• Like
U uni 6 ssb
Upcoming SlideShare
Loading in...5

Thanks for flagging this SlideShare!

Oops! An error has occurred.

U uni 6 ssb


vtu m4 notes

vtu m4 notes

Published in Engineering , Technology , Education
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Be the first to comment
    Be the first to like this
No Downloads


Total Views
On SlideShare
From Embeds
Number of Embeds



Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

    No notes for slide


  • 1. Engineering Mathematics-IV Subject Code: 10Mat41 Part B Unit : VI Probability-I Dr. S. S. Benchalli Associate Professor and Head Department of Mathematics Basaveshwar Engineering College Bagalkot – 587102, Karnataka Email: sbenchalli@gmail.com Mobile:8762644634
  • 2. Introduction: Probability and statistics are concerned with events which occur by chance. Examples include occurrence of accidents, errors of measurements, production of defective and non-defective items from a production line, and various games of chance, such as drawing a card from a well- mixed deck, flipping a coin, or throwing a symmetrical six-sided die. In each case we may have some knowledge of the likelihood of various possible results, but we cannot predict with any certainty the outcome of any particular trial. Probability and statistics are used throughout engineering. In electrical engineering, signals and noise are analysed by means of probability theory, Civil, mechanical, and industrial engineers use statistics and probability to test and account for variations in materials and goods. Chemical engineers use probability and statistics to assess experimental data and control and improve chemical processes, It is essential for today’s engineer to master these tools. Important terms: a) Probability is an area of study which involves predicting the relative likelihood of various outcomes. It is a mathematical area which has developed over the past three or four centuries. One of the early uses was to calculate the odds of various gambling games. Its usefulness for describing errors of scientific and engineering measurements was realized. Engineers study probability for its many practical uses, ranging from quality control and quality assurance to communication theory in electrical engineering. b) Chance is a necessary part of any process to be described by probability. Sometimes that element of chance is due partly or even perhaps entirely to our lack of knowledge of the composition of every part of the raw material used to make bolts, and of the physical processes and conditions in their manufacture, in principle we could predict the diameter of each bolt. But in practice we generally lack that complete knowledge, so the diameter of the next bolt to be produced is an unknown quantity described by a random variation. Under these conditions the distribution of diameters can be described by probability and statistics. If we want to improve the quality of those bolts and to make them more uniform, we will have to look into the causes of the variation and make changes in the raw materials or the production process. But even that, there will very likely be a random variation in diameter that can be described statistically. Fundamental concepts: Probability as a specific term is a measure of the likelihood that a particular event will occur. Just how likely is it that the outcome of a trial will meet a particular requirement? If we are certain that an event will occur, its probability is 1 or 100 %. If it certainly will not occur, its probability is zero. The first situation corresponds to an event which occurs in every trial,
  • 3. whereas the second corresponds to an event which never occurs. At this point we might be tempted to say that probability is given by relative frequency, the fraction of all the trials in a particular experiment that give an outcome meeting the stated requirements. But in general that would not be right. Why? Because the outcome of each trial is determined by chance. Say we toss a fair coin, one which is just as likely to give heads as tails. It is entirely possible that six tosses of the coin would give six heads or six tails, or anything in between, so the relative frequency of heads would vary from zero to one. If it is just as likely that an event will occur as that it will not occur, its true probability is 0.5 or 50 %. As an illustration, suppose the weather man on TV says that for a particular region the probability of precipitation tomorrow is 4o%. Let us consider 100 days which have the same set of relevant conditions as prevailed at the time of the forecast. According to the prediction, precipitation the next day would occur at any point in the region in about 40 of the 100 trials. ( This is what the weather man predicts, but we all know that the weather man is not always right!) Although we cannot make an infinite number of trials, in practice we can make a moderate number of trials, and that will give some useful information. The relative frequency of a particular event, or the proportion of trials giving outcomes which meet certain requirements, will give an estimate of the probability of that event. The larger the number of trials, the more reliable that estimate will be. This is the empirical or frequency approach to probability. (Remember that “empirical” means based on observation or experience.) Example: 260 bolts are examined as they are produced. Five of them are found to be defective. On the basis of this information, estimate the probability that a bolt will be defective. Answer: The probability of a defective bolt is approximately equal to the relative frequency, which is 5/260 = 0.019. Another type of probability is the subjective estimate, based on a person’s experience. To illustrate this, say a geological engineer examines extensive geological information on a particular property. He chooses the best site to drill an oil well, and he states that on the basis of his previous experience he estimates that the probability the well will be successful is 30 %. (Another experienced geological engineer using the same information might well come to a different estimate.) This, then, is a subjective estimate of probability. The executive of the company can use this estimate to decide whether to drill the well. A third approach is possible in certain cases. This includes various gambling games, such as tossing an unbiased coin; drawing a colored ball from a number of balls, identical except for color, which are put into a bag and thoroughly mixed; throwing an unbiased die; or drawing a card from a well-shuffled deck of cards. In each of these cases we can say before the trial that a
  • 4. number of possible results are equality likely. This is the classical or ;”a priori” approach. The phrase “ a priori” comes from Latin words meaning coming from what was known before. This approach is often simple to visualize, so giving a better understanding of probability. In some cases it can be applied directly in engineering. Example: Two fair coins are tossed. What is the probability of getting one heads and one tails? Answer: For a fair or unbiased coin, for each toss of each coin P[heads] =P[tails] = ½ This assumes that all other possibilities are excluded: if a coin is lost that toss will be eliminated. The possibility that a coin will stand on edge after tossing can be neglected. There are two possible results of tossing the first coin. These are heads (H) and tails (T), and they are equally likely. Whether the result of tossing the first coin is heads or tails, there are two possible results of tossing the second coin. Again, these are heads (H) and tails (T), and they are equally likely. The possible outcomes of tossing the two coins are HH, HT, TH and TT. Since the results H and T for the first coin are equally likely, and the results H and T for the second coin are equally likely, the four outcomes of tossing the two coin must be equally likely. These relationships are conveniently summarized in the following tree diagram. In which each branch point (or node) represents a point of decision where two or more results are possible. Outcome H HH H T TT H TH T T TT Figure: Simple Tree Diagram Since there are four equally likely outcomes, the probability of each is 1/4. Both HT and TH correspond to getting one heads and one tails, so two of the four equally likely outcomes give this result. Then the probability of getting one heads and one tails must be 2/4=1/2 or 0.5. P[H]=1/2 P[T]=1/2 P[H]=1/2 P[T]=1/2 P[H]=1/2 P[T]=1/2
  • 5. In the study of probability an event is a set of possible outcomes which meets stated requirements. If a six-sided cube (called a die) is tossed, we define the outcome as the number of dots on the face which is upward when the die comes to rest. The possible outcomes are 1, 2,3,4,5, and 6. We might call each of these outcomes a separate event. Remember that the probability of an event which is certain is 1, and the probability of an impossible event is 0. Then no probability can be more than 1 or less than 0. If we calculate a probability and obtain a result less than 0 or greater than 1, we know we must have made a mistake. If we can write down probabilities for all possible results, the sum of all these probabilities must be 1, and this should be used as a check whenever possible. Sometimes some basic requirements for probability are called the axioms of probability. These are that a probability must be between 0 and 1. These axioms are then used to derive theoretical relations for probability. Definition: Axiomatic Probability: Let S be a sample space of a random experiment and A ⊂ S then the probability of an event A is a set function P(A) satisfying the following axioms. i) Axiom of positiveness, P (A) ≥ 0 ii) Axiom of certainty, P(S) = 1 iii) P(A1 U A2 U A3 U…..) = P(A1) + P(A2) +P(A3) +…. Where A1, A2,…..An are sequence of disjoint subsets of the sample space S Example: A Mathematics class for engineers consists of 25 industrial 10 mechanical, 10 electrical, and 8 civil engineering students. If a person is randomly selected by the instructor to answer a question, find the probability that the student chosen is a) an industrial engineering b) a civil engineering or electrical engineering. Solution: Denote by I, M, E and C the students majoring in industrial, mechanical, electrical, and civil engineering, respectively. The total number of students in the class is 53, all of which are equally likely to be selected. a) Since 25 of the 53 students are majoring in industrial, engineering, the probability of the event I, selecting an industrial engineering at random is
  • 6. P (I) = 25 / 53. b) Since 18 of the 53 students are civil or electrical engineering majors, it follows that P(C U E) = 18 / 53. Basic Rules of Combining Probabilities The basic rules of laws of combining probabilities must be consistent with the fundamental concepts. Addition Rule: This can be divided onto two parts, depending upon whether there is overlap between the events being combined. a) If the events are mutually exclusive, there is no overlap: if one event occurs, other events cannot occur. In that case the probability of occurrence of one or another of more than one event is the sum of the probabilities of the separate events. For example, if I throw a fair six-sided die the probability of any one face coming up is the same as the probability of any other face, or one-sixth. There is no overlap among these six possibilities. Then P [6] =1/6, P [4] =1/6. So P [6 or 4] is 1/6 + 1/6 = 1/3. This, then, is the probability of obtaining a six or a four on throwing one dir. Notice that it is consistent with the classical approach to probability; of six equally likely results, two give the result which was specified. The addition rule corresponds to a logical or and gives a sum of separate probabilities. Often we can divide all possible outcomes into two groups without overlap. If one group of outcomes is event A, the other group is the complement of A and is written _ A or A’. Since A and _ A together include all possible results, the sum of P[A] and P[ _ A ] must be 1. If P[ _ A ] is more easily calculated than P[A], the best approach to calculating P[A] may be by first calculating P[ _ A ]. Example: A sample of four electronic components is taken from the output of a production line. The probabilities of the various outcomes are calculated to be; P [0 defectives]=0.6561. P [1 defective] = 0.2916, P [2 defective] = 0.0486. P [3 defective] = 0.0036. P [4 defective] = 0.0001. What is the probability of at least one defective? Answer: If would be perfectly correct to calculate as follows: P[ at least one defective ] = P[ 1 defective ]+ P[ 2 defective ]+ P[ 3 defective ]+ P[ 4 defective ] = 0.2916 + 0.0486 + 0.0036 + 0.0001 = 0.3439.
  • 7. But is is easier to calculate instead: P [ at least one defective ] = 1- P [ 0 defective ] = 1 – 0.6561 = 0.3439 or 0.344. b) If the events are not mutually exclusive, there can be overlap between them. This can be visualized using a VENN diagram. The probability of overlap must be subtracted from the sum of probabilities of the separate events ( i.e, we must not count the same area on the Vann Diagram twice). Figure: Venn diagram The circle marked A represents the probability (or frequency) of event A, the circle marked B represents the probability (or frequency) of event B, and the whole rectangle represents all possibilities, so a probability of one or the total frequency. The set consisting of all possible outcomes of a particular experiment is called the sample space of that experiment. Thus, the rectangle on the Venn diagram corresponds to the sample space. An event, such as A or B, is any subset of a sample space. Set notation is useful: P[AUB]=P[Occurrence of A or B or Both], the union of the two events A and B P[A∩ B] = P[Occurrence of both A and B ], the intersection of events A and B. Then in above figure, the intersection A∩ B represents the overlap between events A and B. The following Venn diagram representing intersection, union, and completement. The cross- hatched area of Figure (a) represents event A. The cross-hatched area of Figure (b) shows the intersection of events A and B. The union of events A and B is shown on part (c) of the diagram. The cross-hatched area of part ( d ) represents the complement of event A’. A B A∩ B
  • 8. (a) Event A (b) Intersection ( C) Union (d) Complement Figure: Relations on Venn Diagrams If the events being considered are not mutually exclusive, and so there may be overlap between them, the Addition Rule becomes P(AUB) = P[A] +P[B} –P[A∩B] In words, the probability of A or B or both is the sum of the probabilities of A and of B, less the probability of the overlap between A and B. The overlap is the intersection between A and B. Example: A Mathematics class for engineers consists of 25 industrial 10 mechanical, 10 electrical, and 8 civil engineering students. If a person is randomly selected by the instructor to answer a question, find the probability that the student chosen is a) an industrial engineering b) a civil engineering or electrical engineering. Solution: Denote by I,M,E and C the students majoring in industrial, mechanical, electrical, and civil engineering, respectively. The total number of students in the class is 53, all of which are equally likely to be selected. a) Since 25 of the 53 students are majoring in industrial, engineering, the probability of the event I, selecting an industrial engineering at random is A A B A B A’ A
  • 9. P(I) = 25 / 53 . b) Since 18 of the 53 students are civil or electrical engineering majors, it follows that P(C U E ) = 18 / 53 . Addition theorem of Probability: The probability of occurrence of at least one of the two events A and B is given by P(AUB) = P(A) + P(B) – P(A∩B) Proof: To prove the theorem we consider the following Venn diagram From the venn diagram, we have Subtracting, we get P(AUB) – P(B) = P(A) – P(A∩B) ⇒⇒⇒⇒ P(AUB) = P(B) + P(A) – P(A∩B) Hence the theorem. The above theorem can be generalized for more than two events, for instance P(AUBUC) = P(A) + P(B) + P(C) - P(A∩B) - P(B∩C) - P(C∩A)+ P(A∩B ∩C) )A(BB)(ABand)A(BABA __ IUIIUU == )AP(BB)P(AP(B) and)AP(BP(A)B)P(AHence }AxA,x/x{Ai.e AofcomplementtheisAWhere _ _ __ _ IUI IUU = = ∉∈=
  • 10. If the sets are disjoint then the above results reduce to the form P(AUB) = P(A) + P(B) and P(AUBUC) = P(A) + P(B) +P(C). Corollary 1: For any event A, we have A = A U φ since A and φ are disjoint, it follows from the addition theorem that P( A )= P(A U φ) = P(A) + P(φ) ⇒ P(φ) = 0. Corollary 2: Then by definition of probability gives Example : Let us consider the Venn diagram of the following, which concerns the job offers received by engineering college graduates. The letter I and G stand for job offer from industry and a job offer from the government 0.12 Solution: P(I) = 0.18 + 0.12 = 0.30 P(G) = 0.12 +0.24 = 0.36 and P(IUG) = 0.18 + 0.12 + 0.24 = 0.54 we were able to add the various probabilities because they represent mutually exclusive events. S,inEofcomplementtheisEifisthatE,SEIf __ −= )(1 )( )()( )( )( )( _ _ EP SO EOSO SO EO EP −= − ==
  • 11. Here we erroneously used the third axiom of probability to calculate P(IUG), we would have obtained P(I) + P(G) = 0,30+0.36, which exceeds the correct value by 0.12. This error results from adding in P(I∩G) twice, once in P(I)=0.30 and once in P(G)=0.36 and we could correct for it by subtracting 0.12 from 0.66. Thus we get P(IUG) = P(I) + P(G) – P(I ∩ G) =0.30 +0.36 -0.12 =0.54. Example : Ravi 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 the probability that he gets an offer from company B is 0.6. If, on the other hand, 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(AUB) = P(A) + P(B) – P(A ∩ B)=0.8+0.6-0.5 = 0.9 . Example: If the probabilities are, respectively, 0.09, 0.15,0.21, and 0.23 that a person purchasing 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 U W U R U B)=P(G)+P(W)+P(R)+P(B) =0.09+0.15+0.21+0.23=0.68. Theorem: If A and A’ are complementary events then P(A)+P(A’)=1 Proof : Since AUA’ = S and the sets A and A’ are disjoint, then 1 = P(S) =P(AUA’) = P(A) +P(A’). Example: 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?.
  • 12. 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 the above theorem that P(E) = 1- 0.31= 0.69. Example: Suppose the manufacturer specifications of the length of a certain type of computer cable are 2000 ± 10 millimeters. In this industry, it is known that small cables is just as likely to be defective (not meeting specifications ) as large cable. That is , the probability of randomly producing a cable with length exceeding 2010 millimeters 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. i) What is the probability that a cable selected randomly is too large? ii) 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 larger , respectively. Then i) P(M) =0.99 and P(S) =P(L) = (1-0.99) / 2 = 0.005. ii) 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 then P(X ≥ 1990) = P(M) +P(L) =0.995. This also can be solved by using previous theorem P(X ≥1990)+P(X<1990)=1. Thus ,P(X ≥1990)=1-P(S) = 1-0.005 = 0.995. Conditional Probability: In many cases, the probabilities of two or more events depend on one another. That means, the happening of one event depends on the happening of another event. For example: To a merchant of umbrellas, the probability to get profit on a rainy day is more than the probability of getting profit an any other day. Clearly, the event of “ getting profit” depends on the event of raining.
  • 13. Definition: If A,E are any two events of a sample space S, then the event of “happening of E, after the happening of A” is called conditional event and is denoted by P(E/A). However, it is necessary to find the probability of an event E, given the supplementary condition that an event A has preceded it and it has positive probability. Definition:(General definition) If E and A are any events in S, P(A)> 0, the conditional probability of E given A is Example: A die is rolled, If the outcome is an odd number. What is the probability that it is prime? Solution: When a die is rolled, the sample space is S = {1,2,3,4,5,6} Let A = event of getting an odd number = {1,3,5} Let E = event of getting a prime number = {2,3,5} Then P(A)= 3/6 = ½; P(E)=3/6 = ½ and P(getting a prime, already which is an odd number) = P(getting a prime/ getting an odd number) = P(E/A) =(1/3)(2/1)=2/3. Example : The probability that a regularly scheduled flight departs on time is P(D)=0.83; the probability that it arrives on time is P(A) = 0.82; and the probability that it departs and arrives on time is P(D∩A) =0.78. Find the probability that a plane i) arrives on time given that it departed on time, and ii) departed on time given that it has arrived on time. Solution :i) The probability that a plane arrives on time given that it departed on time is P(A/D) = P(D∩A) / P(D) = 0.78/ 0.83=0.94 occurs)AeventP(given occur)AandEeventsP(both )( )( )/( = = AP AEP AEP I }5,3{=AE I 3/16/2)( ==AEP I )( )( AP AEP I =
  • 14. ii) The probability that a plane departed on time given that it has arrived on time is P(D/A) = P(D∩A) / P(A)= 0.78/ 0.82=0.95. Example : The concept of conditional probability has countless applications in both industrial and biomedical applications. Consider an industrial process in the textile industry in which strips or a particular type of cloth are being produced. These strips can be defective in two ways, length and nature of texture. It is known from historical information on the process that 10 % of strips fail the length test, 5% fail the texture test, and only 0.8 % fail both tests. If a strip is selected randomly from the process and a quick measurement identifies it as failing the length test, what is the probability that it is texture defective?. Solution : Consider the events L : length defective T: texture defective Thus, given that the strip is length defective, the probability that this strip is texture defective is given by Thus, the knowledge given by the conditional probability provides considerably more information than merely knowing P(T). Independent Events : Although conditional probability allows for an alteration of the probability of an event in the light of additional material, it also enables us to understand better the very important concept of independence or, in the present context, independent events. In the airport illustration previous example P(A/D) differs from P(A). This suggests that the occurrence of D influenced A and this is certainly expected in this illustration. However, consider the situation where we have events A and B and P(A/B) = (A). In other words, the occurrence of B had no impact on the odds of occurrence of A. Here the occurrence of A is independent of the occurrence of B. Definition : Two events A and B are independent if and only if P(B/A)= P(B) or P(A/B) = P(A). Provided the existences of the conditional probabilities. Otherwise, A and B are dependent. Multiplicative Rule : 0.08 0.1 0.008 P(L) L)P(T P(T/L) === I
  • 15. Multiplying the formula of Conditional probability Definition by P(A), we obtain the following important multiplicative rule, which enables us to calculate the probability that two events will both occur. Theorem: If in an experiment the events A and B can both occur, then provided P(A) > 0. Thus the probability that both A and B occur is equal to the probability that A occur multiplied by the conditional probability that B occurs, given that A occurs. Since the events A∩B and B∩A are equivalent, it follows from above theorem that we can also write P(A∩B) = P(B∩A) = P(B) P(A/B). Example : Suppose that we have a fuse box containing 20 fuses, of which 5 are defective. If 2 fuses are selected at random and removed from the box in succession without replacing the first, what is the probability that both fuses are defective?. Solution: We shall let A be the event that the first fuse is defective and B the event that the second fuse is defective; then we interpret A∩B as, the event that A occurs , and then B occurs after A has occurred. The probability of first removing a defective fuse is ¼; then the probability of removing a second defective fuse from the remaining 4 is 4/19.Hence P(A∩B )=(1/4)(4/19) =1/19. Example: One bag contains 4 white balls and 3 black balls, and a second bag contains 3 white balls and 5 black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from the second bag is black? Solution : Let B1, B2, and W1 represent, respectively, the drawing of a black ball from bag 1, a black ball from 2, and a white ball from bag 1. We are interested in the union of the mutually exclusive events B1∩B2 and W1 ∩B2. The various possibilities and their probabilities are illustrated in the following diagram. P(A)P(B/A)B)P(A =I
  • 16. Now P[(B1 ∩ B2 ) or (W1 ∩ B2)] = P(B1 ∩ B2 ) + P (W1 ∩ B2) = P(B1) P(B2/B1) + P(W1) P(B2/W1) = (3/7) (6/9) + (4/7) (5/9) = 38 / 63. Theorem: Two events A and B are independent if and only if P(A ∩ B) =P(A) P(B) Therefore, to obtain the probability that two independent events will both occur, we simply find the product of their individual probabilities. Example: A small town has one fire engine and one ambulance available for emergencies.The probability that the fire engine is available when needed is 0.98, and the probability that the ambulance is available when called is 0.92. In the event of an injury resulting from a burning building, find the probability that both the ambulance and the fire engine will be available. Bag1 4W 3B B1 3/7 W1 4/7 B2 6/9 P(B1∩B2)=(3/7)(6/9) P(B ∩B )=(3/7)(6/9) W2 3/9 P(B1∩W2)=(3/7)(3/9) B2 5/9 P(W1∩B2)=(4/7)(5/9) W2 4/9 P(W1∩W2)=(4/7)(4/9)
  • 17. Solution: Let A and B represent the respective events that the fire engine and the ambulance are available. Then P(A ∩ B) =P(A) P(B) =(0.98)(0.92) =0.9016. Example: An electrical system consists of four components as illustrated in the following diagram. The system works if components A and B work and either of the components C or D work. The reliability ( probability of working) of each component is also shown in diagram. Find the probability that a) the entire system works and b) the component C does not work, given that the entire system works. Assume that four components work independently. An electrical system Solution: In this configuration of the system, A, B and the subsystem C and D constitute a serial circuit system, whereas the subsystem C and D itself is a parallel circuit system.
  • 18. a) Clearly the probability that the entire system works can be calculated as the following: P(A∩B∩(CUD)) = P(A) P(B) P(CUD) = P(A) P(B) [1 – P(C’ ∩ D ‘ )] = P(A)P(B) [1 – P(C’ )P(D’)] = (0.9)(0.9)[1-(1-0.8)(1-0.8)] = 0.7776. The equalities above hold because of the independence among the four components. b) To calculate the conditional probability in this case, notice that Bayes’ Theorem: The general multiplication rules are useful in solving many problems in which the ultimate outcome of an experiment depends on the outcomes of various intermediate stages. Suppose, for instance, that an assembly plant receives its voltage regulators from three different suppliers, 60 % from supplier B1, 30 % from supplier B2, and 10 % from supplier B3. In other words, the probabilities that any one voltage regulator received by the plant comes from these three suppliers are 0.60, 0.30, and 0.10. If 95% of the voltage regulators from B1, 80 % of those from B2, and 65 % of those from B3 perform according to specifications, what we would like to know is the probability that any one voltage regulator received by the plant will perform according to specifications If A denotes the event that a voltage regulator received by the plant performs according to specifications, and B1,B2, and B3 are the events that it comes from the respective suppliers, we can write A = A∩[B1∪B2 ∪ B3 ] = (A ∩ B1) ∪ (A ∩ B2)∪ (A ∩ B3) and P(A) = P(A ∩ B1) + P(A ∩ B2) + P(A ∩ B3) .1667.0 7776.0 )8.0)(8.01)(9.0)(9.0( works)systemP(the D)'CBP(A works)systemP(the not work)doesCbutworkssystemP(the P = − = = = III
  • 19. Since A ∩ B1 , A ∩ B2 , and A ∩ B3 are mutually exclusive. Then, if we apply the multiplication rules to P(A ∩ B1) , P(A ∩ B2) and P(A ∩ B3) we get P(A) = P(B1) P(A / B1) + P(B2) P(A / B2) + P(B3) P(A / B3) and substitution of the given numerical values yields P(A) = (0.60)(0.95) + (0.30)(0.80) + (0.10)(0.65) = 0.875 for the probability that any one voltage regulator received by the given plant will perform according to specifications. To visualize this result, we have only to construct a tree diagram like that of the following figure, where the probability of the final outcome is given by the sum of the products of the probabilities corresponding to each branch of the tree. Tree diagram for example dealing with 3 suppliers of voltage regulators In the preceding example there were only three alternatives at the intermediate stage, but if there are n mutually exclusive alternatives B1,B2,B3,…….Bn at the intermediate stage, a similar argument will lead to the following result, sometimes called the rule of elimination or the rule of total probability Theorem: If B1,B2,B3,…….Bn are mutually exclusive events of which one must occur, then ∑= = n 1i ii ))P(A/BP(BP(A)
  • 20. To Visualize this result we have only to construct a tree diagram like that of above figure, where the probability of the final outcome is again given by the sum of the products of the probabilities corresponding to each branch of the tree. To consider a problem that is closely related to the one we have just discussed, suppose we want know the probability that a particular voltage regulator, which is known to perform according to specifications, come from supplier B3. Symbolically, we want to know the value of P(B3/A), and to find a formula for this probability we first write Then, substituting P(B3) P(A/B3) for P(A∩B3) and and we get expresses P(B3/A) in terms of given probabilities. Substituting the values Which P(A) )BP(A /A)P(B 3 3 ∩ = P(A)for))P(A/BP(B 3 1i ii∑= ∑= = 3 1i ii 33 3 ))P(A/BP(B ))P(A/BP(B /A)P(B
  • 21. Note that, the probability that a voltage regulator is supplied by B3 decreases from 0.10 to 0.074 once it is known that it perform according to specifications. The method used to solve the preceding example can easily be generalized to yield the following formula called Bayes’ Theorem .Bayes’ Theorem: If A1, A2, A3, …….An are disjoint sets whose union is the sample space S and E is any event in S then Proof: Let S = A1 U A2U A3U …….U An Let E=E∩S = E ∩ (A1 U A2U A3U …….U An) =(E∩A1) U (E∩A2) U (E∩A3) U …. U (E ∩ An) where E∩A1,E∩A2…E ∩ An are disjoint. Hence by addition theorem We get P(E) = P(E∩A1) +P (E∩A2) + P(E∩A3) + …. + P(E ∩ An) Using Multiplication theorem P(E) = P(E/A1) P(A1) +P (E/A2) P (A2) + P(E/A3) P(A3) + ………. + P(E / An) P( An) Example: A factory has three production lines I, II and III contributing 20 %, 30 % and 50 % respectively, to its total output. The percentages of substandard items produced by lines I, II and III are, respectively, 15, 10 and 2. If an item chosen at random from the total output is found to be substandard, what is the probability that the item is from line I?. Solution: We represent the above problem with the help of the following tree diagram 074.0 )65.0)(10.0()80.0)(30.0()95.0)(60.0( )65.0)(10.0( )/( 3 = ++ =ABP ∑= = n 1i ii ii i ))P(E/AP(A ))P(E/AP(A /E)P(A ∑= === n 1i ii iiiii i ))P(E/AP(A ))P(AP(E/A P(E) ))P(AP(E/A P(E) E)P(A /E)P(A I
  • 22. By Bayes formula 069.0 15.02.0 ))P(A/EP(E ))P(A/EP(E /A)P(E 3 1i ii 11 1 = ++ == ∑= xxx x