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ENM 207 Lecture 7 Example: There are three columns entitled “Art” (A) ,“Books” (B) and “Cinema” (C) in a new magazine. Reading habits of a randomly selected reader with respect to choose columns are: read regularly probabilit y A B C A B .14 .23 .37 .08 AC .09 B C .13 A B C .05 (The probability of reading only A column is 14%) What is the probability of column A given that they red column B? P A B 0.08 P A \ B 0.348 PB 0.23 What is the probability of reading A given that they are reading B or C columns? P A \ B C P A B C PB C P A B A C P B P C P B C 0.08 0.09 0.05 0.12 0.255 0.23 0.37 0.13 0.47 What is the probability of reading column A given that they are reading at least one column? P A \ reads at least one P A \ A B C P A A B C P A 0.14 0.286 P A B C P A B C 0.49 A B C The intersection of A with union of ABC is equal A itself. What is the probability of reading A or B columns given that they read C columns? P A B \ C P A B C P A C PB C PC PC P A C P B C P A C B C PC P A C P B C P A B C PC 0.09 0.13 0.05 0.17 0.37 0.37 Example: Mendenhall 3.63 solution • There are five suppliers and a company will choose at least 2 suppliers, therefore, we must find the number of ways to select 2, 3, 4 and 5 suppliers from the 5. • At least 2 means that , a company can choose 2 or 3 or 4 or 5 suppliers from 5. • The total number of obtions are 5 5 5 5 5! 5! 5! 5! 26 2 3 4 5 2! 3! 3! 2! 4! 1! 5! 0! Example:3.65 • Define the following events : (A) Part is supplied by company A ( B) Part is supplied by company B ( C) Part is defective From the problem, P(A)=0.8, P(B)=0.2, P(C\A)= 0.05 , P(C\B)= 0.03 We know the given part is defective and we want to find The probability it come from company A And the probability it come from company B P(A\C)=? , P(B\C)=? P A C P A PC \ A 0.8 0.05 0.04 C is the defective part. It can come from company A or company B. That is We need P(BC) PB C PB PC \ B 0.2 0.03 0.006 PB \ C PB C 0.006 0.13 PC 0.046 As a result we say that the company A is more likely to have supplied the defective part. Total Probability And Bayes Theorem Partitions Total Probability And Bayes Theorem A partition of the sample space may be defined like this. S B1 B2 A B 1 A B 2 A B 3 B3 B4 Definition: If B1,B2,,,,Bk are disjoint subsets of S ( mutually exclusive events ) , and if B1 B2 B3,,,,,,, Bk=S then these subsets are said to form a partition of S. If k events , BI ( I=1,2,,,k) , form a partition and A is an arbitrary event with respect to S, then we may write Total Probability Law A A B1 A B2 .... A Bk So that P A P A B1 P A B2 .... P A Bk Since the events (ABi) are pairwise mutually exclusive events P A k PBi P A \ Bi i 1 is called “total probability law” Bayes Theorem • Another important result of the total probability law is known as “Bayes Theorem”: If B1,B2,,,,Bk constitute a partition of the sample space S and A is an arbitrary event on S , then for r = 1,2, ,k P B r P A \ B r P B r \ A BAYES k P Bi P A \ B i i 1 THEOREM P B r \ A P B r A P B r P A \ B r k P A P Bi P A \ Bi i 1 The numerator is a result of multiplication rule and the denumerator is a result of total probability law. Ex: 2.35 page 56 (Montgomery) A manufacturer of telemetry equipment microprocessors from 3 different facilities. According to the historical data which are recorded by manufacturer company : Supplying facility Fraction defective Fraction supplied by 1 0,02 0,15 2 0,01 0,80 3 0,03 0,05 B1 B2 A B3 The director of manufacturing randomly selects a microprocessor , after test procedure , finds that it is defective . compute the probability that this defective prod came from facility 3. Let A be the event that an item is defective B1 be the event that the item comes from facility 1 B2 be the event that the item comes from facility 2 B3 be the event that the item comes from facility 3 We can find this probability using BAYES THEOREM. PB3 \ A PB3 P A \ B3 PB1 P A \ B1 PB2 P A \ B2 PB3 P A \ B3 PB3 A 0.05 0.03 3 P A 0.15 0.02 0.80 0.01 0.05 0.03 25 Example: • Three machines A,B,C produce respectively 50%, 30%, 20% of the total number of items of a factory . • The percentages of defective output of these machines are 3% , 4% , 5%,respectively. If an item is selected at random, find the probability that the item is defective. • Let X be the event that an item is defective. Which theorem or which rule is used to solve this problem? Total probability law. P X P A P X \ A PB P X \ B PC P X \ C 0.50 0.03 0.30 0.04 0.20 0.05 0.037 Another way to solve this problem: Tree Diagram 0.03 the probability of any part produced by machine A is %50 0.50 D defective of machine A A R reliable 0.04 0.30 D B R 0.20 C 0.05 D R Example: • Consider the factory in the preceeding example. Suppose an item is selected at random and is found to be defective. Find the probability that the item was produced by machine A; • What is the P(A\X)? • Which theorem or which law can be used to solve this problem? P A \ X P A P X \ A P A P X \ A PB P X \ B PC P X \ C 0.50 0.03 15 0.50 0.03 0.30 0.04 0.20 0.05 37 Ex: 4.11 p 62/schaums outline series In a certain college, 25% of the students failed math, 15% of the students failed chemistry, 10% of the students failed both math and chemistry. A student is selected at random. a) If he failed chemistry what is the probability that he failed math? b) If he failed math, what is the probability that he failed chemistry? c) What is the probability that he failed math or chemistry? Let M students who failed math C students who failed chemistry PM 0.25 PC 0.15 PM C 0.10 1 2 3 PM C 0.10 10 PM \ C PC 0.15 15 PC M 0.10 10 PC \ M PM 0.25 15 PM C PM PC PM C 0.25 0.15 0.10 0.30 SYSTEM RELIABILITY System- electronic, mechanical or a combination of both –are composed of components. A component of a system is represented by a capital letter. Two system each composed of tree components A; B; C are shown below. Systems according to their components connections can be classified in two groups. Such as series and parallel. A A B C B A)series system C B)parallel system Definition: • If the system fails when any of the components fails, it is called a series system. A B C •If the system fails only when all of its components fail, it is called a parallel system. A B C D A B C E This system is composed of five components A, B, C, D and E as shown above. Components D and E are from a two-component parallel system. This subsystem is connected in series with A, B and C. subsystem subsystem A B D E F G C Two parallel subsystems are connected in series. The first parallel subsystem contains three componenets such as A, B and C. The second contains two series subsystems: the first composed of D and E, the second composed of F and G. The reliability of a series system is P(system functions ) = P(all componenets function) The multiplicative rule can be applied because the components operate independently of each other in a series system. If there is k components in a series system P(series system functions) = P( A functions) . P(B functions). . P(K functions) PA PB PC .... PK where pI is the probability that ith component functions, I=A,B,,,K The reliability of a parallel system containing k components can be calculated in a similar manner. Since a parallel system will fail only if all components fail, P parallel system fails 1 PA 1 PB ....1 PC Where pI is the probability that the ith component functions, i=A,B,,,,,K. Example: A Given that PA=0.90 PB=0.95 B C PC=0.90 find the reliability of the series system shown in this figure. Psystem functions 0.90 0.95 0.90 0.7695 Summary of system reliability 1. If the system is in series form that all components are connected as a series. Pseries system functions n Peach component i If there are k components like A1,A2,,,Ak Pseries system functions k P Ai i Because a series system will fail if any one of its components fail. functions 2. If the system is parallel and also if it contains k components, k P parallel system fails (1 P Ai ) i Because as you know a parallel system will fail only if all components will fail. From this equation k P parallel system functions 1 i 1 P Ai Note: These formulations can be used to calculate the reliabilities of seriesm systems , parallel systems , or any combinations of them. These systems must satisfy the assumption that the components operate independently. Example: 17.22 : First subcircuits involves comp. A,B,C in parallel A P(first subcircuits)=1-(1-PA)(1-PB)(1-PC) P(A)=0.95, P(B)=0.95 , P(C)=0.90, P(D)=0.90, P(E)=0.98 B C P(second subcircuits)=1-P(D)P(E) D E . If the subcircuits are connected in series subsystem subsystem A B D E F G C P(two subcircuits are connected in series)=P(first subcircuits)P(second subcircuits) If two subcircuits are connected in parallel A B C D E two subcircuits are connected in parallel)=1-(1-P(first scir.))(1-P(sec.scir.))