, applicant, nitrogen, potassium, uranium, oxygen, zinc} and the (a) list the elements of a sample space S; events (b) list the elements of S corresponding to event A that A = {copper, sodium, zinc}, the position of assistant professor is ﬁlled by a male B = {sodium, nitrogen, potassium}, applicant; C = {oxygen}. (c) list the elements of S corresponding to event B that exactly one of the two positions is ﬁlled by a male List the elements of the sets corresponding to the fol- applicant; lowing events: (d) list the elements of S corresponding to event C that (a) A ; neither position is ﬁlled by a male applicant; (b) A ∪ C; (e) list the elements of S corresponding to the event (c) (A ∩ B ) ∪ C ; A ∩ B; (d) B ∩ C ; (f) list the elements of S corresponding to the event (e) A ∩ B ∩ C; A ∪ C; (f) (A ∪ B ) ∩ (A ∩ C). (g) construct a Venn diagram to illustrate the intersec- tions and unions of the events A, B, and C. 2.16 If S = {x | 0 < x < 12}, M = {x | 1 < x < 9}, and N = {x | 0 < x < 5}, ﬁnd 2.12 Exercise and diet are being studied as possi- ble substitutes for medication to lower blood pressure. (a) M ∪ N ; Three groups of subjects will be used to study the ef- (b) M ∩ N ; fect of exercise. Group 1 is sedentary, while group 2 (c) M ∩ N . walks and group 3 swims for 1 hour a day. Half of each of the three exercise groups will be on a salt-free diet. 2.17 Let A, B, and C be events relative to the sam- An additional group of subjects will not exercise or re- ple space S. Using Venn diagrams, shade the areas strict their salt, but will take the standard medication. representing the following events: Use Z for sedentary, W for walker, S for swimmer, Y for salt, N for no salt, M for medication, and F for (a) (A ∩ B) ; medication free. (b) (A ∪ B) ; (a) Show all of the elements of the sample space S. (c) (A ∩ C) ∪ B. 44 Chapter 2 Probability 2.18 Which of the following pairs of events are mutu- (b) region 3; ally exclusive? (c) regions 1 and 2 together; (a) A golfer scoring the lowest 18-hole round in a 72- (d) regions 4 and 7 together; hole tournament and losing the tournament. (e) regions 3, 6, 7, and 8 together. (b) A poker player getting a ﬂush (all cards in the same suit) and 3 of a kind on the same 5-card hand. 2.20 Referring to Exercise 2.19 and the Venn diagram (c) A mother giving birth to a baby girl and a set of of Figure 2.5, list the numbers of the regions that rep- twin daughters on the same day. resent the following events: (d) A chess player losing the last game and winning the (a) The family will experience no mechanical problems match. and will not receive a ticket for a traﬃc violation but will arrive at a campsite with no vacancies. 2.19 Suppose that a family is leaving on a summer (b) The family will experience both mechanical prob- vacation in their camper and that M is the event that lems and trouble in locating a campsite with a va- they will experience mechanical problems, T is the cancy but will not receive a ticket for a traﬃc vio- event that they will receive a ticket for committing a lation. traﬃc violation, and V is the event that they will ar- rive at a campsite with no vacancies. Referring to the (c) The family will either have mechanical trouble or Venn diagram of Figure 2.5, state in words the events arrive at a campsite with no vacancies but will not represented by the following regions: receive a ticket for a traﬃc violation. (a) region 5; (d) The family will not arrive at a campsite with no vacancies. M T 4 5 7 1 2 3 6 8 V Figure 2.5: Venn diagram for Exercises 2.19 and 2.20. 2.3 Counting Sample Points One of the problems that the statistician must consider and attempt to evaluate is the element of chance associated with the occurrence of certain events when an experiment is performed. These problems belong in the ﬁeld of probability, a subject to be introduced in Section 2.4. In many cases, we shall be able to solve a probability problem by counting the number of points in the sample space without actually listing each element. The fundamental principle of counting, often referred to as the multiplication rule, is stated in Rule 2.1. 2.3 Counting Sample Points 45 Rule 2.1: If an operation can be performed in n1 ways, and if for each of these ways a second operation can be performed in n2 ways, then the two operations can be performed together in n1 n2 ways. Example 2.13: How many sample points are there in the sample space when a pair of dice is thrown once? Solution : The ﬁrst die can land face-up in any one of n1 = 6 ways. For each of these 6 ways, the second die can also land face-up in n2 = 6 ways. Therefore, the pair of dice can land in n1 n2 = (6)(6) = 36 possible ways. Example 2.14: A developer of a new subdivision oﬀers prospective home buyers a choice of Tudor, rustic, colonial, and traditional exterior styling in ranch, two-story, and split-level ﬂoor plans. In how many diﬀerent ways can a buyer order one of these homes? Exterior Style Floor Plan h Ranc Two-Story Split -Lev el h Ranc r do Two-Story Tu Split tic -Lev Rus el Col onia h l Ranc Tr Two-Story ad itio Split -Lev na el l h Ranc Two-Story Split -Lev el Figure 2.6: Tree diagram for Example 2.14. Solution : Since n1 = 4 and n2 = 3, a buyer must choose from n1 n2 = (4)(3) = 12 possible homes. The answers to the two preceding examples can be veriﬁed by constructing tree diagrams and counting the various paths along the branches. For instance, 46 Chapter 2 Probability in Example 2.14 there will be n1 = 4 branches corresponding to the diﬀerent exterior styles, and then there will be n2 = 3 branches extending from each of these 4 branches to represent the diﬀerent ﬂoor plans. This tree diagram yields the n1 n2 = 12 choices of homes given by the paths along the branches, as illustrated in Figure 2.6. Example 2.15: If a 22-member club needs to elect a chair and a treasurer, how many diﬀerent ways can these two to be elected? Solution : For the chair position, there are 22 total possibilities. For each of those 22 pos- sibilities, there are 21 possibilities to elect the treasurer. Using the multiplication rule, we obtain n1 × n2 = 22 × 21 = 462 diﬀerent ways. The multiplication rule, Rule 2.1 may be extended to cover any number of operations. Suppose, for instance, that a customer wishes to buy a new cell phone and can choose from n1 = 5 brands, n2 = 5 sets of capability, and n3 = 4 colors. These three classiﬁcations result in n1 n2 n3 = (5)(5)(4) = 100 diﬀerent ways for a customer to order one of these phones. The generalized multiplication rule covering k operations is stated in the following. Rule 2.2: If an operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, and for each of the ﬁrst two a third operation can be performed in n3 ways, and so forth, then the sequence of k operations can be performed in n1 n2 · · · nk ways. Example 2.16: Sam is going to assemble a computer by himself. He has the choice of chips from two brands, a hard drive from four, memory from three, and an accessory bundle from ﬁve local stores. How many diﬀerent ways can Sam order the parts? Solution : Since n1 = 2, n2 = 4, n3 = 3, and n4 = 5, there are nl × n2 × n3 × n4 = 2 × 4 × 3 × 5 = 120 diﬀerent ways to order the parts. Example 2.17: How many even four-digit numbers can be formed from the digits 0, 1, 2, 5, 6, and 9 if each digit can be used only once? Solution : Since the number must be even, we have only n1 = 3 choices for the units position. However, for a four-digit number the thousands position cannot be 0. Hence, we consider the units position in two parts, 0 or not 0. If the units position is 0 (i.e., n1 = 1), we have n2 = 5 choices for the thousands position, n3 = 4 for the hundreds position, and n4 = 3 for the tens position. Therefore, in this case we have a total of n1 n2 n3 n4 = (1)(5)(4)(3) = 60 even four-digit numbers. On the other hand, if the units position is not 0 (i.e., n1 = 2), we have n2 = 4 choices for the thousands position, n3 = 4 for the hundreds position, and n4 = 3 for the tens position. In this situation, there are a total of n1 n2 n3 n4 = (2)(4)(4)(3) = 96 2.3 Counting Sample Points 47 even four-digit numbers. Since the above two cases are mutually exclusive, the total number of even four-digit numbers can be calculated as 60 + 96 = 156. Frequently, we are interested in a sample space that contains as elements all possible orders or arrangements of a group of objects. For example, we may want to know how many diﬀerent arrangements are possible for sitting 6 people around a table, or we may ask how many diﬀerent orders are possible for drawing 2 lottery tickets from a total of 20. The diﬀerent arrangements are called permutations. Deﬁnition 2.7: A permutation is an arrangement of all or part of a set of objects. Consider the three letters a, b, and c. The possible permutations are abc, acb, bac, bca, cab, and cba. Thus, we see that there are 6 distinct arrangements. Using Rule 2.2, we could arrive at the answer 6 without actually listing the diﬀerent orders by the following arguments: There are n1 = 3 choices for the ﬁrst position. No matter which letter is chosen, there are always n2 = 2 choices for the second position. No matter which two letters are chosen for the ﬁrst two positions, there is only n3 = 1 choice for the last position, giving a total of n1 n2 n3 = (3)(2)(1) = 6 permutations by Rule 2.2. In general, n distinct objects can be arranged in n(n − 1)(n − 2) · · · (3)(2)(1) ways. There is a notation for such a number. Deﬁnition 2.8: For any non-negative integer n, n!, called “n factorial,” is deﬁned as n! = n(n − 1) · · · (2)(1), with special case 0! = 1. Using the argument above, we arrive at the following theorem. Theorem 2.1: The number of permutations of n objects is n!. The number of permutations of the four letters a, b, c, and d will be 4! = 24. Now consider the number of permutations that are possible by taking two letters at a time from four. These would be ab, ac, ad, ba, bc, bd, ca, cb, cd, da, db, and dc. Using Rule 2.1 again, we have two positions to ﬁll, with n1 = 4 choices for the ﬁrst and then n2 = 3 choices for the second, for a total of n1 n2 = (4)(3) = 12 permutations. In general, n distinct objects taken r at a time can be arranged in n(n − 1)(n − 2) · · · (n − r + 1) ways. We represent this product by the symbol n! n Pr =. (n − r)! 48 Chapter 2 Probability As a result, we have the theorem that follows. Theorem 2.2: The number of permutations of n distinct objects taken r at a time is n! n Pr =. (n − r)! Example 2.18: In one year, three awards (research, teaching, and service) will be given to a class of 25 graduate students in a statistics department. If each student can receive at most one award, how many possible selections are there? Solution : Since the awards are distinguishable, it is a permutation problem. The total number of sample points is 25! 25! 25 P3 = = = (25)(24)(23) = 13, 800. (25 − 3)! 22! Example 2.19: A president and a treasurer are to be chosen from a student club consisting of 50 people. How many diﬀerent choices of oﬃcers are possible if (a) there are no restrictions; (b) A will serve only if he is president; (c) B and C will serve together or not at all; (d) D and E will not serve together? Solution : (a) The total number of choices of oﬃcers, without any restrictions, is 50! 50 P2 = = (50)(49) = 2450. 48! (b) Since A will serve only if he is president, we have two situations here: (i) A is selected as the president, which yields 49 possible outcomes for the treasurer’s position, or (ii) oﬃcers are selected from the remaining 49 people without A, which has the number of choices 49 P2 = (49)(48) = 2352. Therefore, the total number of choices is 49 + 2352 = 2401. (c) The number of selections when B and C serve together is 2. The number of selections when both B and C are not chosen is 48 P2 = 2256. Therefore, the total number of choices in this situation is 2 + 2256 = 2258. (d) The number of selections when D serves as an oﬃcer but not E is (2)(48) = 96, where 2 is the number of positions D can take and 48 is the number of selections of the other oﬃcer from the remaining people in the club except E. The number of selections when E serves as an oﬃcer but not D is also (2)(48) = 96. The number of selections when both D and E are not chosen is 48 P2 = 2256. Therefore, the total number of choices is (2)(96) + 2256 = 2448. This problem also has another short solution: Since D and E can only serve together in 2 ways, the answer is 2450 − 2 = 2448. 2.3 Counting Sample Points 49 Permutations that occur by arranging objects in a circle are called circular permutations. Two circular permutations are not considered diﬀerent unless corresponding objects in the two arrangements are preceded or followed by a dif- ferent object as we proceed in a clockwise direction. For example, if 4 people are playing bridge, we do not have a new permutation if they all move one position in a clockwise direction. By considering one person in a ﬁxed position and arranging the other three in 3! ways, we ﬁnd that there are 6 distinct arrangements for the bridge game. Theorem 2.3: The number of permutations of n objects arranged in a circle is (n − 1)!. So far we have considered permutations of distinct objects. That is, all the objects were completely diﬀerent or distinguishable. Obviously, if the letters b and c are both equal to x, then the 6 permutations of the letters a, b, and c become axx, axx, xax, xax, xxa, and xxa, of which only 3 are distinct. Therefore, with 3 letters, 2 being the same, we have 3!/2! = 3 distinct permutations. With 4 diﬀerent letters a, b, c, and d, we have 24 distinct permutations. If we let a = b = x and c = d = y, we can list only the following distinct permutations: xxyy, xyxy, yxxy, yyxx, xyyx, and yxyx. Thus, we have 4!/(2! 2!) = 6 distinct permutations. Theorem 2.4: The number of distinct permutations of n things of which n1 are of one kind, n2 of a second kind,... , nk of a kth kind is n!. n1 !n2 ! · · · nk ! Example 2.20: In a college football training session, the defensive coordinator needs to have 10 players standing in a row. Among these 10 players, there are 1 freshman, 2 sopho- mores, 4 juniors, and 3 seniors. How many diﬀerent ways can they be arranged in a row if only their class level will be distinguished? Solution : Directly using Theorem 2.4, we ﬁnd that the total number of arrangements is 10! = 12, 600. 1! 2! 4! 3! Often we are concerned with the number of ways of partitioning a set of n objects into r subsets called cells. A partition has been achieved if the intersection of every possible pair of the r subsets is the empty set φ and if the union of all subsets gives the original set. The order of the elements within a cell is of no importance. Consider the set {a, e, i, o, u}. The possible partitions into two cells in which the ﬁrst cell contains 4 elements and the second cell 1 element are {(a, e, i, o), (u)}, {(a, i, o, u), (e)}, {(e, i, o, u), (a)}, {(a, e, o, u), (i)}, {(a, e, i, u), (o)}. We see that there are 5 ways to partition a set of 4 elements into two subsets, or cells, containing 4 elements in the ﬁrst cell and 1 element in the second. 50 Chapter 2 Probability The number of partitions for this illustration is denoted by the symbol 5 5! = = 5, 4, 1 4! 1! where the top number represents the total number of elements and the bottom numbers represent the number of elements going into each cell. We state this more generally in Theorem 2.5. Theorem 2.5: The number of ways of partitioning a set of n objects into r cells with n1 elements in the ﬁrst cell, n2 elements in the second, and so forth, is n n! = , n1 , n2 ,... , nr n1 !n2 ! · · · nr ! where n1 + n2 + · · · + nr = n. Example 2.21: In how many ways can 7 graduate students be assigned to 1 triple and 2 double hotel rooms during a conference? Solution : The total number of possible partitions would be 7 7! = = 210. 3, 2, 2 3! 2! 2! In many problems, we are interested in the number of ways of selecting r objects from n without regard to order. These selections are called combinations. A combination is actually a partition with two cells, the one cell containing the r objects selected and the other cell containing the (n − r) objects that are left. The number of such combinations, denoted by n n , is usually shortened to , r, n − r r since the number of elements in the second cell must be n − r. Theorem 2.6: The number of combinations of n distinct objects taken r at a time is n n! =. r r!(n − r)! Example 2.22: A young boy asks his mother to get 5 Game-BoyTM cartridges from his collection of 10 arcade and 5 sports games. How many ways are there that his mother can get 3 arcade and 2 sports games? Solution : The number of ways of selecting 3 cartridges from 10 is 10 10! = = 120. 3 3! (10 − 3)! The number of ways of selecting 2 cartridges from 5 is 5 5! = = 10. 2 2! 3! / / Exercises 51 Using the multiplication rule (Rule 2.1) with n1 = 120 and n2 = 10, we have (120)(10) = 1200 ways. Example 2.23: How many diﬀerent letter arrangements can be made from the letters in the word STATISTICS ? Solution : Using the same argument as in the discussion for Theorem 2.6, in this example we can actually apply Theorem 2.5 to obtain 10 10! = = 50, 400. 3, 3, 2, 1, 1 3! 3! 2! 1! 1! Here we have 10 total letters, with 2 letters (S, T ) appearing 3 times each, letter I appearing twice, and letters A and C appearing once each. On the other hand, this result can be directly obtained by using Theorem 2.4. Exercises 2.21 Registrants at a large convention are oﬀered 6 not eat between meals. In how many ways can a person sightseeing tours on each of 3 days. In how many adopt 5 of these rules to follow ways can a person arrange to go on a sightseeing tour (a) if the person presently violates all 7 rules? planned by this convention? (b) if the person never drinks and always eats break- fast? 2.22 In a medical study, patients are classiﬁed in 8 ways according to whether they have blood type AB + , AB − , A+ , A− , B + , B − , O+ , or O− , and also accord- 2.27 A developer of a new subdivision oﬀers a ing to whether their blood pressure is low, normal, or prospective home buyer a choice of 4 designs, 3 diﬀer- high. Find the number of ways in which a patient can ent heating systems, a garage or carport, and a patio or be classiﬁed. screened porch. How many diﬀerent plans are available to this buyer? 2.23 If an experiment consists of throwing a die and then drawing a letter at random from the English 2.28 A drug for the relief of asthma can be purchased alphabet, how many points are there in the sample from 5 diﬀerent manufacturers in liquid, tablet, or space? capsule form, all of which come in regular and extra strength. How many diﬀerent ways can a doctor pre- 2.24 Students at a private liberal arts college are clas- scribe the drug for a patient suﬀering from asthma? siﬁed as being freshmen, sophomores, juniors, or se- niors, and also according to whether they are male or 2.29 In a fuel economy study, each of 3 race cars is female. Find the total number of possible classiﬁca- tested using 5 diﬀerent brands of gasoline at 7 test sites tions for the students of that college. located in diﬀerent regions of the country. If 2 drivers are used in the study, and test runs are made once un- 2.25 A certain brand of shoes comes in 5 diﬀerent der each distinct set of conditions, how many test runs styles, with each style available in 4 distinct colors. If are needed? the store wishes to display pairs of these shoes showing all of its various styles and colors, how many diﬀerent 2.30 In how many diﬀerent ways can a true-false test pairs will the store have on display? consisting of 9 questions be answered? 2.26 A California study concluded that following 7 2.31 A witness to a hit-and-run accident told the po- simple health rules can extend a man’s life by 11 years lice that the license number contained the letters RLH on the average and a woman’s life by 7 years. These followed by 3 digits, the ﬁrst of which was a 5. If 7 rules are as follows: no smoking, get regular exer- the witness cannot recall the last 2 digits, but is cer- cise, use alcohol only in moderation, get 7 to 8 hours tain that all 3 digits are diﬀerent, ﬁnd the maximum of sleep, maintain proper weight, eat breakfast, and do number of automobile registrations that the police may have to check. 52 Chapter 2 Probability 2.32 (a) In how many ways can 6 people be lined up (c) if all the men sit together to the right of all the to get on a bus? women? (b) If 3 speciﬁc persons, among 6, insist on following each other, how many ways are possible? 2.39 In a regional spelling bee, the 8 ﬁnalists consist (c) If 2 speciﬁc persons, among 6, refuse to follow each of 3 boys and 5 girls. Find the number of sample points other, how many ways are possible? in the sample space S for the number of possible orders at the conclusion of the contest for 2.33 If a multiple-choice test consists of 5 questions, (a) all 8 ﬁnalists; each with 4 possible answers of which only 1 is correct, (b) the ﬁrst 3 positions. (a) in how many diﬀerent ways can a student check oﬀ one answer to each question? 2.40 In how many ways can 5 starting positions on a basketball team be ﬁlled with 8 men who can play any (b) in how many ways can a student check oﬀ one of the positions? answer to each question and get all the answers wrong? 2.41 Find the number of ways that 6 teachers can be assigned to 4 sections of an introductory psychol- 2.34 (a) How many distinct permutations can be ogy course if no teacher is assigned to more than one made from the letters of the word COLUMNS? section. (b) How many of these permutations start with the let- 2.42 Three lottery tickets for ﬁrst, second, and third ter M ? prizes are drawn from a group of 40 tickets. Find the number of sample points in S for awarding the 3 prizes 2.35 A contractor wishes to build 9 houses, each dif- if each contestant holds only 1 ticket. ferent in design. In how many ways can he place these houses on a street if 6 lots are on one side of the street 2.43 In how many ways can 5 diﬀerent trees be and 3 lots are on the opposite side? planted in a circle? 2.36 (a) How many three-digit numbers can be 2.44 In how many ways can a caravan of 8 covered formed from the digits 0, 1, 2, 3, 4, 5, and 6 if wagons from Arizona be arranged in a circle? each digit can be used only once? (b) How many of these are odd numbers? 2.45 How many distinct permutations can be made (c) How many are greater than 330? from the letters of the word IN F IN IT Y ? 2.46 In how many ways can 3 oaks, 4 pines, and 2 2.37 In how many ways can 4 boys and 5 girls sit in maples be arranged along a property line if one does a row if the boys and girls must alternate? not distinguish among trees of the same kind? 2.38 Four married couples have bought 8 seats in the 2.47 How many ways are there to select 3 candidates same row for a concert. In how many diﬀerent ways from 8 equally qualiﬁed recent graduates for openings can they be seated in an accounting ﬁrm? (a) with no restrictions? (b) if each couple is to sit together? 2.48 How many ways are there that no two students will have the same birth date in a class of size 60? 2.4 Probability of an Event Perhaps it was humankind’s unquenchable thirst for gambling that led to the early development of probability theory. In an eﬀort to increase their winnings, gam- blers called upon mathematicians to provide optimum strategies for various games of chance. Some of the mathematicians providing these strategies were Pascal, Leibniz, Fermat, and James Bernoulli. As a result of this development of prob- ability theory, statistical inference, with all its predictions and generalizations, has branched out far beyond games of chance to encompass many other ﬁelds as- sociated with chance occurrences, such as politics, business, weather forecasting, 2.4 Probability of an Event 53 and scientiﬁc research. For these predictions and generalizations to be reasonably accurate, an understanding of basic probability theory is essential. What do we mean when we make the statement “John will probably win the tennis match,” or “I have a ﬁfty-ﬁfty chance of getting an even number when a die is tossed,” or “The university is not likely to win the football game tonight,” or “Most of our graduating class will likely be married within 3 years”? In each case, we are expressing an outcome of which we are not certain, but owing to past information or from an understanding of the structure of the experiment, we have some degree of conﬁdence in the validity of the statement. Throughout the remainder of this chapter, we consider only those experiments for which the sample space contains a ﬁnite number of elements. The likelihood of the occurrence of an event resulting from such a statistical experiment is evaluated by means of a set of real numbers, called weights or probabilities, ranging from 0 to 1. To every point in the sample space we assign a probability such that the sum of all probabilities is 1. If we have reason to believe that a certain sample point is quite likely to occur when the experiment is conducted, the probability assigned should be close to 1. On the other hand, a probability closer to 0 is assigned to a sample point that is not likely to occur. In many experiments, such as tossing a coin or a die, all the sample points have the same chance of occurring and are assigned equal probabilities. For points outside the sample space, that is, for simple events that cannot possibly occur, we assign a probability of 0. To ﬁnd the probability of an event A, we sum all the probabilities assigned to the sample points in A. This sum is called the probability of A and is denoted by P (A). Deﬁnition 2.9: The probability of an event A is the sum of the weights of all sample points in A. Therefore, 0 ≤ P (A) ≤ 1, P (φ) = 0, and P (S) = 1. Furthermore, if A1 , A2 , A3 ,... is a sequence of mutually exclusive events, then P (A1 ∪ A2 ∪ A3 ∪ · · · ) = P (A1 ) + P (A2 ) + P (A3 ) + · · ·. Example 2.24: A coin is tossed twice. What is the probability that at least 1 head occurs? Solution : The sample space for this experiment is S = {HH, HT, T H, T T }. If the coin is balanced, each of these outcomes is equally likely to occur. Therefore, we assign a probability of ω to each sample point. Then 4ω = 1, or ω = 1/4. If A represents the event of at least 1 head occurring, then 1 1 1 3 A = {HH, HT, T H} and P (A) = + + =. 4 4 4 4 Example 2.25: A die is loaded in such a way that an even number is twice as likely to occur as an odd number. If E is the event that a number less than 4 occurs on a single toss of the die, ﬁnd P (E). 54 Chapter 2 Probability Solution : The sample space is S = {1, 2, 3, 4, 5, 6}. We assign a probability of w to each odd number and a probability of 2w to each even number. Since the sum of the probabilities must be 1, we have 9w = 1 or w = 1/9. Hence, probabilities of 1/9 and 2/9 are assigned to each odd and even number, respectively. Therefore, 1 2 1 4 E = {1, 2, 3} and P (E) = + + =. 9 9 9 9 Example 2.26: In Example 2.25, let A be the event that an even number turns up and let B be the event that a number divisible by 3 occurs. Find P (A ∪ B) and P (A ∩ B). Solution : For the events A = {2, 4, 6} and B = {3, 6}, we have A ∪ B = {2, 3, 4, 6} and A ∩ B = {6}. By assigning a probability of 1/9 to each odd number and 2/9 to each even number, we have 2 1 2 2 7 2 P (A ∪ B) = + + + = and P (A ∩ B) =. 9 9 9 9 9 9 If the sample space for an experiment contains N elements, all of which are equally likely to occur, we assign a probability equal to 1/N to each of the N points. The probability of any event A containing n of these N sample points is then the ratio of the number of elements in A to the number of elements in S. Rule 2.3: If an experiment can result in any one of N diﬀerent equally likely outcomes, and if exactly n of these outcomes correspond to event A, then the probability of event A is n P (A) =. N Example 2.27: A statistics class for engineers consists of 25 industrial, 10 mechanical, 10 electrical, and 8 civil engineering students. If a person is randomly selected by the instruc- tor to answer a question, ﬁnd the probability that the student chosen is (a) an industrial engineering major and (b) a civil engineering or an electrical engineering major. Solution : Denote by I, M , E, and C the students majoring in industrial, mechanical, electri- cal, and civil engineering, respectively. The total number of students in the class is 53, all of whom are equally likely to be selected. (a) Since 25 of the 53 students are majoring in industrial engineering, the prob- ability of event I, selecting an industrial engineering major at random, is 25 P (I) =. 53 (b) Since 18 of the 53 students are civil or electrical engineering majors, it follows that 18 P (C ∪ E) =. 53 2.4 Probability of an Event 55 Example 2.28: In a poker hand consisting of 5 cards, ﬁnd the probability of holding 2 aces and 3 jacks. Solution : The number of ways of being dealt 2 aces from 4 cards is 4 4! = = 6, 2 2! 2! and the number of ways of being dealt 3 jacks from 4 cards is 4 4! = = 4. 3 3! 1! By the multiplication rule (Rule 2.1), there are n = (6)(4) = 24 hands with 2 aces and 3 jacks. The total number of 5-card poker hands, all of which are equally likely, is 52 52! N= = = 2,598,960. 5 5! 47! Therefore, the probability of getting 2 aces and 3 jacks in a 5-card poker hand is 24 P (C) = = 0.9 × 10−5. 2, 598, 960 If the outcomes of an experiment are not equally likely to occur, the probabil- ities must be assigned on the basis of prior knowledge or experimental evidence. For example, if a coin is not balanced, we could estimate the probabilities of heads and tails by tossing the coin a large number of times and recording the outcomes. According to the relative frequency deﬁnition of probability, the true probabil- ities would be the fractions of heads and tails that occur in the long run. Another intuitive way of understanding probability is the indiﬀerence approach. For in- stance, if you have a die that you believe is balanced, then using this indiﬀerence approach, you determine that the probability that each of the six sides will show up after a throw is 1/6. To ﬁnd a numerical value that represents adequately the probability of winning at tennis, we must depend on our past performance at the game as well as that of the opponent and, to some extent, our belief in our ability to win. Similarly, to ﬁnd the probability that a horse will win a race, we must arrive at a probability based on the previous records of all the horses entered in the race as well as the records of the jockeys riding the horses. Intuition would undoubtedly also play a part in determining the size of the bet that we might be willing to wager. The use of intuition, personal beliefs, and other indirect information in arriving at probabilities is referred to as the subjective deﬁnition of probability. In most of the applications of probability in this book, the relative frequency interpretation of probability is the operative one. Its foundation is the statistical experiment rather than subjectivity, and it is best viewed as the limiting relative frequency. As a result, many applications of probability in science and engineer- ing must be based on experiments that can be repeated. Less objective notions of probability are encountered when we assign probabilities based on prior informa- tion and opinions, as in “There is a good chance that the Giants will lose the Super 56 Chapter 2 Probability Bowl.” When opinions and prior information diﬀer from individual to individual, subjective probability becomes the relevant resource. In Bayesian statistics (see Chapter 18), a more subjective interpretation of probability will be used, based on an elicitation of prior probability information. 2.5 Additive Rules Often it is easiest to calculate the probability of some event from known prob- abilities of other events. This may well be true if the event in question can be represented as the union of two other events or as the complement of some event. Several important laws that frequently simplify the computation of probabilities follow. The ﬁrst, called the additive rule, applies to unions of events. Theorem 2.7: If A and B are two events, then P (A ∪ B) = P (A) + P (B) − P (A ∩ B). S A AB B Figure 2.7: Additive rule of probability. Proof : Consider the Venn diagram in Figure 2.7. The P (A ∪ B) is the sum of the prob- abilities of the sample points in A ∪ B. Now P (A) + P (B) is the sum of all the probabilities in A plus the sum of all the probabilities in B. Therefore, we have added the probabilities in (A ∩ B) twice. Since these probabilities add up to P (A ∩ B), we must subtract this probability once to obtain the sum of the probabilities in A ∪ B. Corollary 2.1: If A and B are mutually exclusive, then P (A ∪ B) = P (A) + P (B). Corollary 2.1 is an immediate result of Theorem 2.7, since if A and B are mutually exclusive, A ∩ B = 0 and then P (A ∩ B) = P (φ) = 0. In general, we can write Corollary 2.2. 2.5 Additive Rules 57 Corollary 2.2: If A1 , A2 ,... , An are mutually exclusive, then P (A1 ∪ A2 ∪ · · · ∪ An ) = P (A1 ) + P (A2 ) + · · · + P (An ). A collection of events {A1 , A2 ,... , An } of a sample space S is called a partition of S if A1 , A2 ,... , An are mutually exclusive and A1 ∪ A2 ∪ · · · ∪ An = S. Thus, we have Corollary 2.3: If A1 , A2 ,... , An is a partition of sample space S, then P (A1 ∪ A2 ∪ · · · ∪ An ) = P (A1 ) + P (A2 ) + · · · + P (An ) = P (S) = 1. As one might expect, Theorem 2.7 extends in an analogous fashion. Theorem 2.8: For three events A, B, and C, P (A ∪ B ∪ C) = P (A) + P (B) + P (C) − P (A ∩ B) − P (A ∩ C) − P (B ∩ C) + P (A ∩ B ∩ C). Example 2.29: John is going to graduate from an industrial engineering department in a university by the end of the semester. After being interviewed at two companies he likes, he assesses that his probability of getting an oﬀer from company A is 0.8, and his probability of getting an oﬀer from company B is 0.6. If he believes that the probability that he will get oﬀers from both companies is 0.5, what is the probability that he will get at least one oﬀer from these two companies? Solution : Using the additive rule, we have P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = 0.8 + 0.6 − 0.5 = 0.9. Example 2.30: What is the probability of getting a total of 7 or 11 when a pair of fair dice is tossed? Solution : Let A be the event that 7 occurs and B the event that 11 comes up. Now, a total of 7 occurs for 6 of the 36 sample points, and a total of 11 occurs for only 2 of the sample points. Since all sample points are equally likely, we have P (A) = 1/6 and P (B) = 1/18. The events A and B are mutually exclusive, since a total of 7 and 11 cannot both occur on the same toss. Therefore, 1 1 2 P (A ∪ B) = P (A) + P (B) = + =. 6 18 9 This result could also have been obtained by counting the total number of points for the event A ∪ B, namely 8, and writing n 8 2 P (A ∪ B) = = =. N 36 9 58 Chapter 2 Probability Theorem 2.7 and its three corollaries should help the reader gain more insight into probability and its interpretation. Corollaries 2.1 and 2.2 suggest the very intuitive result dealing with the probability of occurrence of at least one of a number of events, no two of which can occur simultaneously. The probability that at least one occurs is the sum of the probabilities of occurrence of the individual events. The third corollary simply states that the highest value of a probability (unity) is assigned to the entire sample space S. Example 2.31: If the probabilities are, respectively, 0.09, 0.15, 0.21, and 0.23 that a person pur- chasing a new automobile will choose the color green, white, red, or blue, what is the probability that a given buyer will purchase a new automobile that comes in one of those colors? Solution : Let G, W , R, and B be the events that a buyer selects, respectively, a green, white, red, or blue automobile. Since these four events are mutually exclusive, the probability is P (G ∪ W ∪ R ∪ B) = P (G) + P (W ) + P (R) + P (B) = 0.09 + 0.15 + 0.21 + 0.23 = 0.68. Often it is more diﬃcult to calculate the probability that an event occurs than it is to calculate the probability that the event does not occur. Should this be the case for some event A, we simply ﬁnd P (A ) ﬁrst and then, using Theorem 2.7, ﬁnd P (A) by subtraction. Theorem 2.9: If A and A are complementary events, then P (A) + P (A ) = 1. Proof : Since A ∪ A = S and the sets A and A are disjoint, 1 = P (S) = P (A ∪ A ) = P (A) + P (A ). Example 2.32: If the probabilities that an automobile mechanic will service 3, 4, 5, 6, 7, or 8 or more cars on any given workday are, respectively, 0.12, 0.19, 0.28, 0.24, 0.10, and 0.07, what is the probability that he will service at least 5 cars on his next day at work? Solution : Let E be the event that at least 5 cars are serviced. Now, P (E) = 1 − P (E ), where E is the event that fewer than 5 cars are serviced. Since P (E ) = 0.12 + 0.19 = 0.31, it follows from Theorem 2.9 that P (E) = 1 − 0.31 = 0.69. Example 2.33: Suppose the manufacturer’s speciﬁcations for the length of a certain type of com- puter cable are 2000 ± 10 millimeters. In this industry, it is known that small cable is just as likely to be defective (not meeting speciﬁcations) as large cable. That is, / / Exercises 59 the probability of randomly producing a cable with length exceeding 2010 millime- ters is equal to the probability of producing a cable with length smaller than 1990 millimeters. The probability that the production procedure meets speciﬁcations is known to be 0.99. (a) What is the probability that a cable selected randomly is too large? (b) What is the probability that a randomly selected cable is larger than 1990 millimeters? Solution : Let M be the event that a cable meets speciﬁcations. Let S and L be the events that the cable is too small and too large, respectively. Then (a) P (M ) = 0.99 and P (S) = P (L) = (1 − 0.99)/2 = 0.005. (b) Denoting by X the length of a randomly selected cable, we have P (1990 ≤ X ≤ 2010) = P (M ) = 0.99. Since P (X ≥ 2010) = P (L) = 0.005, P (X ≥ 1990) = P (M ) + P (L) = 0.995. This also can be solved by using Theorem 2.9: P (X ≥ 1990) + P (X < 1990) = 1. Thus, P (X ≥ 1990) = 1 − P (S) = 1 − 0.005 = 0.995. Exercises 2.49 Find the errors in each of the following state- 2.51 A box contains 500 envelopes, of which 75 con- ments: tain $100 in cash, 150 contain $25, and 275 contain (a) The probabilities that an automobile salesperson $10. An envelope may be purchased for $25. What is will sell 0, 1, 2, or 3 cars on any given day in Febru- the sample space for the diﬀerent amounts of money? ary are, respectively, 0.19, 0.38, 0.29, and 0.15. Assign probabilities to the sample points and then ﬁnd the probability that the ﬁrst envelope purchased con- (b) The probability that it will rain tomorrow is 0.40, tains less than $100. and the probability that it will not rain tomorrow is 0.52. 2.52 Suppose that in a senior college class of 500 stu- (c) The probabilities that a printer will make 0, 1, 2, dents it is found that 210 smoke, 258 drink alcoholic 3, or 4 or more mistakes in setting a document are, beverages, 216 eat between meals, 122 smoke and drink respectively, 0.19, 0.34, −0.25, 0.43, and 0.29. alcoholic beverages, 83 eat between meals and drink (d) On a single draw from a deck of playing cards, the alcoholic beverages, 97 smoke and eat between meals, probability of selecting a heart is 1/4, the probabil- and 52 engage in all three of these bad health practices. ity of selecting a black card is 1/2, and the proba- If a member of this senior class is selected at random, bility of selecting both a heart and a black card is ﬁnd the probability that the student 1/8. (a) smokes but does not drink alcoholic beverages; (b) eats between meals and drinks alcoholic beverages 2.50 Assuming that all elements of S in Exercise 2.8 but does not smoke; on page 42 are equally likely to occur, ﬁnd (c) neither smokes nor eats between meals. (a) the probability of event A; (b) the probability of event C; 2.53 The probability that an American industry will (c) the probability of event A ∩ C. locate in Shanghai, China, is 0.7, the probability that / / 60 Chapter 2 Probability it will locate in Beijing, China, is 0.4, and the proba- 2.60 If 3 books are picked at random from a shelf con- bility that it will locate in either Shanghai or Beijing or taining 5 novels, 3 books of poems, and a dictionary, both is 0.8. What is the probability that the industry what is the probability that will locate (a) the dictionary is selected? (a) in both cities? (b) 2 novels and 1 book of poems are selected? (b) in neither city? 2.61 In a high school graduating class of 100 stu- 2.54 From past experience, a stockbroker believes dents, 54 studied mathematics, 69 studied history, and that under present economic conditions a customer will 35 studied both mathematics and history. If one of invest in tax-free bonds with a probability of 0.6, will these students is selected at random, ﬁnd the proba- invest in mutual funds with a probability of 0.3, and bility that will invest in both tax-free bonds and mutual funds (a) the student took mathematics or history; with a probability of 0.15. At this time, ﬁnd the prob- (b) the student did not take either of these subjects; ability that a customer will invest (c) the student took history but not mathematics. (a) in either tax-free bonds or mutual funds; (b) in neither tax-free bonds nor mutual funds. 2.62 Dom’s Pizza Company uses taste testing and statistical analysis of the data prior to marketing any 2.55 If each coded item in a catalog begins with 3 new product. Consider a study involving three types distinct letters followed by 4 distinct nonzero digits, of crusts (thin, thin with garlic and oregano, and thin ﬁnd the probability of randomly selecting one of these with bits of cheese). Dom’s is also studying three coded items with the ﬁrst letter a vowel and the last sauces (standard, a new sauce with more garlic, and digit even. a new sauce with fresh basil). (a) How many combinations of crust and sauce are in- 2.56 An automobile manufacturer is concerned about volved? a possible recall of its best-selling four-door sedan. If (b) What is the probability that a judge will get a plain there were a recall, there is a probability of 0.25 of a thin crust with a standard sauce for his ﬁrst taste defect in the brake system, 0.18 of a defect in the trans- test? mission, 0.17 of a defect in the fuel system, and 0.40 of a defect in some other area. 2.63 According to Consumer Digest (July/August (a) What is the probability that the defect is the brakes 1996), the probable location of personal computers or the fueling system if the probability of defects in (PC) in the home is as follows: both systems simultaneously is 0.15? Adult bedroom: 0.03 (b) What is the probability that there are no defects Child bedroom: 0.15 in either the brakes or the fueling system? Other bedroom: 0.14 Oﬃce or den: 0.40 2.57 If a letter is chosen at random from the English Other rooms: 0.28 alphabet, ﬁnd the probability that the letter (a) What is the probability that a PC is in a bedroom? (a) is a vowel exclusive of y; (b) What is the probability that it is not in a bedroom? (b) is listed somewhere ahead of the letter j; (c) Suppose a household is selected at random from (c) is listed somewhere after the letter g. households with a PC; in what room would you expect to ﬁnd a PC? 2.58 A pair of fair dice is tossed. Find the probability 2.64 Interest centers around the life of an electronic of getting component. Suppose it is known that the probabil- (a) a total of 8; ity that the component survives for more than 6000 (b) at most a total of 5. hours is 0.42. Suppose also that the probability that the component survives no longer than 4000 hours is 0.04. 2.59 In a poker hand consisting of 5 cards, ﬁnd the probability of holding (a) What is the probability that the life of the compo- nent is less than or equal to 6000 hours? (a) 3 aces; (b) What is the probability that the life is greater than (b) 4 hearts and 1 club. 4000 hours? / / Exercises 61 2.65 Consider the situation of Exercise 2.64. Let A two of these individuals purchase an electric oven. be the event that the component fails a particular test What is the probability that at least three purchase and B be the event that the component displays strain the electric oven? but does not actually fail. Event A occurs with prob- (b) Suppose it is known that the probability that all ability 0.20, and event B occurs with probability 0.35. six purchase the electric oven is 0.007 while 0.104 is (a) What is the probability that the component does the probability that all six purchase the gas oven. not fail the test? What is the probability that at least one of each (b) What is the probability that the component works type is purchased? perfectly well (i.e., neither displays strain nor fails the test)? 2.69 It is common in many industrial areas to use (c) What is the probability that the component either a ﬁlling machine to ﬁll boxes full of product. This fails or shows strain in the test? occurs in the food industry as well as other areas in which the product is used in the home, for example, 2.66 Factory workers are constantly encouraged to detergent. These machines are not perfect, and in

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