 As I said, turning to the next homework instead of that pile of paper, just, you know, just handed it to me electronically and I'm, should be nice. How do I turn this off? Okay. Sorry about that. I don't want you guys to read my mail. It can cook. It can do everything you think of. It does. No, it doesn't. All right. So I think I'm ready to start the tape now. Okay. Thanks. Oh, I'm not. Let's see. So before I just want to mention one thing about the, this kind of probably models we haven't really start talking about seriously. So we should hopefully today. But I do want to point out one of the books that I listed in the syllabus. Well, there are lots of places you can find. Well, let me, let me say this. You can find, you can make any model. You can include a probabilistic, a random kind of a non deterministic component and then make it, you know, stochastic probabilistic model. But the ones that are kind of a good, good, good sample of those or some, some example of such are in this one book, mathematical modeling for life sciences. So if you go there, you will see a few of the chapters starting with chapter five. It talks about Markov chains and diffusion. These are two topics we're going to actually cover. Chapter six has some more, you know, random processes and also has a little bit of statistics. So I'm going to refer to some of this. I want to start talking about Markov chains and diffusion. But, you know, feel free to browse through this before that. Okay, so last time I kind of gave you an example of a random variable that has some distribution. It's the wrong one. By the way, this comes with a stylus too. So I'm going to be confused with stylus to use all the devices you have. Okay, so, so if I have X to be a random variable, which again in the discrete case, it's just like a function from the sample space to the real numbers. So this would be a real random variable, rv for short. Then we talked about expectation, right? And expectation really means kind of an average. And we're going to be defining this for continuous random variables as an integral. But if it's discrete random variable, then we know what this means is, so discrete is simply a summation of over all possible, you know, you can say over all possible numbers, all possible values that this random variable takes of K times the probability that X takes the value K. Okay? And then depending on the experiment, depending on the random variable that you're observing out of that experiment, you know, oftentimes you can actually compute this explicitly. So this is for discrete and we'll see what it means for continuous. The next thing is the variance. Let's say I think that's, I mean notation may change slightly. I think I just use the one that's in the book. The variance of a discrete random variable, let's say, is actually the expectation of, if you want of a new random variable, define as X minus its expectation squared. Okay? So again, for now just think of discrete random variables where you have X takes a discrete set of values. Okay? So if X is, let's say, what's an even simpler, a simpler, well, let's say that the experiment is passing one die. Okay? And X is a function that assigns zero if the die is even and one if the die is odd. Okay? Yeah. So, I mean, this, okay, this is probably over simple, I mean too simple to have any meaning. But maybe it's, think about the one with two dies and take the sum of the two. But if you were like this, if you toss one die and you take this, then obviously what would be the expectation? Well, there's only two possibilities, zero with the possibility for the outcome, right? Zero, so this is zero times the probability of X being zero, and that's always zero. Plus one times the probability that X is one, and that's a half, right? So the expectation is a half, and what's the variance? So that's, well, what is the, to compute the variance, it's enough to compute what is the X minus its expectation, right? And what is it going to be? Negative a half and one half, right? If the number, the die appearing on the, the number appearing on the die is even and a half if it's odd, right? Yep. So, so what's the E minus X minus C of X squared? Well, it's always one quarter, right? Well, if you have a random variable that's constant, obviously the expectation of that random variable is, is that constant, right? So it's, so the random, the variance of this random variable would be one quarter. And so again, this is maybe too simplistic, but I just wanted to kind of to, you know, show what that formula means. And so that's the variance, and of course there is the so-called standard deviation, which says, which is defined as the square root of the variance. So this is the variance, maybe just use sigma, sigma of X. So this is the square root of the variance. So it is, you know, the formula becomes maybe a little bit harder to memorize, but it's expectation of X minus expectation of X squared, everything to one half. Okay, so it's a number. So in this example would be one half, right? In that simple example, it would be one half, meaning that, so you would say that the random variable has expectation one half. And variance one half, right? And standard deviation one half. Okay, so you can actually do this obviously for other distributions. So for example, plus one distribution. So if I have a random variable such that the probability, it being taken the value n is, what was it last time we wrote down? e to the minus lambda n, lambda to the n over n factorial. No, e to the minus lambda, excuse me. Yes, question? I was just wondering what is the application of the variance? Is that a very good answer? So what does that mean in this simple example? Nothing. I mean, think about it. The standard deviation as being the... Oh, the ultimate rule. Yeah, I mean standard deviation has a meaning, so we know that. Okay, so for a random variable having this kind of distribution, right? So we're not saying what the experiment is. We're not saying what the random variable is. We're saying any random variable that has this distribution, right? Has, and we said last time, has the expectation to be... Well, it's simply the summation of overall n, positive n, I guess starting with zero. So n starting from zero, one, two. Of n times probability of x being equal to n, and this ended up being equal to lambda. So this parameter lambda was, this is with rate lambda, signifies exactly the expected value of this random variable. And let's see, if you are to compute... Well, okay, so there are two ways to go about competing the standard deviation of the variance and then take the square root, right? So one way would be to... Let's see, by the way, the variance is the square of the standard deviation. So oftentimes you just write like this instead of the variance, and it's going to be expectation of x minus expectation of x squared. And it's not maybe that hard to see, but it's not easy either to see what the expectation of... So applying this formula directly would actually make the computation a little bit harder because we know this is lambda, right? And then you would have to compute the expectation of x minus lambda squared, right? So using this row, it would actually look like it's going to be a summation of what? Somebody tell me, squared, probability that x minus lambda squared is n minus lambda squared. So it's not undoable, but, you know, so again, this probability is the same as the probability that x is n, right? So then you plug it in and you do the... You compute the series and what you're going to get is, again, this is lambda, okay? So the square of the standard deviation is, again, the variance is also going to be lambda. I'm going to leave this as an exercise if you don't object, right? Because it's e to the minus lambda and, I'm sorry, lambda to the n over n factorial, right? You're experts in computing series, right? From calculus, too. So you know the computer series, you would just... e to the minus lambda is a common factor, right? So here what you would do is square the n minus lambda, right? And then you're going to have three series, okay? Do the math and it's going to look like it's going to simplify to just being lambda, okay? So this Poisson distribution has this very special feature that you have variance and expectation is, again, equal to lambda. Now there's an alternate way to compute the variance or the square of the standard deviation. And that is that the expectation becomes expectation of x squared minus the expectation of x squared, okay? Now you can actually see this if you do the same. So why is this? Because what was the definition? It was x minus the expectation of x. By the way, sometimes we're going to use e of x without parenthesis. This term is expectation of x, right? So what's the expectation of? What's this random variable that we take the expectation of? It's x minus a constant square. So you can square this. Then minus twice x times e of x plus e of x squared. So this is for a general. This computation shows is a valid for any random variable. So even if it's not discrete, but if it's discrete, expectation of a sum of random variables is a sum of the expectations. So there are certain properties that one uses, and those can be formalized very carefully. It's linear in the random variable, yes. It's like being an integral, right? So that's why it's good to think about it as an integral, even though one is discrete is really a sum. But think about a sum as an integral, then you know that when you integrate a sum of two random variables is the sum of the integrals. So this is also a constant comes outside of the integral sign, right? So you have the expectation of x times the expectation of x. Yep. And the last one is expectation of a constant random variable is that constant, right? So you see how you end up with expectation of x squared minus twice the expectation of x, the square of the expectation of x, so it's minus. You pretty much have to know when to put parenthesis and when you don't have to be least confusing, I guess. But obviously this confirms that. So an alternative way would be to, let's say for this Poisson random variable would be, that's for the Poisson distribution. It would lead to not a huge simplification, but sum, it would tell you that the variance of the square of the standard deviation is the expectation of the square minus the square of the expectation of x. And again, each of this is a series, but the one for x squared is slightly simplified. It's n squared, but times the probability that x squared is n squared, which is the same as probability of x equal to n, right? And we know the probability of x squared, we know this is already lambda. Yep. So you see it just kind of simplifies a little bit because, I mean, I'm saying this thing, but maybe it's not clear. So the probability that x squared is n squared is the same as probability that x is n. Say it in a sentence. Yeah. So why are these probabilities the same? Is the probability of what? Okay, when we say probability, it's probability of what? Yeah. And what is that? That is an event. Okay, it's the probability of an event. In an event, the event that x squared is n squared is identical with the event that x is n, right? Of course, assuming x is positive, x only takes a positive, right? So these are the exact same events. That's why, so in probability theory, you can always kind of, it's like taking the squares, but you don't take the square roots, right? You just say, well, this event can be written like this. It's slightly different, right? It's the same event. It's rewriting the event. It's an empower. Because you can rewrite the event, the actual event doesn't matter. Well, and you're going to write it in the simplest way possible. You know? So an event could be written like in a very convoluted way, right? Or in a simplified fashion. And oftentimes you can actually, you can pursue the computation by rephrasing the event you're looking at. That's, you know, yeah. Right, so the event that this happens is identical with the event that this happens. They're the same events. They're the same subsets of the sample space. When you run your, you know, runs, what do you say? When you run it runs. When you run that experiment several times, right? The outcome that x squared is n squared is the same as the outcome that x is n, right? It's just rephrasing of the event. Yeah. Yeah, I mean, okay, it is specialized to this, but I wanted to make sure that, because oftentimes in this computation we'll just rewrite the event and we're going to say probability of this equals the probability of something else. And I know people have, you know, sometimes a hard time saying, how come I just take the square root? What is that taking the square root? You know, how come, right? It's just re-saying that same event in different words. And then, of course, this is, now it's e to the minus lambda, lambda to the n over n factorial. Okay, so you see the slight advantage here? You said this is one series compared to the before you had three series coming from the squaring the n minus lambda squared. Not a big advantage, but some, yeah. In general, is that what's going to happen if your function that you're doing in the x and the n is the reverse function over your set? Say it again, if... Is this going to happen in general if the function you're doing in your dash has a reverse function on your set? Yeah, yeah, you can... Right, I mean, you're talking about this, right? The fact that events are the same. Yeah, yeah. Yeah, I mean, the square is not important. It's true that it's because of one to one. So, you know, it has an inverse. But, okay, we'll see this again. That's why I want to kind of emphasize it. So, okay, so when you do this, it's going to, you know, this lambda square is going to be, you get, you know, eaten by some term here that's going to be lambda squared. So, what comes out is just lambda, okay? So, again, for the... For such distributions, for Poisson distribution, this is what you need to remember, is that if it is a Poisson distribution with rate lambda, then the expected value of that random variable is precisely lambda and the standard deviation is squared of lambda for any X random variable with Poisson distribution. Okay. So, why are these things important? Well, here's one thing that's important to write down and to understand. It's called the strong law of large numbers. So, it says that if I have a sequence of random variables with same distribution and they are independent, that's going to be a new concept in independent random variables, and of course, moreover, well, okay, then this average over the first n, so this is called a Cesare average, but it doesn't matter how it's called. It's the average of the first n random variables. This in itself is a random variable. You agree with that? Because what is a random variable? It's a function. Okay, in this good case, right? So, you can add two random variables. You're going to get a random variable. You can get n random variables, so that'd be a random variable, right? So, this is now a new sequence and this sequence converges to a constant. The constant being the expected value of each of the random variables, therefore of all the random variables, because the random variables we're starting with is all have the same distribution. So, since they have the same distribution, how come they have the same average or expected value? I want to make another. We want it to be finite, so of course the example we talked about is clear that it's finite, but this might be applied. You could try to apply it for... There are random variables that have infinite expectation, so... Yeah. What you just wrote says even capital X... I. Oh, I. So, basically what I'm saying is that because they all have the same distribution, probability distribution, they all have the same expectation. Because expectation is only depending on the probability distribution, right? So, they all have the same number. The expected value is the same for all random variables. Call it E of X, right? This is E of X1, E of X2, E of Xn. All of them are the same. Well, so a strong law of large number says that as n goes to infinity, on average, if you want to think out like this, the average of these random variables converges to this expected value. Now, the key ingredient here is that these random variables are independent. So, what does it mean that we have independent random variables? So, independence is kind of the foundation of the probability theory. And we're not going to do much of it, I mean theoretically, but just to say that if I have two events, so two events, A and B, of course these are subsets of the same sample space, are called independent if the probability of both occurring at the same time, and that's the event of A intersect with B, right? The event of A intersect with B contains all the outcomes that are both in A and in B, right? So, if I remember this sample space, right, and I have two events, then I'm going to have, if the experiment, if a run of an experiment ends up with an outcome here, then both events are occurring, right? So, we say that these two are independent if the probability of the intersection is the product of the probabilities. This is very different than the probability of the union, for instance. The probability of the union of two events is, we talked about it earlier, right? We said that if the two events are disjoint, so they are incompatible, right? So, all this wording is kind of, it takes a little bit to digest, but if two events are incompatible, then the probability of the, either of two occurring is the sum of the probabilities, right? If they have an overlap, then the sum of the two probabilities minus the probability of the intersection. So, this is actually different now. It says it's a property of the intersection only, which of course would spill in the probability of the union, but it's a specific probability of the intersection. So, for instance, if I have two events that are incompatible, are they going to be independent? No intersection was probability of the empty set. Zero. So, for them to be independent, one of them has to have probability zero, right? So, if I have two events that are incompatible, and neither has probability zero, right? Then are they independent? No, because it doesn't satisfy zero, it's going to be not equal to the product of the two. And that makes sense because it says that if they're incompatible and one happens, the other doesn't happen. So, they're not independent in that sense, right? If one happens, the other doesn't happen. Now, of course, there will be cases when there are events that are overlap and they're still independent, right? It just has to have that, this property has to hold. Okay? Yeah? Yes, you can have, yeah, certainly. Right. And you can, I mean, again, we can cook examples of that. Throwing two dice and looking at, I don't know, probability that one has, I don't know. It's negative. You're following? Yeah. It's just come up with two events for which there's an intersection and the intersection is, either has this problem or doesn't have this property, right? Yeah? That's the way you get an event itself, provided your first die is included. Right. So that the sum is even and the other event would be that the sum is, no, that would be incompatible, right? The other one is that the, one of them is even, right? Right? Or, I don't know. So, let's see. So, they can overlap and have both. So, they could be independent or they could overlap and they could be not independent, right? But let's see. Is there a possibility that one is a subset of the other? If one is a subset of the other, the intersection is the smaller of the two, right? So, how can a probability of the smaller of the two equal the product of the probabilities of the two? Well, the probability of the larger one has to be one. So, the larger one has to be the whole set. So, that's, it's also not a possible to have two events, one a subset of the other, unless the big, well, and the big one should not be the whole set, right? Then they're not independent. And again, it makes sense because it says if the smaller one happens, then the larger one happens, right? So, it's not an independent pair. Okay. But, again, we could spend a whole lecture on just the significance of the independence. And this, unfortunately, we're not going to have the time, but it is an extremely important, well, it's the most important concept and probability of independence. Now, what it has to do with the random variables, though? Well, so here's what it has to do. So, two random variables, X and Y, are independent if the probability that of the event where one takes values within a range and the other takes values within another range. So, by the way, this is just saying that this is one event, right? This is not an event. And when we say this and this, we mean the intersection of the two, right? So, that's the same as the probability that the product of the probability is, okay? So, the random, independence of two random variables reduces to really the independence of the events when the random variables are within a certain range, okay? Which, in the discrete case, it would be even, it would reduce to even more, you know, more restrictive case. For instance, the probability that X is n and Y is m, let's imagine of two, so, for discrete random variables. By the way, this, the one above, it's really true for continuous also. But for discrete random variables, when you say that something is in a range, and you can narrow that range until you only kind of, in your range, there's only one value, right? So, it says the probability that X takes a certain value and the probability that, and Y takes another value is the same as the probability that X, sorry, sometimes I'm using accolades, sometimes I'm using square brackets, sometimes I'm using brackets. But this is what independent random variables mean, okay? And this is if there are two of them, right? And now you can think about probability that you can talk about independence of not two, but any number of random variables, right? So, the formalizing of this concept is, again, it's very important in probability and it's something we're not going to have the luxury to go into much detail. But I want to show this to you on a practical problem called, let's say it's the, well, it's a manufacturing problem, it's a problem coming from manufacturing diodes or any sort of electronic components and it's kind of at the beginning of chapter seven. So, let me just summarize that. The problem is quality control, this product that is being manufactured. So, there's a testing for faulty diodes and one way to test something like, you know, you manufacture the same product over and over again, is you take individual diode, you can take each of them and test it, right? If it's good or bad, right? If it's bad, you throw it out. If it's good, you keep it, right? So, certainly that's going to do the job, right? But it's not going to be the most efficient way. So, just simply because of the costs that are going to be incurred. So, the testing for faulty diodes will be done as follows. So, it can be done in two different ways. One is, as I said individually, at a cost of five cents each diode. Or it can be done in groups of diodes. And here's how that's done. So, let me show you individually. You produce one diode, then you test it. Then you produce the next one and you test it, right? And there's going to be some of them that are going to be faulty in each of them. And the ones that are faulty, you just label, you put them on the side, right? So, you can see the cost. If you have a thousand diodes producing and you do it individually, it's going to be a thousand times five cents each, right? So, it's very simple. The question though is, is there a cheaper way of doing it? And the answer will be yes. So, the groups of diodes is going to be the following. So, if I, for instance, decide I'm going to do groups of three diodes, for an example. I'm saying, I'm just picking an example. So, if I'm picking groups of three and I test them, I test these groups as an entity, each as an entity, right? And I, when I put this, whatever, in series, so if I, if I test a group and the group ends up being good, then the conclusion is that each member of that group is good, right? If the group is bad, then what is next thing to do? Then we know that at least one is wrong, faulty. So, then we're going to test individuals in that group, okay? So, the whole problem is figure out which one is the more efficient way and we should know what is the cost of testing a group. So, this at a cost of four cents. A group. Okay? Yeah, three or four times. To help a really large group identify a bad group, break it down and then break that into smaller groups. Sure, sure. But that's not, that's not our problem. Our problem is between these two, if it is to decide on a strategy, the strategy is how many, how large a group should we use the first time? Yeah. Four questions at once, okay. Something must be wrong. Yeah, good question. So, you're right. So, maybe I'll fix this because I'm sure you've read the problem before, right? So, this is really four plus n. But yes, if this is faulty, then you do it individually and it's going to be five each. Okay? Four plus five n. Okay, let me just set this up here so you can see. So, it's okay. So, first group, n is the number of diodes. So, this picture represents the corresponding n equals three. Okay, so, what is the problem? The problem is the goal, determine an optimal strategy. That is, what should be the number of diodes to be tested at once in a group, such that the total cost is minimized. And you, excuse me. And of course, you still want to exhaust like every single one. You want to pass the good ones and identify the bad ones. Yeah. What's the cost of faulty diodes? Well, you don't know in ahead of time. Until you test it, you don't know if it's faulty or not. Well, yeah, I know that if there is a faulty diode, what's the consequences of cost and money if it's faulty? Again, that would be an additional thing in the problem. I mean, all of this, right. I mean, if you see yourself with a hat in that factory, yes, you're going to ask all those questions, yeah. But the point is you don't know. You don't know your ratio. You know your ratio. But what you know, okay. So what you know is you know the probability. Let's see. So I'm missing something. You're right. So I'm missing the probability that a manufactured diode is faulty. Right? So the probability, so that's an assumption, is the probability that, let me just say how they say it here. So the assumption is that 0.3% of all diodes are faulty. Okay? And this can come from whatever. Right? The previous experience. Okay. So here's kind of where the strong law of large number comes in the picture. So think about the first group of diodes. What would be the cost of testing that group? The cost is going to be either 4 plus n if whole group is good. I mean, each member in the group is good. It's 4 plus n plus 5 times n if at least one diode in the group is faulty. The cost. I'm saying the cost. So here's the thing. C1 is a random variable. Okay? And maybe you don't want C1. Let's call it X1. So I'm saying this is going to be X1. I define it to be this random variable, which represents the cost. Now, it's a function that takes for the first group of diodes, right, any outcome of the manufacturing and testing process, right, and it assigns a number to it. Okay? Now, why is that a random variable? Why is this random? Is it 4 plus n or 4 plus n plus 5? Well, it's true. So each will have some probability of occurring, right? Yeah. So the event, so underlying this, there is this sample space, right? So the sample space is unpicking n diodes and they may all be good. Some may be wrong, right? So with some probability. Now, do you agree that this is a discrete random variable? It only has two possible values, right? Now, the sample space may not actually be discrete or finite, right? It's even hard to define what the sample space is, right? The sample space is an outcome in that experiment is you pick a group of diodes that are only good, some may be bad, right? That's the best you can do in describing the sample space, yeah? But the random variable is very simple to write. And now you do the same with the second group. And it's going to be exactly the same. Now, it's not only that it's going to be exactly the same distribution, right? But can we say that these two random variables are independent? Well, I don't know. You can argue not, but... I was going to say that when two diodes of the second group are faulty, would not. Exactly. A direct outcome. Unless you pick the second group from the faulty ones from the first group. Like if there's a replacement there, right? Yeah, if you don't do that, right? You just have like a sea of diodes, right? And you just pick groups, right? Those are, I don't know, intuitively they're independent, right? Yeah. That's good enough for us. Okay. For me, at least. It's the same distribution. And this is the same with the k-th group, okay? And that's where I should have... Anyway, they call it x. Can I change it to xk? Doesn't matter. N was for a number of diodes, but... Okay, so... So xk... Now, why do we even say these are distinct random variables? Why not just say this is the one I wanted? I don't know, because the time sequence, you take the first diode, right? You test it, then the second diode. So they are... They are not the same... They're different random variables, but with the same distribution, and they're independent. Yeah. Yeah, they're also going to be different. Okay, anybody has anything against this? They're not the same random variables? Although they have the same distribution? Because they actually physically... You observe different things, right, in different experiments. Yep. I mean, it gets a little bit... If you're really analytical about this, you can have issues, but we don't want to have too many issues with this thing. The point of this exercise is that we want to draw a practical conclusion from these assumptions, okay? And that is very simple to do, so the distribution xi is going to be always the same. xi is 4 plus n with probability, with a certain probability, and 4 plus n plus 5n with a certain probability. Now, here's where we have to talk about probability of the event in which the whole group is good. I have n diodes, and all of them are good. This means what? This means that each of them in that group is good with a probability of what? 1 minus 0.0003. Yeah, that's 0.3%, right? And because each diode is good with this probability, how can all n of them be good with that probability? Again, assuming they're independent, then this would be the product. So it's the product of this. And what's the opposite? Well, the complement is going to be with probability. So if I call this p, then it's going to be probably 1 minus p, okay? So that's how we find the distribution. And then what is the expected value of this? It's simply taking the values that each of this random variable takes, 4 plus n times p plus 4 plus n plus 5n times 1 minus p. Now, this is not pleasant, but it can be written explicitly. So 1 minus 0.0003 to the power n plus, that's 4 plus 6n, 1 minus 1 minus 0.0003n, which is 1 minus that power of n. Okay, so take a look at what we've gotten. We've just simply gotten a function of n. n is a number. It can be 1 to a million, right? And we have the sequence of random variables. By the strong law of large number, we can say that on average, when I have lots of groups of such, so when this k is large enough, when k is large, the x1 plus xk over k goes to this, or let me say it's close to, we're not taking a limit. We're just saying when k is large, this average is this number, right? Now, in words, what does this average mean? x1 was for the first group, x2 for the second group, xk was for the kth group. So this is simply saying what is the average cost, right? For testing a group of dials. So this is the average cost for testing the first k, the groups of dials. Okay, so if k is large, this is actually what you want, is you want this quantity, so the objective is you want to minimize the cost, what did I say? We minimize. Go to determine options such as total cost is minimized. Wrong, well, let's see. Is it total cost or is it the, it's not the total cost, so if I could change it so you can see, is the cost per diode. So the point is you want to minimize the average cost for a group of diodes divided by the number of diodes in that group. So this is the average cost per diode. Until now, x was the average cost per, so this is the cost per group of diodes, right? Okay, so this is all kind of setting it up. What is then the actual solving? So that's the step in which you solve the model. So you have to minimize 1 over n times this messy thing, which was 4 plus n, 0.0997 to the n plus 4 plus 6n, 1 minus 0.997 to the power n. What is going to be your favorite tool? Is that again? Random? There's nothing random at this stage. You see, you just need to minimize a function of n. I mean, you can actually, yeah, you can actually simulate, you can actually show that this is the average cost, but in the end, it's just an optimization problem of one variable, right? It just differentiates how to do it. Yeah, so if you could differentiate set derivative equal to 0 and check whatever, fine. The problem here is going to be there's powers n in the power n in the infront, right? So it's not going to be an easy way to differentiate, right? So you can try symbolically, but what I would say is just plot. Or you can actually run a search, you know, through the possible values, but it's a non-linear optimization, one-dimensional optimization, right? And whichever way you do it, you will see that 17, again, it may be 17 or 18, and then you have to decide which one. 17 is optimal. That is the minimum number, the minimum cost on average, right? So the optimal strategy will be to take groups of 17, and if a group is faulty, then you're going to test it individually, right? And then you can figure out what the minimum cost is. Okay? So a big part in this model is actually just setting it up, figuring out what are the random variables, how can you use the tools of probability to actually reduce it. But I'll tell you that this little secret in this whole business that we do for the next three or four weeks, we're going to actually end up with a function to maximize or minimize. Okay? So the whole thing is, how can you actually bring it down to that level, and then just use an optimization tool to draw your conclusion. Okay? So, and again, we're going to learn other things about probability, but not everything. We're going to learn next Wednesday all about continuous random variables and how you can use that for modeling certain processes. Okay? Hopefully I'm going to have the iPad working by Wednesday. It will work better. My lectures will be...