Section 2.2 Sample Space and Events 23 4.If the experiment consists of tossing two dice.then the sample space consists of the 36 points S={i:i,j=1,2,3,4,5,6 where the the other die. tcome(i)is said to occur ifiappears on the leftmost die and jon 5.If the experiment consists of measuring (in hours)the lifetime of a transistor, then the sample space consists of all nonnegative real numbers;that is, S=r:0sx<o∞ stin nother word ment is contained in E.then we say that E has occurred.Following are some examples of events. In Example 2,if E=fall outcomes in Sstarting with a 3) thinge)t the eveat that a bead appear on that horsa wi the first coin. .).)).(6.1).then E is the event that thempl 5.f5).then is the event that the transistor does odFofmve dee to event EU F will occur if either Eor Foccurs.For instance,in Example 1.if event E=g)and F=(b),then E U F=lg.bl P格5 e wole ample space3.iE=H,队.以.D列and E U F=((H,H).(H,T).(T.H) Similarly or anyddene the neeven s that are if h occur.For instance.in Example))(i the event that at head occurs and F=((T).(T.(T.T))is the event that at least 1 tail EF={(H,T),(T,H1 is the event that exactly 1 head and 1 tail occur.In example 4,if E=((1.6).(2,5) does not contar
Section 2.2 Sample Space and Events 23 4. If the experiment consists of tossing two dice, then the sample space consists of the 36 points S = {(i, j): i, j = 1, 2, 3, 4, 5, 6} where the outcome (i, j) is said to occur if i appears on the leftmost die and j on the other die. 5. If the experiment consists of measuring (in hours) the lifetime of a transistor, then the sample space consists of all nonnegative real numbers; that is, S = {x: 0 . x < q} Any subset E of the sample space is known as an event. In other words, an event is a set consisting of possible outcomes of the experiment. If the outcome of the experiment is contained in E, then we say that E has occurred. Following are some examples of events. In the preceding Example 1, if E = {g}, then E is the event that the child is a girl. Similarly, if F = {b}, then F is the event that the child is a boy. In Example 2, if E = {all outcomes in S starting with a 3} then E is the event that horse 3 wins the race. In Example 3, if E = {(H, H),(H, T)}, then E is the event that a head appears on the first coin. In Example 4, if E = {(1, 6),(2, 5),(3, 4),(4, 3),(5, 2),(6, 1)}, then E is the event that the sum of the dice equals 7. In Example 5, if E = {x: 0 . x . 5}, then E is the event that the transistor does not last longer than 5 hours. For any two events E and F of a sample space S, we define the new event E ∪ F to consist of all outcomes that are either in E or in F or in both E and F. That is, the event E ∪ F will occur if either E or F occurs. For instance, in Example 1, if event E = {g} and F = {b}, then E ∪ F = {g, b} That is, E ∪ F is the whole sample space S. In Example 3, if E = {(H, H),(H, T)} and F = {(T, H)}, then E ∪ F = {(H, H),(H, T),(T, H)} Thus, E ∪ F would occur if a head appeared on either coin. The event E ∪ F is called the union of the event E and the event F. Similarly, for any two events E and F, we may also define the new event EF, called the intersection of E and F, to consist of all outcomes that are both in E and in F. That is, the event EF (sometimes written E ∩ F) will occur only if both E and F occur. For instance, in Example 3, if E = {(H, H),(H, T),(T, H)} is the event that at least 1 head occurs and F = {(H, T),(T, H),(T, T)} is the event that at least 1 tail occurs, then EF = {(H, T),(T, H)} is the event that exactly 1 head and 1 tail occur. In example 4, if E = {(1, 6),(2, 5), (3, 4),(4, 3),(5, 2),(6, 1)} is the event that the sum of the dice is 7 and F = {(1, 5),(2, 4), (3, 3),(4, 2),(5, 1)} is the event that the sum is 6, then the event EF does not contain
24 Chapter 2 Axioms of Probability any outcomes and hence could not occur.To give such an event a name,we shall refer moe.at oente n en E anc ilar manner If E1,E2.are events,then the union of these events,denoted by En,is defined to be that event which oucomes that are one value of n=1,2.Similarly,the intersection of the events En,denoted by En,is Finally,for any event E,we define the new event E,referred to as the com- a622504352 01 n the m of the dice followthot that because the experiment must result in some outcome,it any two event and F.if all of the which we sometimes say as F is a superset of E).Thus,if EF,then the occurrence of Eimplies the occurrence of F.IfE C Fand FCE.we say that Eand Fare equal A graphical representation that is useful for illustrating logical relations amo eventsthe Venn diagram.The sample spacerepreted snof .G.are represent s con (a)Shaded region:E U (b) (c)Shaded region:E FIGURE 2.1:Venn Diagrams
24 Chapter 2 Axioms of Probability any outcomes and hence could not occur. To give such an event a name, we shall refer to it as the null event and denote it by Ø. (That is, Ø refers to the event consisting of no outcomes.) If EF = Ø, then E and F are said to be mutually exclusive. We define unions and intersections of more than two events in a similar manner. If E1, E2, . are events, then the union of these events, denoted by q n=1 En, is defined to be that event which consists of all outcomes that are in En for at least one value of n = 1, 2, . Similarly, the intersection of the events En, denoted by q n=1 En, is defined to be the event consisting of those outcomes which are in all of the events En, n = 1, 2, . Finally, for any event E, we define the new event Ec, referred to as the complement of E, to consist of all outcomes in the sample space S that are not in E. That is, Ec will occur if and only if E does not occur. In Example 4, if event E = {(1, 6),(2, 5),(3, 4),(4, 3),(5, 2),(6, 1)}, then Ec will occur when the sum of the dice does not equal 7. Note that because the experiment must result in some outcome, it follows that Sc = Ø. For any two events E and F, if all of the outcomes in E are also in F, then we say that E is contained in F, or E is a subset of F, and write E ( F (or equivalently, F ) E, which we sometimes say as F is a superset of E). Thus, if E ( F, then the occurrence of E implies the occurrence of F. If E ( F and F ( E, we say that E and F are equal and write E = F. A graphical representation that is useful for illustrating logical relations among events is the Venn diagram. The sample space S is represented as consisting of all the outcomes in a large rectangle, and the events E, F, G, . are represented as consisting of all the outcomes in given circles within the rectangle. Events of interest can then be indicated by shading appropriate regions of the diagram. For instance, in the three Venn diagrams shown in Figure 2.1, the shaded areas represent, respectively, the events E ∪ F, EF, and Ec. The Venn diagram in Figure 2.2 indicates that E ( F. E F E F S S (a) Shaded region: E F. (b) Shaded region: EF. S (c) Shaded region: Ec . E FIGURE 2.1: Venn Diagrams
Section 2.2 Sample Space and Events 25 FIGURE 2.2:E C F cerThcOotiomiaofbocnsotraieter9camiandomlkemieofoentsobey The Commutative laws EUF=FUE EF=FE Associative laws (EUF)UG=EU(FUG)(EF)G=E(FG) Distributive laws (EUF)G=EGUFG EFUG=(EUG)(FUG) These relatic CentonheoIen5aeof6dguahewntkgoaomineihheeemtonhenght ontained in the (c)Shaded region:(EUF)G. FIGURE 2.3:(EUF)G-EGUFG
Section 2.2 Sample Space and Events 25 S F E FIGURE 2.2: E ( F The operations of forming unions, intersections, and complements of events obey certain rules similar to the rules of algebra. We list a few of these rules: Commutative laws E ∪ F = F ∪ E EF = FE Associative laws (E ∪F) ∪ G = E ∪(F ∪ G) (EF)G = E(FG) Distributive laws (E ∪F)G = EG ∪FG EF ∪ G = (E ∪ G)(F ∪ G) These relations are verified by showing that any outcome that is contained in the event on the left side of the equality sign is also contained in the event on the right side, and vice versa. One way of showing this is by means of Venn diagrams. For instance, the distributive law may be verified by the sequence of diagrams in Figure 2.3. E F (a) Shaded region: EG. G E F (b) Shaded region: FG. G E F (c) Shaded region: (E F )G. G FIGURE 2.3: (E ∪F)G = EG ∪ FG
26 Chapter 2 Axioms of Probability The following useful relationships between the three basic operations of forming unions,intersections,and complements are known as DeMorgan's laws: )-s =1 0-gs To prove DeMorgan's laws,suppose first thatsan outcome ofEThen is not contained inE which means thatx is not contained in any of the events E,i=1.2.n,implying that x is contained in E for all i=1.2.n and thus is contained inE.Togo the other way,suppose that x is an outcome ofE.Then xis contained in E for alli=12.which means that x is not contained in E for anyi=1.2.implying that x is not contained in E in turn implying that x is contained inE.This proves the first of DeMorgan's laws. To prove the second of DeMorgan's laws,we use the first law to obtain -0r which,since ()=E.is equivalent to (0s-s Taking complements of both sides of the preceding equation yields the result we seek. namely. 0s-s) 2.3 AXIOMS OF PROBABILITY For eac the event E,is defined as
26 Chapter 2 Axioms of Probability The following useful relationships between the three basic operations of forming unions, intersections, and complements are known as DeMorgan’s laws: ⎛ ⎝n i=1 Ei ⎞ ⎠ c = n i=1 Ec i ⎛ ⎝n i=1 Ei ⎞ ⎠ c = n i=1 Ec i To prove DeMorgan’s laws, suppose first that x is an outcome of n i=1 Ei c . Then x is not contained in n i=1 Ei, which means that x is not contained in any of the events Ei, i = 1, 2, . , n, implying that x is contained in Ec i for all i = 1, 2, . , n and thus is contained in n i=1 Ec i . To go the other way, suppose that x is an outcome of n i=1 Ec i . Then x is contained in Ec i for all i = 1, 2, . , n, which means that x is not contained in Ei for any i = 1, 2, . , n, implying that x is not contained in n i Ei, in turn implying that x is contained in n 1 Ei c . This proves the first of DeMorgan’s laws. To prove the second of DeMorgan’s laws, we use the first law to obtain ⎛ ⎝n i=1 Ec i ⎞ ⎠ c = n i=1 (Ec i) c which, since (Ec)c = E, is equivalent to ⎛ ⎝n 1 Ec i ⎞ ⎠ c = n 1 Ei Taking complements of both sides of the preceding equation yields the result we seek, namely, n 1 Ec i = ⎛ ⎝n 1 Ei ⎞ ⎠ c 2.3 AXIOMS OF PROBABILITY One way of defining the probability of an event is in terms of its relative frequency. Such a definition usually goes as follows: We suppose that an experiment, whose sample space is S, is repeatedly performed under exactly the same conditions. For each event E of the sample space S, we define n(E) to be the number of times in the first n repetitions of the experiment that the event E occurs. Then P(E), the probability of the event E, is defined as P(E) = limn→q n(E) n
Section 2.3 Axioms of Probability 27 That is,P(E)is defined as the (limiting)proportion of time that Eoccurs.It is thus the limiting frequency of E. possible sequence of repe periment to b do we know asn gets large?Also,even if it does converge to some valuc,how do we know that. limiting proprtionfy pertormed a second time,we shall obtain the same obection by stating that the convergence oftoa constant limiting value isan assumption,or an ariom,of the system.However,to assume that n(will neces quency exists,it does not at all seem to be a priori evident that this need be the to assume a set of simpler and mor stant limiting frequency does in some sense exist?The lattera roach is the modern we shal ime that, ample space sts a va satisfy a certain set of axioms.which,we hope the reader will agree,is in accordance wito tuitive notion of probability. space axioms: Axiom 1 0≤P(E≤1 Axiom 2 P(S=1 Axiom 3 PUE-∑PE We refer to P(E)as the probability of the event E. 1.the outcome will be a point in the sample space S.Axiom 3 states that,for any
Section 2.3 Axioms of Probability 27 That is, P(E) is defined as the (limiting) proportion of time that E occurs. It is thus the limiting frequency of E. Although the preceding definition is certainly intuitively pleasing and should always be kept in mind by the reader, it possesses a serious drawback: How do we know that n(E)/n will converge to some constant limiting value that will be the same for each possible sequence of repetitions of the experiment? For example, suppose that the experiment to be repeatedly performed consists of flipping a coin. How do we know that the proportion of heads obtained in the first n flips will converge to some value as n gets large? Also, even if it does converge to some value, how do we know that, if the experiment is repeatedly performed a second time, we shall obtain the same limiting proportion of heads? Proponents of the relative frequency definition of probability usually answer this objection by stating that the convergence of n(E)/n to a constant limiting value is an assumption, or an axiom, of the system. However, to assume that n(E)/n will necessarily converge to some constant value seems to be an extraordinarily complicated assumption. For, although we might indeed hope that such a constant limiting frequency exists, it does not at all seem to be a priori evident that this need be the case. In fact, would it not be more reasonable to assume a set of simpler and more self-evident axioms about probability and then attempt to prove that such a constant limiting frequency does in some sense exist? The latter approach is the modern axiomatic approach to probability theory that we shall adopt in this text. In particular, we shall assume that, for each event E in the sample space S, there exists a value P(E), referred to as the probability of E. We shall then assume that all these probabilities satisfy a certain set of axioms, which, we hope the reader will agree, is in accordance with our intuitive notion of probability. Consider an experiment whose sample space is S. For each event E of the sample space S, we assume that a number P(E) is defined and satisfies the following three axioms: Axiom 1 0 . P(E) . 1 Axiom 2 P(S) = 1 Axiom 3 For any sequence of mutually exclusive events E1, E2, . (that is, events for which EiEj = Ø when i Z j), P ⎛ ⎝q i=1 Ei ⎞ ⎠ = q i=1 P(Ei) We refer to P(E) as the probability of the event E. Thus, Axiom 1 states that the probability that the outcome of the experiment is an outcome in E is some number between 0 and 1. Axiom 2 states that, with probability 1, the outcome will be a point in the sample space S. Axiom 3 states that, for any sequence of mutually exclusive events, the probability of at least one of these events occurring is just the sum of their respective probabilities