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手机:13870643020 QQ:386067194 QQ学习群: 489193537 Email:[email protected] Li Jie Probability and Statistics For Engineering and the Sciences Fifth Edition JAY L.DEVORE Li Jie Birthday • How many students in your class? • The same birthday (at least two classmates)? Li Jie What Is Probability • Probability is a way of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory, that is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems. • http://en.wikipedia.org/wiki/Probability Li Jie 2 Probability Introduction The term probability refers to the study of randomness and uncertainty. In any situation in which one of a number of possible outcomes may occur, the theory of probability provides methods for quantifying the chances, or likelihoods, associated with the various outcomes. The language of probability is constantly used in an informal manner in both written and spoken contexts. Li Jie 2 Probability Introduction In this chapter, we introduce some elementary probability concepts, indicate how probabilities can be interpreted, and show how the rules of probability can be applied to compute the probabilities of many interesting events. The methodology of probability will then permit us to express in precise language such informal statements as those given above. Li Jie 2.1 Sample Spaces and Events Random Experiment An experiment is any action or process that generates observations. For examples 1.tossing a coin once or several times 2.selecting a card or cards from a deck 3.weighing a loaf of bread 4.ascertaining the commuting time from home to work on particular morning 5.obtaining blood types from a group of individuals, etc. throw six-sided die Li Jie Sample Space and Sample Point Definition : The sample space of an experiment, denoted by S (Ω), is the set of all possible outcomes of that experiment, and sample point of the sample space, denoted by ω, is a outcome of the experiment. 8 5 1 9 4 6 7 2 3 10 Li Jie Example 2.1 : The simplest experiment to which probability applies is one with two possible outcomes. One such experiment consists of examining a single fuse to see whether it is defective. The sample space for this experiment can be abbreviated as Ω={N,D}, where N represents not defective, D represents defective, and the braces are used to enclose the elements of a set. Another such experiment would involve tossing a thumbtack and noting whether it landed point up or point down, with sample space Ω={U,D}, and yet another would consist of observing the sex of the next child born at the local hospital, with Ω={M,F}. Li Jie Example 2.2 : driving to work , a commuter passes through a sequence of three intersections with traffic lights. At each light, she either stops, s, or continues, c, the sample space is the set of all possible outcomes : Ω ={ccc, ccs, css, csc, sss, ssc, scc, scs } Where csc for example denotes the outcomes that the commuter continues through the first light,stops at the second light,and continues through the third light. Li Jie • Example 2.3 Two gas stations are located at a certain intersection. Each one has six gas pumps. Consider the experiment in which the number of pumps in use at a particular time of day is determined for each of the stations. • An experimental outcome specifies how many pumps are in use at the first station and how many are in use at the second one. One possible outcome is (2,2), another is (4,1). • The 49 outcomes is Ω are displayed in the companying table. The sample space for the experiment in which a six-sided die is thrown twice results from deleting the 0 row and 0 column from the table, given 36 outcomes Li Jie Li Jie Example 2.4 The number of jobs in a print queue of a mainframe computer may be modeled as random . Here the sample space can be taken as Ω= 0,1,2,3,... That is ,all the nonnegative integers. In practice there is probably an upper limit,N,on how large the print queue can be ,so instead the sample space might be defined as Ω= 0,1,2,..., N Li Jie Example 2.5 Earthquakes exhibited very erratic behavior,which is sometimes modeled as random. For example, the length of time between successive earthquakes in a particular region that are greater in magnitude than a given threshold may be regarded as an experiment. Here Ω is the set of all nonnegative numbers: Ω= t t 0 Li Jie Random Event Definition : An event, are usually denoted by italic uppercase letters, is any collection (subset) of outcomes contained in the sample space Ω. An event is said to be simple if it consists of exactly one outcome and compound if it consists of more than one outcome. Remark: When an experiment is performed, a particular event A is said to occur if the resulting experimental outcome is contained in A. In general, exactly one simple event will occur, but many compound events will occur simultaneously. The event that the commuter stops at the first light is the subset of Ω denoted by A ={sss, ssc, scc, scs} Li Jie Example 2.5 : Consider an experiment in which each of three automobiles taking a particular freeway exit turns left (L) or right (R) at the end of the exit ramp. The eight possible outcomes that comprise the sample space are LLL,RLL,LRL,LLR,LRR,RLR, RRL, and RRR. Thus, there are eight simple events, among which are E1={LLL} and E5={LRR}. Some compound events include A={RLL,LRL,LLR}=the event that exactly one of the three cars turns right B={LLL,RLL,LRL,LLR}=the event that at most one of the cars turns right C={LLL,RRR}=the event that all three cars turn in the same direction Li Jie Suppose that when the experiment is performed, the outcome is LLL. Then the simple event E1 has occurred and so also have the events B and C (but not A). Some Relations From Set Theory An event is nothing but a set, so that relationships and results from elementary set theory can be used to study events. The following concepts from set theory will be used to construct new events from given events. Li Jie Union(∪) The algebra of set theory carries over directly into probability theory. The union of two events , A and B,is the event C that either A occurs or B occurs or both occur:C=A∪B. For example,if A is the event that commuter stops at the first light and if B is the event that she stops at the third light, B= sss , scs, ccs, css Then C is the event that he stops at the first light or stops at the third light and consists of the outcomes that are in A or in B or in both : C= sss, ssc, scc, scs, ccs, css Li Jie Intersection (∩) The intersection of two events, C=A∩ B ,is the event that both A and B occur. If A and B are as listed previously,then C is the event that the commuter stops at the first light and stops at the third light and thus consists of those outcomes that are common to both A and B: C= sss, scs Li Jie Complement( A ) A, A c The complement of an event , A ,is the event that A does not occur and thus consists of all those elements in the sample space that are not in A ,the complement of the event that the commuter stops at the first light is the event that she continues at the first light: A = ccc, ccs, css, csc Li Jie You may recall from previous exposure to set theory that rather mysterious set called the empty set,usually denoted by Ф. The empty set with no elements; it is the event with no outcomes. For example,if A is the event that the commuter stops at the first light and C is the event that she continues through all three lights , c=CCC then A and C have no outcomes in common, and we can write: A C In such cases ,A and C are said to be mutually exclusive or disjoint. Li Jie Venn diagrams B A Ω A A B B A A A B A B A B A∩B A B A-B Li Jie The following are some lows of set theory. Commutative laws: A B B A A B B A Associative laws: A B C A B C A B C A B C Distributive laws: A B C A C B C A B C A C B C De Morgan laws: A B A B A B A B Li Jie 2.2 Axioms, Interpretations And Properties Of Probability Given an experiment and a sample spaceΩ, the objective of probability is to assign to each event A a number P(A), called the probability of the event A, which will give a precise measure of the chance that A will occur. To ensure that the probability assignments will be consistent with our intuitive notions of probability, all assignments should satisfy the following axioms : 1.ΡΩ 1 2.If A Ω, then Ρ A 0 3.If A1 and A2 are disjoint ,then Ρ A1 A2 Ρ A1 Ρ A2 Li Jie More generally ,if then 1 , 2 , n are mutually disjoint , Ρ Αi Ρ Αi i 1 i 1 The first two axioms are rather obvious. Since consists of all possible outcomes. ΡΩ 1 The second axiom simply states that a probability is nonnegative. Li Jie The third axiom states that if A and B are disjoint that is have no outcomes in common then Ρ Α Β Ρ Α ΡΒ and also that this property extends to limits. For example the probability that the print queue contains either one or three jobs is equal to the probability that it contains one plus the probability that it contains three. Li Jie Properties Of Probability The following properties of probability measures are consequences of the axioms. Property 1 ΡΑ 1 Ρ Α this property follows since A and are disjoint with Α Α Ω and thus ,by the first and third axioms, Ρ Α Ρ Α 1 in words this property says that the probability that an event does not occur equals one minus the probability that it does occur. Li Jie Property 2 Ρφ 0 this property follows from property 1 since φΩ In words this says that the probability that there is no outcome at all is zero. if A B Φ then P( A B) 0 Li Jie Property 3 if Α Β then Ρ Α ΡΒ this property follows since B can be expressed as the union of two disjoint sets: Β Α Β Α Then from the third axiom, And thus ΡΒ Ρ Α ΡΒ Α Ρ Α ΡΒ ΡΒ Α ΡΒ This property states that if B occurs whenever A occurs, then Ρ Α ΡΒ . For example if whenever it rains (A) it is cloudy (B), then the probability that it rains is less than or equal to the probability that it is cloudy. Li Jie Property 4 addition law Ρ Α Β Ρ Α ΡΒ Ρ Α Β to see this ,we decompose Α Β into three disjoint subsets, as shown in figure2.1 : C ΑΒ D ΑΒ E ΒΑ Li Jie FIGURE 2.1 Venn diagram illustrating the addition law. A B C D E We then have from the third axiom, Ρ Α Β ΡC ΡD ΡE Li Jie Also , Α C D and C and D are disjoint ; so Ρ Α ΡC ΡD similarly, ΡΒ ΡD ΡE putting these results together ,we see that Ρ Α ΡΒ ΡC ΡE 2ΡD Ρ Α Β ΡD or Ρ Α Β Ρ Α ΡΒ ΡD this property is easy to see from the venn diagram 2.1. if P(A) and P(B) are added together , Ρ Α Β is counted twice. Li Jie The probability of a union of more than two events can be computed analogously. For three events A, B, and C, the result is P( A B C ) P( A) P( B) P(C ) P( A B) P( A C ) P( B C ) P( A B C ). P(( A B) C ) P( A B) P(C ) P(( A B) C ) P( A) P( B) P(C ) P( A B) P( A C ) P( B C ) P( A B C ). P( A B) P( A AB) P( A) P( AB) Li Jie Equally Likely Outcomes In many experiments consisting of N outcomes, it is reasonable to assign equal probabilities to all N simple events. With p=P(Ei) for every i, N N 1 1 P( Ei ) p pN so p N i 1 i 1 That is, if there are N possible outcomes, then the probability assigned to each is 1/N. Now consider an event A, with N(A) denoting the number of outcomes contained in A. Then P( A) P( Ei ) Ei in A number of ways A can occur 1 N ( A) total number of outcomes N Ei in A N Li Jie Example 2.6 : When two dice are rolled separately, there are N=36 outcomes. If both the dice are fair, all 36 outcomes are equally likely, so P(Ei)=1/36. Then the event A={sum of two numbers=7} consists of the six outcomes (1,6), (2,5), (3,4), (4,3), (5,2), and (6,1), so N ( A) 6 1 P ( A) N 36 6 Li Jie 2.3 Counting Techniques When the various outcomes of an experiment are equally likely, then the task of computing probabilities reduces to counting. In particular, if N is the number of outcomes in a sample space and N(A) is the number of outcomes contained in an event A, then P( A) number of ways A can occur N ( A) total number of outcomes N (2.1) If a list of the outcomes is available or easy to construct and N is small, then the numerator and denominator of Equation (2.1) can be obtained without the benefit of any general counting principles. Li Jie Example 2.7: a black urn(瓮、缸) contains 5 red and 6 green balls and a white urn contains 3 red and 4 green balls. You are allowed to choose an urn and then choose a ball at random from the urn. If you choose a red ball ,you get a prize. Which urn should you choose to draw from? If you draw from the black urn ,the probability of choosing a red ball is 0.455. If you choose to draw from the white urn, the probability of choosing a red ball is 0.429,so you should choose to draw from the black urn. Li Jie Now consider another game in which a second black urn has 6 red and 3 green balls and a second white urn has 9 red and 5 green balls. If you draw from the black urn, the probability of red ball is 6/9. whereas if you choose to draw from the white urn ,the probability is 9/14. so ,again you should choose to draw from the black urn. In the final game, the contents of the second black urn are added to the first black urn and the contents of the second white urn are added to the first white urn .again you can choose which urn you can draw from .which should you choose? Intuition says choose the black urn, but let’s calculate the probabilities. Li Jie The black urn now contains 11 red and 9green balls, so the probability of drawing a red ball from it is 11/20=0.55 .The white urn now contains 12 red and 9 green balls, so the probability of drawing a red ball from it is 12/21=0.571 . so, you should choose the white urn. This counterintuitive result is an example of Simpson’s paradox. Li Jie The Product Rule For Ordered Pairs If one experiment has m outcomes and another experiment has n outcomes ,then there are m×n possible outcomes for the two experiments. Proof: Denote the outcomes of the first experiment by a1 , a2 ,, am and The outcomes of the second experiment by b1 , b2 ,bn the outcomes for the two experiments are the ordered pairs ai , b j these pairs can be exhibited as the entries of an m × n rectangular array, in which the pair ai , b j is in the ith row and the jth column. There are mn entries in this array. Li Jie Example 2.8: A family has just moved to a new city and requires the services of both an obstetrician and a pediatrician. There are two easily accessible medical clinics, each having two obstetricians and three pediatricians. The family will obtain maximum health insurance benefits by joining a clinic and selecting both doctors from that clinic. In how many ways can this be done? Denote the obstetricians by O1,O2,O3, and O4 and the pediatricians by P1,…,P6. Then we wish the number of pairs (Oi, Pj) for which Oi and Pj are associated with the same clinic. Because there are four obstetricians, n1=4, and for each there are three choices of pediatrician, so n2=3. Applying the product rule gives N= n1 n2=12 possible choices. Li Jie Tree diagrams P1 P2 O1 O2 O3 O4 P3 P1 P2 P3 P4 P5 P6 P4 P5 P6 Li Jie A More General Product Rule If there are p experiments and the first has n1 possible outcomes, the second n2 , and the pth n p possible outcomes ,then there are a total of n1 n2 n p possible outcomes for the p experiments. Example 2.9: an 8-bit binary word is a sequence of 8 digits, of which each may be either 0 or 1. How many different 8-bit words are there? There are two choices for first bit, two for the second, etc, and thus there are 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2= 28=256 such words. Li Jie Permutations(排列) Definition : Any ordered sequence of k objects taken from a set of n distinct objects is called a permutation of size k of the objects. The number of permutations of size k that can be constructed from the n objects is denoted by Pnk . Pnk nn 1n 2 n k 1 Definition : For any positive integer m, m! is read “m factorial” and is defined by m! mm 1m 22 1 so n! P (n k )! k n Li Jie Combinations (组合) Definition : Given a set of n distinct objects, any unordered subset of size k of the objects is called a combination. The number of combinations of size k that can be formed from the n distinct objects will be denoted by C nk . k P n! k n Cn k! k!(n k )! Cnm Cnn m Cnm Cnm1 Cnm11 Li Jie Example A suppose that a fair coin is thrown twice and the sequence of heads and tails is recorded. The sample space is hh, ht , th, tt As in Example A of previous section, we assume that each outcome in has probability 0.25. let A denote the event that at least one head is throw. Then hh, ht , th And P(A)=.75 Li Jie Example Consider the experiment of selecting five cards from a deck of 52 cards. Let A={the five cards number different each other} B={two cards have the same number, and other three cards also have homology number} C={two cards have the same number, the other three cards have different number} D={five cards have four kinds of designs} 5 C13 45 P( A) 5 C52 1 2 1 C C C C P( B) 13 4 512 4 C52 1 2 3 C13 C4 C12 43 P(C ) 5 C52 3 1 2 1 1 1 C C C C C P( D) 4 13 13 5 13 13 C52 Li Jie Example There are m boys and n girls, they stand randomly to one row, Let A={girls stand together} B={girls stand together, and boys stand together,too} C={there is at least one boy in any two girls} Solution: (m 1)!n! P( A) (m n)! m!n! P22 P( B) (m n)! m! Pmn1 P(C ) (m n)! Li Jie Li Jie 2.4 Conditional probability Example 2.10: Complex components are assembled in a plant that uses two different assembly lines, A and A . Line A uses older equipment than A , so it is somewhat slower and less reliable. Suppose on a given day line A has assembled 8 components, of which 2 have been identified as defective (B) and 6 as nondefective (B), whereas A has produced 1 defective and 9 nondefective components. This information is summarized in the accompanying table. Condition Line A A B 2 1 B 6 9 Li Jie Unaware of this information, the sales manager randomly selects 1 of these 18 components for a demonstration. Prior to the demonstration N ( A) 8 P(line A component selected ) P( A) N 18 However, if the chosen component turns out to be defective, then the event B has occurred, so the component must have been 1 of the 3 in the B column of the table. Since these 3 components are equally likely among themselves after B has occurred, 2 2 / 18 P( A B) P( A | B) 3 3 / 18 P( B) Li Jie The Definition Of Conditional Probability Definition : For any two events A and B with P(B)>0, the conditional probability of A given that B has occurred is defined by P( A B) P( A | B) P( B) Li Jie Example 2.11: Suppose that of all individuals buying a certain personal computer, 60% include a word processing program in their purchase, 40% include a spreadsheet program, and 30% include both types of programs. Consider randomly selecting a purchaser and let A={word processing program included} and B={spreadsheet program included}. Then P(A)=0.6, P(B)=0.4, and P(both included)=P(A∩B)=0.3. Given that the selected individual included a spreadsheet program, the probability that a word processing program was also included is P( A B) 0.3 P( A | B) 0.75 P( B) 0.4 Li Jie Example In one year, if the first city has 20% rainy days, the second city has 18% rainy days, raining in both cities at the same time has 12%. Determine the probability of A,B A={at least one city has raining} B={if known one city has raining, then the other city has raining too} Solution: A1={the first city is raining}, A2={the second city is raining} P( A1 A2 ) P( A1 ) P( A2 ) P( A1 A2 ) 0.20 0.18 0.12 0.26 P(A2|A1)=P(A1A2)/P(A1) =0.12/0.2=0.6 P(A1|A2)=P(A1A2)/P(A2) =0.12/0.18=0.67 Li Jie Example A batch of accessories is 100, in which there are 5 defectives. Each time select one accessory to inspect, then put it aside. Select two accessories continuously to inspect. Determine the probabilities of the following events. a. the first inspected accessory is good. b. If the first inspected accessory is good, the second inspected accessory is good too. c. the accessories inspected in two times are all good. Solution: A1={the first inspected accessory is good}, A2={the second inspected accessory is good } 95 (a) P(A ) 0.95 100 94 (b) P(A | A ) 0.9495 99 1 2 1 (c ) P(A A ) P(A ) P(A | A ) 0.9020 1 2 1 2 1 Li Jie .Multiplication Law Let A and B be events and assume P(B) 0.Then P(A B)= P(A|B)P(B) The multiplication law is often useful in finding the probabilities of intersections, as the following examples illustrate. Li Jie Example An urn contains three red balls and one blue ball. Two balls are selected without replacement. What is the probability that they are both red? Let R 1and R 2 denote the events that a red ball is drawn on the first trial and on the second trial, respectively. From the multiplication law, P(R1 R 2 ) P(R1 )P(R2 |R1 ) P(R ) 1 is clearly 3 , and if a red ball has been 4 removed on the first trial, there are two red balls and one blue ball left. Therefore, 1 2 P( R2 | R1 ) . Thus, P( R1 R2 ) . 3 2 Li Jie Example Suppose that if it is cloudy (B), the probability that it is raining (A) is .3, and that the probability it is cloudy is P( B) .2 The probability that it is cloudy and raining is P( A B) P( A | B) P( B) .3 .2 .06 Another useful tool for computing probabilities is provided by the following law. Li Jie Corollary: P( A1 A2 A 3) P( A1 ) P( A2 | A1 ) P( A3 | A1 A2 ) P( A1 A2 An ) P( A1 ) P( A2 | A1 ) P( A3 | A1 A2 ) P( An | A1 A2 An 1 ) Li Jie Example A fair coin is tossed three times in an honest manner. Given that the head shows at least once, what is the probability that it shows all three times? Solution: suppose A={the head shows all three times} B={the head shows at least once} P( AB ) P( A) 1/ 8 1 P( A | B) P( B) P( B) 1 1 / 8 7 Li Jie Example There are ten same size coins in one bag, in which 7 coins is 1 jiao, 3 coins is 5 jiao. Sampling with non-replacement. Please determine: (1) The probability of 5 jiao coin only selected in the third time. (2) The probability of 5 jiao coin at least selected once in three selections. Solution: suppose Ai={5 jiao coin be selected in ith time}, i=1,2,3 5 jiao coin only selected in the third time A A A 1 2 3 Li Jie 5 jiao coin at least selected once in three selections A1 A2 A3 (1) Due to multiplication law P( A1 A2 A3 ) P( A1 )P( A2 | A1 )P( A3 | A1 A2 ) 7 6 3 7 0.175 10 9 8 40 (2) Due to dual law A1 A2 A3 A1 A2 A3 Li Jie P( A1 A2 A3 ) 1 P ( A1 A2 A3 ) P( A1 A2 A3 ) 1 P ( A1 A2 A3 ) 1 P( A1 )P( A2 | A1 )P( A3 | A1 A2 ) 7 1 10 5 17 8 24 Li Jie Li Jie The law of total probability Definition: if A1,A2,…,An satisfied: (1) P( Ai ) 0, i 1,2,n (2) A1 , A2 ,, An be mutually exclusive (3) A1 A2 An exhaustive events A1 An B A2 A3 A4 Then we named A1,A2,…,An is a dividing of sample space Li Jie Theorem: suppose A1,A2,…,An is a dividing of sample space, then n P( B) P( Ai ) P( B | Ai ) i 1 Proof: P( B | A1 ) B B B( A1 A2 An ) BA1 BA2 BAn BA1 , BA2 ,, BAn are mutex and P( BAi ) P( Ai ) P( B | Ai ) B P( A1 ) P( A2 ) P( A3 ) P( An ) P( B) P( BA1 ) P( BA2 ) P( BAn ) n P( Ai ) P( B | Ai ) A1 B P( B | A2 ) A2 B P( B | A3 ) A3 B An P( B | An ) B Tree diagram i 1 Li Jie Example2.13: Suppose that occupations are grouped into upper (U), middle (M), and lower (L) levels. U1will denote the event that a father’s occupation is upper-level; U 2 will denote the event that a son’s occupation is upper-level, etc. U2 M2 L2 U1 .45 .48 .07 M1 .05 .70 .25 L1 .01 .50 .49 Li Jie Such a table, which is called a matrix of transition probabilities, is to be read in the following way: If a father is in U, the probability that his son is in U .45, the probability that his son is in M is .48, Examination of the table reveals that there is more upward mobility from L into M than from M into U. Suppose that of the father’s generation, 10% are in U, 40% in M, and 50% in L. What is the probability that a son in the next generation is in U? Applying the law of total probability, we have P(U 2 ) P(U 2 | U1 ) P(U1 ) P(U 2 | M 1 ) P( M 1 ) P(U 2 | L1 ) P( L1 ) 0.45 0.10 0.05 0.40 0.01 0.50 0.07 Li Jie Example: Urn A has three red balls and two white balls. and urn B has two red balls and five white balls. A fair coin is tossed; if it lands heads up, a ball is drawn from urn A and otherwise a ball is drawn from urn B. a. What is the probability that a red ball is drawn? b. If a red ball is drawn, what is the probability that the coin landed heads up? Li Jie Bayes’ theorem In the discussion of conditional probability, we indicated that revising probabilities when new information phase of probability analysis. Often, we begin the analysis with initial or prior probability estimates for specific events if interest. Then, from sources such as a sample, a special report, or a product test, we obtain additional information about the events. Given this new information, we update the prior probability values by calculating revised probabilities, referred to as posterior probabilities. Bayes’ theorem provides a means for making these probability calculations. Li Jie Example2.14: As an application of Bayes’ theorem, consider a manufacturing firm that receives shipments of parts from two different suppliers A1 and A2. Let P(A1)=0.65, P(A2)=0.35. The quality of the purchased parts varies with the source of supply. We let G denote the event that a part is good and B denote the event that a part is bad, P(G|A1)=0.98 P(B|A1)=0.02, P(G|A2)=0.95, P(B|A2)=0.05 Suppose now that the parts from the two suppliers are used in the firm’s manufacturing process and that a machine breaks down because it attempts to process a bad part. Given information that the part is bad , what is the probability that it came from supplier 1 and what is the probability that it came from supplier 2? Li Jie P(G|A1)=0.98 P(A1)=0.65 P(A2)=0.35 P(B|A1)=0.02 P(G|A2)=0.95 P(B|A2)=0.05 P(A1G)=P(A1)P(G|A1)=0.637 P(A1B)=P(A1)P(B|A1)=0.013 P(A2G)=P(A2)P(G|A2)=0.3325 P(A2B)=P(A2)P(B|A2)=0.0175 Solution: we will determine P(A1|B), P(A2|B) P( A1B) P( A1B) P( B | A1 ) P( A1 ) P( A1 | B) P( B) P( B) P( B | A1 ) P( A1 ) P( B | A2 ) P( A2 ) P( A1B) P( B | A1 ) P( A1 ) P( A1 | B) P( B) P( B | A1 ) P( A1 ) P( B | A2 ) P( A2 ) Li Jie Baye’s theorem (Two-event case) P( A1 B ) P( B | A1 )P( A1 ) P( A1 | B ) P( B ) P( B | A1 )P( A1 ) P( B | A2 )P( A2 ) P( A2 B ) P( B | A2 )P( A2 ) P( A2 | B ) P( B ) P( B | A1 )P( A1 ) P( B | A2 )P( A2 ) Corollary: if A1,A2,….,An are mutually exclusive and their union is the entire sample space. Then P( Ai B ) P( Ai )P( B| Ai ) P( Ai | B ) P( B ) P( B ) P( Ai )P( B| Ai ) n P( A )P( B| A ) k 1 k k Li Jie Example 2.15: In digit communication, on account of random factors, the signal received would not be the same signal send out. If the digits send out are ‘0’ and ‘1’, then the receiving digits will ‘0’, ‘1’, and ‘*’. if ‘0’ would be send out by probability 0.6, then the probability of receiving ‘0’, ‘1’, ‘*’are 0.8, 0.05,0.15, if ‘1’would send out by probability 0.4, then the probability of receiving ‘0’, ‘1’, ‘*’are 0.9, 0.05,0.05. If now we have received signal ‘*’, what are the probability of original send signal ‘0’, and ‘1’. receive send 0 0.8 0.05 0.15 0 1 * receive send 1 0 0.05 0.9 0.05 1 * Li Jie Solution: A={send ‘0’}, AC={send ‘1’}, B={receive ‘*’} receive 0 C|B) We asked to determine P(A|B), P(A 1 send 0.8 0 0.05 0.15 * P( AB ) P( B | A) P( A) P( A | B) C C P ( B ) P ( B | A ) P ( A ) P ( B | A ) P ( A ) receive send 0.6 0.15 0 0.05 0.8182 1 0.6 0.15 0.4 0.05 0.9 1 0.05 C C C P ( A B ) P ( B | A ) P ( A ) C * P( A | B) P( B) P ( B | A) P( A) P ( B | AC ) P( AC ) 0.4 0.05 0.1818 0.6 0.15 0.4 0.05 Li Jie Example: A factory runs three shifts. In a given day, 1% of the items produced by the first shift are defective, 2% of the second shift’s items are defective, 5% of the third shift’s items are defective. If the shifts all have the same probability, what percentage of the items produced in a day are defective? If an item is defective, what is the probability that it was produced by the third shift? Example: There are 12 new ping-pong balls in one bag. We take 3 balls at each match, then replace its to the bag after play. What is the probability that the 3 balls are all new at the third match. Li Jie Example: There are 12 new ping-pong balls in one bag. We take 3 balls at each match, then replace its to the bag after play. What is the probability that the 3 balls are all new at the third match. Bi {i new balls at the 2nd match}, i 0,1,2,3 i 9 3 i 3 3 12 CC P( Bi ) C , i 1,2,3 C93i P(C3 Bi ) 3 , i 1,2,3 C12 3 P(C3 ) P( Bi ) P(C3 Bi ) i 0 Li Jie 2.5 Independence Definition: Two events A and B are independent if P(A|B)=P(A) and are dependent otherwise. Remark: The definition of independence might seem “unsymmetric” because we do not demand that P(B|A)= P(B) also. However, using the definition of conditional probability and the multiplication rule, we have P(A|B)=P(A) if and only if P(B|A)=P(B) P(B|A)=P(AB)/P(A) = P(A|B) P(B)/P(A) so P(B|A)=P(B) Li Jie P(A∩B) When Events Are Independent Proposition: A and B are independent if and only if P(A∩B)=P(A)·P(B) and P( A B) P( A) P( B) P( A B) P( A ) P( B) P( A B ) P( A) P( B ) P( A B ) P( A ) P( B ) If A and B are independent, then so are the following pairs of events: (1) A' and B (2) A and B ' (3) A' and B ' Li Jie If two events are mutually exclusive , they cannot be independent. Example 2.32 Let A and B be any two mutually exclusive events with P(A)>0. For example, for a randomly chosen automobile, let A={the car has four cylinders}, B={the car has six cylinders}. A and B mutually exclusive, P(A︱B)=0≠P(A), The message here is that if two events are mutually exclusive , they cannot be independent. Li Jie Example2.16: A card selected randomly from a deck. Let A denote the event that it is an ace and D the event that it is a diamond. Knowing that the card is an ace gives no information about its suit. Checking formally that the events are independent, we have P(A) 4 1 and P ( D ) 1 Also, A D is the 52 13 4 event that the card is the ace of diamonds 1 and P(A D) 52 . Since P(A)P(D) ( 1 ) ( 1 ) 1 ,the 4 13 52 events are in fact independent. Li Jie Example2.17: Suppose that a system consists of components connected in series,so the system fails if any one component fails. If there are n mutually independent components and each fails with probability p, what is the probability that the system will fail? It is easier to find the probability of the complement of this event;the system works if and only if all the components work, and this situation has probability (1-p)n .The probability that the system fails is then 1-(1-p)n. For example, if n=10 and p=0.05,the probability that the system work is only 0.9510=0.6, and the probability that the system fails is 0.40. Suppose, instead, that the components are connected in parallel, so the system fails only when they all fail. In this case, the probability that the system fails is only 0.0510=9.8×10-14. Li Jie Independence Of More Than Two Events Definition: Events A1,…,An are mutually independent if for every k (k=2,3,…,n) and every subset of indices i1,i2,…ik, P( Ai1 Ai2 Aik ) P( Ai1 ) P( Ai2 ) P( Aik ) ABC mutually independent , Then P(AB)=P(A)P(B), P(AC)=P(A)P(C), P(BC)=P(B)P(C), P(ABC)=P(A)P(B)P(C) Li Jie Normal tetrahedron regular tetrahedron 1 red 2 white 3 black 4 red white black regular octahedron 1234 red 1235 white 1678 black A={ red } B={white} C={ black} Li Jie Example B A system is designed so that fails only if a unit and a backup unit both fail. Assuming that these failures are independent and that each unit fails with probability p .the system fails with probability p2 .If, for example ,the probability that any unit fails during a given year is .1, then the probability that the system fails is .01, which represent a considerable improvement in reliability. Li Jie Things become more complicated when we consider more than two events. For example, suppose we know that events A, B, and C are pairwise independent (any two are independent ). We would like to be able to say that they are all independent based on the assumption that knowing something about two of the events does not tell us anything about the third, for example, P(C | A B) P(C) . But as the following example shows, pairwise independence does not guarantee mutual independence. Li Jie Example C A fair coin is tossed twice. Let A denote the event of heads on the first toss, B the event of heads on the second toss, and C the event that exactly one head is thrown. A and B are clearly independent, and P(A)=P(B)=P(C)=.5. To see that A and C are independent, we observe that P(C|A)=.5. But P(A B C) 0 P(A)P(B)P(C) Li Jie Example A fair die is repeatedly rolled in an honest manner until one spot shows for the first time. Determine the probability that ten rolls are required. Solution: Li Jie