 Now, having discussed the reliability of a component or a device, where we are all assuming that it is a single piece, a single component. Let me now talk about reliability of systems, where there are more than one component. So, here again the treatment will be simple. So, that is why I have studied in the beginning only. The very simple cases will be considered, but once you learn the basic technique, then you can always divide, you know, break up a complex device into smaller systems and then you can try to compute the reliability of the whole system. So, let us see, we can, now here I am just beginning with this simple case that two components are hooked up in series. So, this is how they are, C 1 and C 2, they are two components and they are in series and so, in order for the system to work, both components must be functioning. They are in series, may be performing different tasks for the whole device and so, they both have to function. If any one of them fails, then the system will fail. So, this is the whole idea, they are working in series. Now, we also make the assumption and of course, this is important otherwise things will get complicated and we have learnt methods for handling dependence also, but right now we just assume the independence to show you the, how to develop, how to compute the reliability of the system. So, if they are functioning independently, then the reliability R T of the system can be obtained in terms of the reliability R 1 T and R 2 T of the two components. So, I am just denoting the reliability of the first component by R 1 and the reliability of the second component. I should probably write just this small t to choose the value of the, yeah. So, therefore, I can compute the reliability of the system in terms of the reliability R 1 T and R 2 T of the two components and this is a simple computation, we have already derived so many times. So, here you are asking for probability T greater than T and this would be, this can, this what it means is that your T 1 should be get that means, T 1 is the lifetime of the first component and T 2 is the lifetime of the second component. Then, we are asking for probability T 1 greater than T and probability T 2 greater than T, both must be functioning up to time t. If we are saying that the system for the system, the functioning time is greater than or equal to t, fine. So, now, because of independence, this joint probability can be written as probability T 1 greater than T into probability T 2 greater than T, right. This is because we assume that the two components are functioning independently and so, this is R 1 T into R 2 T, right. Now, because these are probabilities, so they are each of the numbers is less than 1. So, therefore, this product would be less than the smaller of the two, right. Because if R 2 T is less than R 1 T, then I am multiplying R 2 T by number less than 1. So, the whole product is still less than R 2 T, right. Similarly, if R 1 T is the minimum of the two, then I am multiplying R 1 T by a number which is less than 1 and therefore, the product is again less than R 1 T. So, essentially what we are saying is that the reliability, that function R T, here again I should use small t. So, R T is less than or equal to minimum of this. So, that means the reliability goes down, right. If you have components hooked up in series, then the reliability of the system goes down, right. Because this is less than or equal to minimum of the two. So, whatever the numbers, we do two numbers here, this will be R T will be smaller than the minimum of the two numbers at any time t, right. Now, one can now generalize this result and to say that if n components functioning independently are connected in series and if the ith component has reliability R i t, then the reliability R T, I have this habit of writing capital T, I do not know. Then reliability R T of the system is given by the product of the individual reliability, because we are assuming that the components are functioning independently of the other components. So, therefore, you have this general formula and you see the moment you have the components hooked up in series as many components, the lower the reliability, because you require all of them to be functioning for the system to function. Now, consider the case when n is equal to 2 and the failure law for the ith component is exponential. So, then your reliability for the system and two components will be e raise to minus alpha 1 t into e raise to minus alpha 2 t, which is this is e raise to minus alpha 1 plus alpha 2 t. And therefore, the pdf for the system for the failure time of the system would be minus r prime t, which is alpha 1 plus alpha 2 e raise to minus alpha 1 plus alpha 2 t. So, this is negative exponential with parameter alpha 1 plus alpha 2. So, again just repeating what we have already learned that if you have two exponentials distributions and the corresponding random variables are independent, then the sum would be sorry I mean no here we are asking for you are taking the. So, the pdf becomes negative exponential where the parameters get added up. So, this is exponential essentially this is. So, I have written it down here alpha 1 plus alpha 2 into e raise to minus alpha 1 plus alpha 2 t. So, this is what happens if you have two components and both exponentials both have the failure law exponential failure law and they are functioning independently. Then for the system if you want to make the computation for the reliability then it will be the parameters added up. So, it would be alpha 1 plus alpha 2 t and e raise to minus this and the pdf for the time to failure for the system would be exponential negative exponential distribution with parameters alpha 1 plus alpha 2. So, now we can look at some more examples and then we will look at some more another kind of system which is when you have components arranged in parallel. I have taken this example you know from a book which got published in 1961. So, this is Bazewski reliability theory and practice Prentice Hall. So, may be the book is not available now, but I have basically chosen the example because you know these figures are not easily available you know the failure rate for a silicon transistor composition transistor and so on. So, just for that reason and I wanted to show you the numbers because you see what we are saying here is. So, let me read out the problem first. So, you consider an electronic circuit consisting of 4 silicon transistors, 10 silicon diodes, 20 composition resistors and 10 ceramic capacitors in continuous series operation. So, they are all hooked up in series and suppose under certain stress conditions that is prescribed voltage, current and temperature each of the items has the following constant failure rate. So, that means the failure law is exponential. So, for silicon diodes it is in hours. So, the parameter is 0.0002 that means if you convert this into that means the mean failure time will be how much 1, 2, 3, 4, 5, 6. So, you will have to write 1, 0.0002. So, it is a very fairly large number. So, this exactly so I thought that since you know Bazowski has somewhere got this data from and so we can use it and actually the example appears in Meyer's book which for which the reference I will give you at the end of the lecture. So, anyway so the mean failure time is you know millions of hours will be there 2, 4, 6 yes. So, thousands of hours you can say fine. So, similarly these are the various numbers. So, the parameters that means each has a exponential failure law follows exponential failure law and we are assuming that they are hooked up they are they are failure I mean they are functioning independent of each other. So, therefore, just now as we saw that we just add up the parameters to get the distribution for the parameter for the failure law for the whole system and that will also be exponential. We just saw it can be easily shown that of course I showed it you for 2, but the same thing will easily can be shown for any number of. So, if you have lot of components many components hooked up in series each is following an exponential failure law then the when you look up failure law for the whole system then that will be simply again exponential with the parameters added up. So, you have how many you have 10 silicon diodes and so the parameter is this. So, therefore, 10 times the parameter for the silicon diodes and silicon transistors they are 4 of them 4 silicon transistors. So, 4 times this parameter which is 0.001 plus 20 times you have 20 composition transistors. So, this 20 times this plus 10 ceramic capacitors 10 times this. So, the parameter for the exponentially distributed see the time to failure for the entire circuit is exponentially distributed and the parameter will be equal to since they are 4 of them. So, 10 into see the numbers are given to you. So, this adds up to 0.001. So, earlier in the computation I had written 4 0s, but actually when you do the addition multiplication and addition it will come out to be 0.001. So, that is your mu and thus for a 10 hour period of operation the probability that the circuit will not fail will be e raise to minus mu into 10. So, mu 10 right. So, the time period whatever the time period the parameter gets multiplied by that for the corresponding parameter during that period. So, e raise to minus 0.001 into 10 which is e raise to minus 0.001. So, the final answer was given correctly which is 0.999 and therefore, your e t will be 10000 hours. So, therefore, probability is very high obviously, because these diodes and capacitors have life time mean life time in 1000s of hours. So, obviously, for 10 hour period you do not expect the system to fail. So, the probability is very high. So, therefore, we would expect the system to with high probability the system will continue to function for 10 hours without any failure this is the idea. So, again you may say simple examples, but just to drive home the point that this is kind of things you can consider. Now, other system that we would like to consider is a parallel system. So, here the system fails to function only if the of all the components fail. So, you know so diagrammatically you can depict this for two components. If you have the components arranged in parallel then you see it is like this. So, the input comes and then you have either it can go this way or it can go this way. So, the system will fail to function only if both of them fail. Because as long as one of them is functioning the things can be the input can go this if this fails then this will go this way and then it will go out this way. So, still the operation will be performed and if this fails then it will go this way and so the operation will be still be performed. So, therefore, for the system to fail both of them have to fail. So, that is what we mean when we say that all the components have to fail. And again under independence. So, if we are saying that they function independent of each other and that is what is expected if you have in parallel then each component to functions parallely independently of the other. So, then if you want to compute the I do not know why I keep writing capital T here. So, this is if you want to compute the reliability for the system then this is probably T greater than T which is 1 minus probability T less than or equal to T. So, then in that case this is what you want that 1 minus probability T 1 less than or equal to T and T 2 less than or equal to T both of them should not be functioning by time T. So, therefore, because of independence you would write this as a product and therefore, this becomes. So, probability T 1 less than or equal to T is 1 minus r 1 T because r 1 T is probability T 1 greater than T. So, this will be 1 minus r 1 T into 1 minus r 2 T. And now when you open out multiply the two terms and then you get this and so this reduces to r 1 T because 1 minus 1 cancels out r 1 T plus r 2 T minus r 1 T into r 2 T. So, this is the expression for the and of course. So, once we get this expression which I can write because 1 cancels with the minus 1. So, you will be left with r 1 T plus r 2 T minus r 1 T r 2 T. Now, you see that this is the probability this is the reliability for the two systems because this is r 1 T into r 2 T and we are assuming that the two systems function independently. So, therefore, you see that this is a number which is less than r 1 T and r 2 T both. So, considering see for example, r 1 T is larger than r 2 T then this whole number is bigger than r 1 T. So, if r 1 T is the maximum of r 2 and r 1 then this number this whole number because r 2 minus r 1 T r 2 T is something non negative. And therefore, r 1 T plus something non negative this is going to come out to be and hence I can immediately conclude. So, that is what I have written that since r 1 T r 2 T is less than or equal to both r 1 T and comma r 2 T. Therefore, it follows that your reliability when you have you know system two components working in parallel. See, I have shown you that the two systems are working in parallel. So, in that case the reliability of the system become is greater than or equal to max of the reliability of the two components. And so that immediately shows that systems composed of components functioning independently in parallel the reliability will be higher than the reliability of the of each of the components that are in parallel. And so parallel components are often used to increase reliability. And if you want to now generalize it to n components which are functioning in parallel then this will be r T into 1 minus of 1 minus r 1 T into 1 minus r 2 T into 1 minus r n T. So, the same principle will be used and you can show that. So, this is the important thing that when two components. So, that means here we are considering the case when two components are working in parallel right. And the system has to fail only when both of them fail because all components have to fail in that case the reliability of the system will be greater than or equal to reliability of both the components. So, therefore, you know working in parallel having components in parallel and having components in series. So, this is the basic the way you make up the devices. And then you can as I said you can decompose them and you know into smaller these things where you can consider components arranged in parallel and components arranged in series then put them together. So, I will try to show you some more examples of you know systems you know system of components arranged in different orders. So, let us see take consider the example where two components are in parallel and each of whose failure law is exponential distribution right. And of course, we are assuming that the first component as parameter alpha 1 the other one as alpha 2 then the reliability of the system and since they are in parallel. So, the reliability is given by the formula we just obtained r 1 t plus r 2 t minus r 1 t into r 2 t because they are functioning independent of each other. And therefore, this will be your reliability function and then the E t the expected time to failure for the system would be because you know when you take the expectation you will be integrating each of them separately this t into d t this t into d t the integral of this 0 to infinity t into this each of them is exponential distribution. So, it will be 1 upon alpha 1 plus 1 upon alpha 2 minus 1 upon alpha 1 plus alpha 2. And you can now since you have the whole machinery with you you can do all any computation that you want to do once you know the functional form of the you know of the reliability function you can make these computations. Now, again just to drive home the point that parallel arrangements of components definitely increase the reliability of the system. And I have taken this example from Meyer's book which again is a very old one, but a very good one and so see the thing is that yes. So, anyway I have just given the reference I will give you the reference, but the book may not be easily available does not matter. Suppose three units are operated in parallel assume that each has the same constant failure rate alpha equal to 0.01. So, they are identical components. So, all have the exponential failure law with parameter 0.01. Now, hence reliability of each unit for a period of 10 hours is e raise to minus 0.01 into 10 which is e raise to minus 0.1 which is 0.905 or about 90 percent. So, if each component is functioning you know by itself then the reliability is for in 10 hours period that it will not fail is you know 90 percent. Now, how much of an improvement can be achieved in terms of increasing reliability of the system by operating three such units in parallel. And so by our formula this is of course I am not looking at the expanded form this is 1 minus 1 minus p probability of t less than t raise to 3. So, that is what will be this is what you see that I got after opening up the brackets, but if you do not do it see this whole thing you wrote as 1 minus 1 minus 1 minus r 1 t into 1 minus r 2 t. So, this was the formula which then we opened up and then the 1 1 got cancelled and so on. So, I am just and since they are identical. So, it would be the same function. So, it will be 1 minus r 1 t square. So, in our case it will be cube and so this is 1 minus 0.905 because r this came out to be 0.905. So, 1 minus of that raise to 3 and that comes out to be this and therefore, this is equal to 0.99914 and therefore, the reliability has gone up to 99.9 percent. So, the numbers drive home the point and that is why it is important that you know individually if you just had one component in the system then for the probability of its operating for 10 hours would be only 90 percent. You expect 90 percent of the time it will be functioning still at the end of 10 hours period, but if you have 3 in parallel then you will almost be sure that the device with 3 parallel components will still be functioning at the end of 10 hour period. So, this is the idea and therefore, the reliability can be improved upon, but as I said that it has to be versus reliability versus cost, reliability versus the volume of the device and so on. So, you cannot just go on having components in parallel. Now, see the thing is that as I was saying that I discussed the very basic arrangements for you basic systems and you can have things like you know series parallel. So, you have these 3 components here in series, these 3 components in series and then parallel. So, there is no problem because you can compute the reliability for this and for this and then you know how to compute the reliability for the parallel, because when they 2 are parallel. So, you just have to you know iteratively do this arrangement you compute this which will be the product and this will be the product and then it will be you know 1 minus of. So, what we have been doing so far right. Similarly, if you have a arrangement parallel in series then these are parallel. So, you can compute the reliability of this you can compute the reliability of this. So, this will be form your R 1 and this will form your R 2 and then you are doing it in series. So, that is what I meant that all complex devices can be broken up into you know whether either they are series parallel, parallel series and so on and then you want to put them together. So, most of the time you should be able to come up with a reasonable functional form for the reliability of the whole device and you know things can just and of course I will be discussing a few problems like this in the exercise which will follow. So, that exercise on problems related to whatever we have discussed about reliability theory. So, let me give you the references that are the books that I have been referring to all along in this course and yes I agree that some of them are out of print. But in any case the idea is that even though like this is a 76 edition and that one is 71 edition of course Hillier and Lieberman 8th edition has just come out and I think even the 9th one may be ready. But so anyway the thing is to get the basic material I had to refer to these old books, but certainly substitutes will be there and now lot of material is available on the net. So, you just have to type the word that you would want whatever subject matter you want and lot of things come out. So, therefore, but in any case I just want to refer to these books because I have used material from these books. So, the first one is introduction to probability models 10th edition by Sheldon Ross. This is Elsevier academic press and this I think is a 2010 edition and may be see this one will keep coming out with new edition. So, therefore, no problem getting a copy of this book. Operations research principle and practice Don T. Philips, A. Ravindran and James Holberg. This is 76 book, but some treatments are some topics have been treated very well and so I have used lot of examples also from here and I have whenever I use figures I have referred to that part also. So, this is 76 book then introduction to operations research by Hillier and Lieberman 8th edition McGraw Hill and yes I do not remember the edition for this particular year, but this has been coming out with new edition. So, therefore, no problem and this is a McGraw Hill. So, cheap edition should be available easily and the fourth one that I have used is introductory probability and statistical applications. Second edition Paul Mayer again this is a classic book and very neatly and simply the material has been presented. So, reliability theory portion I have used this book and this is an Edison-Wesley 1971 book. So, in any case this is what my sources were and now you can as I told you Google search for any subject matter that you need and I hope you get interested enough in the topic to read more. Now, let me just discuss the last exercise 11 with you which is based on probability theory. Let us just look at question 1. Suppose that T the time to failure of an item is normally distributed with E t as 90 that is mu hours and standard deviation 5 hours. In order to achieve reliability of 0.9, 0.95 and 0.99 how many hours of operation may be considered. So, now you know the reliability function this is 1 minus phi of T minus mu upon sigma. So, this would be. So, for example, you put it equal to 0.9 and you know you do not know mu or mu is given to you sigma is also given to you. So, now, from the normal tables you are looking for phi T minus mu I have done problems with you like this is equal to 0.01 or sorry 0.1 1 minus 0.9 is 0.1. So, then given mu and sigma you can look up the tables and for corresponding to 0.1 what is the value here and then corresponding value of T will be available. So, similarly when you put 0.95, 0.99 you can accordingly get the values of T. So, how many hours of operation may be considered you can answer for all the three values of the reliable level. Question 2, suppose that the life length of an electronic device is exponentially distributed it is known that the reliability of the device for a 100 hour period of operation is 0.9. How many hours of operation may be considered to achieve a reliability of 0.95. So, first you the first data that is given to you you will compute the parameter for the exponentially for the exponential failure law and then once you get the alpha then you can compute the time corresponding to the reliability level of 0.95. Question 3, suppose that the life length of a device has constant failure rate C naught for 0 less than T less than T naught and a different constant failure rate C 1 for T greater than or equal to T naught. Obtain the pdf of T the time to failure and sketch it the sketching part I will leave to you, but see here all that is saying is that you have. So, up to T naught you have one failure law and then you after T naught you have another failure law. So, therefore, but at the point T naught the too much must meet right. So, therefore, you what you will say that C naught. So, you see the required density function will be C naught e raise to minus C naught T as long as T is between 0 and T naught because this is the failure law. So, the rate of failure is C naught and this exponential failure law. So, therefore, for the T less than or equal to T naught that means lying between 0 and T naught you will write this and for T greater than T naught it will be C 1 e raise to minus C naught T naught plus. So, this is e raise to I should yeah and then you see C 1 T naught minus T. So, since T is greater than T naught. So, when you write it as T minus T naught then it will be minus here. So, anyway so this is now this will be the probability density function. So, to show that f T is a pdf we have to show that it is integral from 0 to infinity would be equal to 1 and in particular. So, the first part we will integrate from 0 to T naught and the second part of the function f T we will integrate from T naught to infinity and therefore, the calculation shows that the integral comes out to be equal to 1. So, this is the required pdf right for question 3 and then you can try to sketch it. Notice that the failure rate Z is given by so now 4 is a special case of 3. So, here your time between 0 and A the failure rate is 0 and for T greater than A it is C. So, it is constant. So, again it is the same thing as 3 except that now your C naught is 0 and for C 1 is C and so therefore, you can because we have obtained the form for the failure law in 3. So, now just substituting the special values you can compute you can compute the so let me just see. So, here you have to find the pdf associated with T the time to failure. So, that you can find out because I have already obtained it for you and now you have to put in the values of C naught and C and of course, your T naught is A and evaluate E T. So, then you can also evaluate by actual integration. Now, let us look at question 5. Suppose that the failure law of a component has the following pdf. So, this is F T is r plus 1 into A r plus 1 divided by A plus T raise to r plus 2 T greater than 0. So, for what values of A and r is the above a pdf? We can look at it I can look at the. So, suppose that the failure law of a component has the following question 5. Suppose that the failure law of a component has the following pdf F T is r plus 1 into A raise to r plus 1 divided by A plus T raise to r plus 2 T greater than 0. So, it should not be difficult because all you have to do is to and so you have to say that for what values of A and r is the above a pdf and I think if my this thing is right, then it really does not matter it is a pdf for all A and r because you see you simply have to write the integral as A plus T raise to r minus r minus 2 and then integrate from 0 to infinity and then you just have I think the values of A will cancel out and r will. So, you will get the integral as equal to 1 without specifying any values for A and r. Anyway, so now you can do this and then you can obtain an expression for the reliability function and hazard function and show that the hazard function is decreasing in T. So, I will let you do this problem. Now, question 6 suppose that the failure law of a component is a linear combination of k exponential failure laws that is the pdf of the failure time is given by f t is equal to sigma j varying from 1 to k C j beta j e raise to minus beta t T and beta j beta j for all j is positive. So, for what values of C j is the above a pdf. So, now when you integrate because this is a finite sum. So, I can take the integral inside and so when you integrate beta j integration integral of 0 to infinity beta j e raise to minus beta it should be there is a missing. Yeah, let me write it down I think your f t should be sorry this happens the typing errors they creeping. So, this is j varying from 1 to k and this is C j beta j e raise to minus beta j t this is for t greater than 0 and beta j greater than 0 for all j. So, this is beta j make the correction and so when you integrate this from 0 to infinity d t you will be integrating all separate one this is 0 to t d t. So, therefore, now this is equal to 1. So, therefore, this whole thing will add up to sigma C j j varying from 1 to k. So, this is the condition that all and of course, C j's have to be non-negative and then you say that sigma C j well actually it is a linear combination. So, I will just to be on the safe side say that C j's be non-negative and they add up to 1. So, in that case so this becomes a convex combination in that case of all these different exponential laws and so this is also again a PDF this will be a PDF. Then obtain an expression for the reliability function and hazard function obtain an expression for the mean time to failure. So, see here again because it is a summation. So, you will have to integrate if you have to compute this integral the same principle you will use for the reliability thing you will have to. So, r t when you do t to infinity so you again can integrate separately and so it will be a convex combination of or the separate reliability functions r 1 r 2 r k. So, it will be C 1 r 1 plus C 2 r 2 plus C k r k. So, straight forward there is not and then answer b and c beta j is equal to beta for all j and of course obtain an expression for the mean time to failure. So, the mean time to failure would be see for each of them it is 1 by beta j. So, it will be summation. So, the mean time to failure that means your e t will be summation C j beta j j into 1 to k. Now, let us go to the next problem then a question 7 expected life time is 3 by 2 years. So, which means that lambda is 2 by 3 again exponential failure law and probability that it is still functioning after 2 years will be e raise to minus. So, it will be e raise to minus lambda t which is minus 4 by 3 after 2 years. So, this will be e raise to minus 4 by 3. Now, you want the probability that 2 still functioning after 2 years. So, 2 still functioning after 2 years at least so that means you may have after 2 years either 2 of them are functioning or 3 of them. So, when 2 of them are functioning it will be 3 C 2 e raise to minus 4 by 3 into 2 because 2 of them are functioning and 1 of them is not functioning. So, it will be 1 minus e raise to minus 4 by 3 and or you have all 3 of them functioning. So, it will be e raise to minus 4 by 3 into 3 that means here this is so actually this will be e raise to minus 12 by 3. So, this will be the required probability. Now, this figure refers to problem 8, I think. Three independently functioning components are connected into a single system as indicated in figure above. So, two are parallel and then one is series. Suppose, that reliability for each of the components for an operational period of t hours is given by e raise to minus 0.03 t. So, now, by now we have discussed all this. So, therefore, you have to treat these two as parallel. So, then you compute the reliability of this component and then this together with this component C 3 in series. So, now, you can do it. You will have to first and they are all identically distributed the failure laws for the three components. So, therefore, you first compute these two in parallel and then it will be. So, which will be, I have given you the formula for this and then that into C 3. So, that will be multiplication. So, you can do that and let us see what else is asked. So, if t is the time to failure of the entire system, what is the pdf of t that you can find out? Well, you will first find out the reliability function or you can try to find the pdf directly. What is the reliability of the system? How it compares with e raise to 0.03 t. So, obviously, I think even your guess should be that it will definitely improve the reliability, because there are two components in parallel. So, definitely and even though, see the thing is that the reliability of this component consisting of C 1 and C 2 will go up. But, C 3 will have the same and since when you take the combination and series and you hook them up in series, then your reliability is the minimum of the two. So, therefore, you cannot say much. It will almost be the same. In fact, it will not be better than e raise to 0.0 minus 0.03 t. In fact, it will not be better than e raise to 0.0 minus 0.03 t. Now, I have given you a big system here. So, this is suppose that n components are hooked up in a series arrangement. Then k such series connections are hooked up in parallel to form an entire system. So, see the figure below is given. If each component has the same reliability say r for a certain period of operation, find an expression for the reliability of the entire system for that same period of operation. So, you should enjoy doing it. Just sit down, patiently write down the first for this series connection, you write the reliability function for this and this. It will be the same and then you do the parallel thing. So, there are k in parallel and you have how many n in series. So, just patiently sit down and you can write down the expression for the reliability function for the whole system should be able to do it. So, now I am saying that suppose that each of the above component obey an exponential failure law with failure rate 0.05. Suppose, further more that the time of operation is 10 hours and that n is 5. So, now I have given you the numbers and determine the value of k in order that the reliability of the entire system equals 0.99. So, you will enjoy doing this problem because so k is not given to you. So, but then when you write down the reliability function, it will be a function of k and then you put the value for 10 hours. So, your t is 10 hours. So, then you substitute that in the expression and put the whole thing equal to 0.99 then you will get the value of k. So, interesting problem I am sure you will enjoy doing it. Now, component has reliability 0.9 when used for a particular purpose. Component B which may be used in place of component A has a reliability of only 0.75. So, A has reliability of 0.9 for a certain period and component B which may be used in place of component A has a reliability only 0.75. So, what is the minimum number of components of type B that would have to be hooked up in parallel. So, now you can write down the reliability function suppose you take k number of B components and you hook them up in parallel. You know the reliability function and then you want to say that its reliability should be equal to 0.9. So, that also again you can using the formula that I have given you for computing the reliability when they are hooked up in parallel then and you know 0.75 reliability of 1 component and how many should be there so that the reliability goes up to 0.9. So, these are the kind of questions which I am sure you will that is it. So, I think this is the last lecture and effort has been made to acquaint you with probability theory basic probability theory I would say and then its applications and I have tried to through exercises try to give you a good insight into the subject and I hope you enjoy doing these exercises.