What's up everybody? Jason here for jaspyscasebrick.com. We just sold out 2021 Panini Immaculate Collegiate Football five box case break random number block number two And again guys you get a random number from zero through nine and we go by the first number on the left side of the serial number for the break So here's the dice roll. Here's the customer names Todd down to Brandon zero down to nine roll One and six lucky number seven seven times one two three four five six seven Diane down to Diane seven times one two three four five six seven eight down at three Diane with eight and zero and in our redemptions PJ with nine and seven Brandon with two David with five Todd any one of ones will be yours Vipple with four PJ with six and Diane with three So I'll give you guys a quick 30 seconds. I don't know if you guys want to try to trade numbers But if not we'll close it We could go through a couple orders really quick When a close well, so Chris congratulations Great first experience. I know I can't get better any better than that if you buy your first time sell a pack and then you get in Yeah, so basically the next sell a pack is gonna sell out right now Three left. We have to do one more after that. I expected to sell it at some point tonight. So You got to do one more after this one that sells out so we did three total and sold 14 spots straight up So eventually at some point, I'm sure we'll sell that out tonight Including cup hockey is the same kind of a them both four left plus halfway through the filler And I guess I for NT three left one more after that and that's it. I got a C But we may have one more either R&B or pick a team But I will let you guys know Once we get done with pick a team number three, I got to see how many cases we would have left and I know There's at least one more for Instagram as well. So awesome. Yeah, so That pin right there that schedule is right that Google dog You can definitely save that or whenever you want click that pen If you notice that the sell a packs are sold out and everything sold out You can always click that break schedule to see when it's supposed to break but You know, we just got to do that third filler, which I'm sure we will Sell that out tonight at some point No problemo. All right, let's pull some fire guys. Let's make it for the last case. Although last case wasn't that bad I guess I was just being a little harsh, but we just had some really great cases so far And again, this is R&B style guys, so it doesn't matter What team you have all that matters is the number on the car? Good luck everybody So first one was a DK Mac if I said yeah 87 out of 99 spots having on a PJ. We have seven out of 25 another seven That is Stefan digs out of Maryland Another one for PJ in spot seven Michael Carter eight out of 99 Spot eight going to Diane Redemption automatically to Diane as well Ramon Dre Stevenson premium patch rookie autographs Very nice that was rookie right there for the Patriots yesterday broke one off like 91 yards There you go nice patch Seth Williams three out of ten very nice patch Auburn three going to Diane and Trevor Lawrence 63 out of 99 three color patch That is a beauty Spot three going to Diane Back to back hits right there for you Diane Let's go AJ Brown for the rebels 44 out of 49 they pull with spot four Wow, what a pass Michael Carter 97 out of 120 PJ with spots seven little gloves Greg or so 43 out of 99 Caleb Farley to 25 64 They pull with that one redemption Demetric felton premium patch rookie autographs another one for spot zero and Diane and four out of ten Travis ETN his teammate Trevor Lawrence Four out of ten look at that patch That the one that he's wearing right here on this jersey Might be spot four going to Vipple. You know, man Yeah, the one I got for you was one of them and I think that is my second. Yeah, I Think so. Yeah I don't think I'm better than a third to 39 Derek Carr 14 spot for Vipple. I feel like you had spot four earlier PJ. I feel like spot four was great one out of ten Mari Rogers nice jumbo patch Todd was spot one Another redemption Micah Parsons premium patch Ricky autographs Diane racking up those Redemptions in this case Zach Wilson 28 out of 99 On card as well Spot a Diane baby. It's your break Diane. Man. Oh another great bull patch 2-2 at wealth five out of ten nice bull Patrick there for Louisville five going to David and 17 out of 25 Ron Dale Moore He's on the Cardinals, right? It's about seven going to PJ CD lamb 29 out of 49 About nine going to PJ Justin Fields little piece of the glove 12 out of 86 spot two going to brand it last spot mojo There you go, Brian on the board Lin Bowden jr. That's actually an update from 2020 11 out of 99 Todd with that one Devonta Smith fly goes fly 34 Vipple my first Devonta Ramon J. Stevenson 15 out of 99 RPA David with that one and Kellan Mond 95 out of 99 RPA Aggies five David. Hey guys What am I? Yeah, dude, there's some really really nice Relics in this. All right. Good luck everybody. Another Derek Carr 33 Hey, I'm barely what's about three about three spot three Sammy Walkins 71 out of 99 spot one going to top Edwards. What's up, man? How's life been up there in Canada? Patrick Surtain 50 out of 99 zero going to Diane redemption Najee Harris signature moves Another one for spot zero. Yeah, killing it Killing it. Good man. I'm good over here too, man Living life grinding grinding these breaks out 17 out of 49 Elijah Moore Out of 49 37 spot seven going to PJ last one For my Eagles 39 out of 99 Devonta Smith. I'm a trophy winner Two-color patch and autograph spot nine PJ. Yeah You might have to sell me this PJ And there you go, everybody that was R&B number two and again, I'll double check and see we could post a third Picker chain number three though is pretty close. I've not sold out already So quick little recap some of the biggest Devonta Smith beautiful to to at well patch autograph Zach Wilson 99 Travis ETN patch and Trevor Lawrence to 99 And of course do some other great hits as well Think for the most part I feel like everybody at least got one card for their spot So that makes me a little happy. So at least everybody should be walking up with something But again, this collegiate is amazing again, I'll check it out. See we have some more R&B supposed up Jasper's case breaks calm


In this video we are going to learn about corrosion. Imagine four objects, an aluminium toy, an iron now, a copper pipe and a gold ring are left out in the open for many years. Which of the four materials do you think will be in the best condition when they are rediscovered? Pause, think and continue when ready. If you think the gold ring is likely to be in the best condition, you're right. The copper pipe and the iron now will appear in the worst condition, but the aluminium toy is also preserved very well. Most likely the iron now will appear in the worst condition. This is because of corrosion. Corrosion is described as formation of compounds on the surface of a metal when it is exposed to air and or water or an electrolyte like salt water. Typically once a metal corrodes it forms compounds known as oxides or hydrated oxides. The now made of iron rusted. This by the way is the only time you should use the term rusting. All other metals are considered as having been corroded when they react with water and or oxygen it in the air. But why would the now be more corroded than the aluminium toy, the copper pipe or the gold ring? Have you any ideas? Pause the video and continue when ready. The reason is the metals all have different reactivities. Iron corrodes faster because it is more reactive than the copper and the gold. But you may be wondering why didn't the aluminium toy corrode the fastest? The truth is the aluminium is more reactive than the iron so it reacts fastest with the oxygen from the air forming aluminium oxide. Aluminium oxide is very unreactive because once the oxide layer is formed on the surface it binds very tightly to the surface of the metal. So no cracks can form and cause further corrosion. Here's a challenge. How might you use this idea to protect the iron now? The answer is that by covering a metal like iron with another more reactive metal like zinc is that the zinc forms the zinc oxide layer quickly. The iron underneath is protected. This process is called galvanizing and can last up to 100 years. If you think of the metal on a bike corrosion can be prevented by painting the frame a static part or greasing the metal parts like the chain which moves or by covering it with a thin layer of a less reactive metal like tin, copper or gold. But that would be very expensive.


If you don't necessarily succeed right away, don't give up. Because I've tried for many Olympic trials to make this team, and there were many days where I wanted to give up, but with the support of my husband and friends and family and just my own determination to make it, I didn't give up, I kept trying, kept getting better, and here I am, going to the Olympics. See you in London.


Welcome traders to the TITMEL chart, I just want to update the NASDAQ, you were tracking this quality objective down into the 14,538 and this ascending trend line resistance. Saw some nice demand come through there yesterday, long positions now running with just in the 300 points of profit at the moment so I'm going to move my stop now to entry 14,654 and we'll let this run. Coming into this pivot cluster here this could be a bit sticky, I want to see a strong move through there, through the descending trend line resistance at 15,254 and then that opens up this 16,000 test. As always traders plan the trade, trade the planner most importantly manage your risk. Until next time thanks very much


As Salaamu Alaikum, welcome to lecture number 22 of the course on statistics and probability. You will recall that in the last lecture, I discussed with you the concept of independent events and we also considered the special case of the multiplication theorem in case of independent events. We did an interesting example concerning the gender of the newborn babies and the nature of birth, life born versus still born and through this example, we also noticed that there is an important concept called marginal probability. In today's lecture, I will begin with the Bayes theorem. This theorem deals with conditional probabilities in an interesting manner. As you now see on the screen, the Bayes theorem states that if the events a 1, a 2, so on up to a k form a partition of a sample space S that is the events a i are mutually exclusive and their union is S and if b is any other event of S such that it can occur only if one of the a i occurs, then for any i p of a i given b is equal to p of a i into p of b given a i divided by the summation of p of a i into p of b given a i for i going from 1 to k. Students, I am sure that you are saying that this theorem is very difficult. Actually, it is not like this. If you see this step by step, as I have stated so many times earlier, everything will fall in place. The formula that I just presented to you is for the case of k mutually exclusive and exhaustive events. It can be written in the manner that you now see on the screen. Probability of a i given b is equal to p of a i into p of b given a i divided by p of a 1 into p of b given a 1 plus p of a 2 into p of b given a 2 plus so on up to p of a k into p of b given a k. If you open the denominator or open it, then this is the formula. If we take the case when k is equal to only 2, then the formula becomes p of a i given b is equal to p of a i into p of b given a i over p of a 1 into p of b given a 1 plus p of a 2 into p of b given a 2. Let me explain this to you with the help of an example. In a developed country where cars are tested for the emission of pollutants, 25 percent of all cars emit excessive amounts of pollutants. When tested, 99 percent of all cars that emit excessive amounts of pollutants failed, but 17 percent of the cars that did not emit excessive amounts of pollutants also failed. What is the probability that a car that fails the test actually emits excessive amounts of pollutants? Students, I will analyze this problem step by step and see how the formula of base theorem that I have just presented to you with it. The first thing is that you have seen that 25 percent of the cars emit excessive amounts of pollutants. So, it is obvious that 75 percent of the cars are such which do not emit excessive amounts. Now, what was our condition for the base theorem that a 1, a 2 form a partition of the sample space, a 1 represents the event that the car does emit excessive amounts of pollutants whereas, a 2 represents the event that it does not. So, together these two events are mutually exclusive and exhaustive and hence they do form a partition of the sample space. This is the first thing. The next thing is that in base theorem it is said that if B is another event which occurs only if one of the AI occurs. So, in this example, the second event is that the car fails this test. That means, the car either emits excessive amounts or does not. But, if the car fails, it might still fail the test. Therefore, this is the event which will be represented by B. Now, when we have identified a 1, a 2 and a 3, then let us look at that formula again. As you see on the screen, the formula for the case of two mutually exclusive events a 1 and a 2 a, probability of a i given B is equal to probability of a i into probability of B given a i divided by P of a 1 into P of a 1 into P of B given a 1 plus P of a 2 into P of B given a 2. Students, this may, the denominator, you see that there is no problem. That is, a 1 or a 2 or B is mentioned and we have identified them. But, you may be confused that what is this A i? This formula will be applied, it can be applied two times. If you keep A i as 1, then naturally the formula will be read as you now see on the screen. P of a 1 given B is equal to P of a 1 into P of B given a 1 over the same denominator as before. And if we put i equal to 2, we have P of a 2 given B is equal to P of a 2 into P of B given a 2 divided by the same denominator. Now, the question is that in this particular example, we want to compute P of a 1 given B or P of a 2 given B. If we go back to the question of this example, it said what is the probability that a car that fails the test actually emits excessive amounts of pollutants? What is the probability that the car will emit excessive amounts of pollutants given the information that it has failed the test? What is the probability that a car that fails the test actually emits excessive amounts of pollutants given the information that it has failed the test? What is the probability that a car that fails the test actually emits excessive amounts of pollutants? This can also be stated as what is the probability that the car emits excessive amounts of pollutants given that it has failed the test. Let a 1 be the event that the car does emit excessive amounts and a 2 the event that the car does not emit excessive amounts of pollutants. That is probability of a 1 given B. We are wanting the probability of the event that the car emits excessive amounts of pollutants given that it has failed the test. So, applying the formula that I presented to you, probability of a 1 given B is equal to probability of a 1 into probability of B given a 1 over P of a 1 into P of B given a 1 plus P of a 2 into P of B given a 2. In this question, we have available to us the values of P of a 1 P of B given a 1 and P of B given a 2. As you see on the screen once again, the statement was when tested 99 percent of all cars that emit excessive amounts of pollutants failed, but 17 percent of the cars that do not emit excessive amounts of pollutants also failed. These two statements can also be read as follows. The probability that a car will fail given that it emits excessive amounts is 99 percent whereas the probability that a car will fail given that it does not emit excessive amounts of pollutants is 17 percent. In other words, P of B given a 1 is 0.99 and P of B given a 2 is 0.17. So, students we can now substitute these values in the formula and as you see on the screen probability of a 1 given B is equal to 0.25 into 0.99 over 0.25 into 0.99 plus 0.75 into 0.17 and this comes out to be 0.25 into 0.17. So, the probability that a car will fail given that it emits excessive amounts is 25 percent so naturally the probability that the car does not emit excessive amounts will be 75 percent because the two events are the complements of each other or baaki saari explanation abhi hum discuss kar chuke hain. So, having applied the formula then the final result is 0.66 yani the probability is 66 percent that a car that fails the test actually emits excessive amounts of pollutants. Students yejo example have me consider kia this was the simplest case when k is equal to 2 only, but of course it can be extended to the case when k is equal to 3, 4 or more. Let us consider another example in a bolt factory 25 percent of the bolts are produced by machine A 35 percent are produced by machine B and the remaining 40 percent are produced by machine C of their outputs 2 percent 4 percent and 5 percent respectively of the bolts are defective if one bolt is selected at random and it is found to be defective what is the probability that it came from machine A students is example me ab dekhye kye humare paas teen mutually exclusive or exhaustive events hain yani either the bolt is being produced by machine A or by B or by C or is ke lawa koi machine nahi hain or 100 percent of the bolts yani tamamter bolts do hain they are covered by these three machines 25 percent by A 35 percent by B and 40 percent by C. So, humare jo bunyadi requirement tina k A 1, A 2, A 3, waga jo hain they should be mutually exclusive and exhaustive aap note karein ke yeh shat yaha pe puri ho rahi hain. A bolt can either come from machine A or from machine B ye to nahi hosakta hain ke baik vaakth hum kaha hain ke wo machine A se bha raha hain aur B se bha raha hain wo ek bolt zair hai ke ek baik main koi ek bolt jam uthain ke it will be coming from one of the three machines exhaustive isliye hua ke 25 percent plus 35 percent plus 40 percent is 100 percent that is 1. So, students yeh jo humare basic concept tha that A 1, A 2, A 3 should form a partition of the sample space yeh baat yaha pe puri ho rahi. Aur wo jo dosra event hai, jis ko hum B kahenge is example mein that is that the bolt is defective yani jo bolt uthaya gaya at random that came out to be defective. Aur hum nikalna kya chahate hain? Ke what is the probability that the bolt came from machine A given this fact that it is defective? To aap dekhye ke isska to yeh matlab hain hain, ke we are trying to compute P of A 1 given B the probability that the bolt has come from the first machine given the information that it is defective. Now, I would like to encourage you to apply the Bayes theorem to this example and to solve it yourself in order to determine this particular probability. This brings us to the end of the discussion of some of the very basic and important concepts of probability. I will now begin with you the next very very important topic and that is probability distributions. As I mentioned in the very beginning, there are two types of quantitative variables the discrete variable and the continuous variable and accordingly there are two types of probability distributions the discrete probability distribution and the continuous probability distribution. I will first discuss with you the case of discrete probability distributions and in this regard the first thing that we will do is to consider what is meant by the term random variable. As you now see on the screen such a numerical quantity whose value is determined by the outcome of a random experiment is called a random variable. It is an any experiment in which the outcome is unpredictable. For example, if we toss a pair of dice we know that there are 36 possible outcomes. We cannot say which of those 36 outcomes is going to occur. If we consider the case of tossing three coins simultaneously, students you will recall that the sample space of this particular random experiment consists of eight outcomes head, head, head, head, head, tail, head, tail, head, tail, head, head and so on. If we are not interested in the sequence but are only interested in what is the number of heads that appear students then we are talking about a random variable. The random variable X denotes the number of heads that appear and in this example this number can be 0, 1, 2 or 3. If we get tail, tail, tail it is obvious that X is equal to 0, n e number of heads is 0 and if we have for example head, head, tail how much is X? Obviously X is equal to 2. Now that the basic concept of a random variable is clear the next question is what kind of a random variable are we dealing with? Obviously in this particular example that I just presented X is a discrete random variable because it can take the value 0, no head, 1, 1 head, 2 or 3 but obviously we cannot have 2.9 heads or 1.7 heads. So because we are dealing with a discrete random variable hence the associated probability distribution will be called a discrete probability distribution and you will be interested in noticing as we study this topic that there you will find a number of similarities between the discrete probability distribution and the discrete frequency distribution that we dealt with in an earlier lecture. Let us now study this concept in detail with the help of an example. If a biologist is interested in the number of petals on a particular flower this number may take the values 3, 4, 5, 6, 7, 8, 9 and each one of these numbers will have its own probability. Suppose that upon observing a large number of flowers of this particular species say 1000 flowers the following results are obtained. The X values are 3, 4, 5, 6, 7, 8, 9 and the frequencies are 50, 100, 200, 300, 250, 75 and 25. Students, abhi jo table aapne screen pe dekhi obviously this is a frequency distribution aur aap kahange ke phir ye toh wo concept hai johan bohot pehle uswag study kar rehte johan frequencies deal kar rehte. But if you remember the relative frequency definition of probability and you relate that definition with this particular example you will realize that the total number of flowers that have been inspected in this problem that is 1000 and 1000 is a reasonably large number for us to treat the relative frequencies for the various values of X as probabilities. After all relative frequency definition of probability ye hi thi na ke if an experiment is repeated again and again and again and again a large number of times then the proportion in which a certain event occurs that proportion can be regarded as the probability. So, in this example if I compute the relative frequencies or in other words the proportion of flowers that had 3 petals, 4 petals, 5 petals and so on these proportions are eligible to be treated as probabilities. So, as you now see on the screen we can say that the probability that X is equal to 3 is 50 over 1000 that is 0.05, the probability that X is equal to 4 is 100 over 1000 and that is 0.10 and similarly all the other probabilities. In other words the probability that a flower of this particular species contains 3 petals is only 5 percent, the probability that a flower of this species contains 4 petals is 10 percent, 5 petals 20 percent probability, 6 petals 30 percent probability, 7 25 percent probability, 8 petals only 0.075 is the probability of this event and for 9 petals it is only 0.025 that is 2.5 percent. Students, this table is an example of a discrete probability distribution. These three words that I have used, let us look at them. First of all why are they called as discrete? That is most obvious. The number of petals on a flower will be 3, 4, 5, 5 petals. Now, you will argue that it is possible that there were 6 petals, but out of that 3, 4th part of a petal is a petal, so now there are 5 petals left. It can be argued in this manner, but this is not the way that we will be interpreting it from the mathematical point of view. It will always be a whole number. After this, students, note that the total probability 1 has been distributed among the various values of x. As you once again see on the screen, 5 percent of it has been allocated to x equal to 3, 10 percent of it has been allocated to x equal to 4 and so on. And this is exactly the reason why we say probability distribution. Like the frequency distribution discussion, total frequency has been distributed among the various classes. Students, a discrete probability distribution has two basic properties. As you now see on the screen, the first property is that the probability of x i lies between 0 and 1 for each x i. And the second property states that the sum of the probabilities is equal to 1. In the example that we just considered, you can notice that each of these properties is being fulfilled as will be the case for any discrete probability distribution. Let us now consider the graphical representation of a discrete probability distribution. Of course, in the case of the continuous variable, we talked about the histogram and the frequency polygon and the frequency curve. But, if you remember, in the case of the frequency distribution of a discrete variable, I proposed that you draw a line chart. This is exactly the chart that we will be drawing for the discrete probability distribution as well. And for the example that we just considered, we have the following situation. Since x 1 is equal to 3 and the probability of this x value is 0.05, hence the length or the height of the first line is 0.05. Similarly, the height of the second line is 0.10, the 1 for the third is 0.20 and so on. Hence, we get the line chart that you now see on the screen. And I think that you will agree with me that we can say that this particular discrete probability distribution is approximately symmetric. You have noted that I gave you the concept of symmetry here. That means, the same concept that we were discussing at that time when we were discussing discrete distributions. But, symmetry or skewness or kurtosis was discussed later. First we mentioned mean and then spread. This means that in the case of discrete probability distribution, we may measure skewness later. First, we measure its mean or its central value or its spread or its standard deviation. As you see on the screen once again, this distribution is approximately symmetric. And because of that, you might agree with me that we can say that the mean number of petals for this particular example is 6. And we are saying this because the central value of our random variable X in this problem is 6. Students, Zahirek ye to ek estimate hai of the center because I have not computed the mean according to the proper formula. How will I do that if I wish to compute the exact mean of this particular random variable? In order to understand this, I would like to draw your attention to the fact that the formula is very similar actually to the formula that we have in the case of a frequency distribution. As you will remember, the formula for X bar was sigma F X over sigma F and this can also be written as sigma X F over sigma F. Now, the next point to realize is that in that situation we had a column of frequencies, but now we have a column of probabilities. So, instead of F I can write P and my formula will become sigma X P over sigma P. But sigma P, the sum of the probabilities is 1 and hence the formula becomes sigma X P over 1. That is sigma X P and this is exactly the formula for the mean of a discrete probability distribution. Formally speaking, the mean of a discrete probability distribution is equal to sigma X into P of X and in this particular example, if we multiply the X column with the column of P of X, we obtain 0.15, 0.40, 1.00, 1.80 and so on. And upon adding this particular column, the mean of this particular probability distribution comes out to be 5.925. So, you have seen that our thought that this distribution is almost symmetric as the line chart was clear. This answer is validating because 5.925 is very close to 6 and if this distribution was exactly symmetrical the answer exactly 6. You may be confused that when I explained you first, I said that sigma X into P and now I am saying sigma X into P of X, well that was only to explain the point that F is being replaced by P, the probability. But formally probability of X is always denoted as P of X. In fact, in many situations we do not even write P of X rather we write F of X or if you see this notation in a book, then please do not mix F of X with F. If F is written, then this means frequency. But F of X, if any discrete probability distribution is written, then it represents probability. Next, we consider the spread of the probability distribution. As you see on the screen, in case of a frequency distribution, the formula for the standard deviation is the square root of sigma F X square over sigma F minus sigma F X over sigma F whole square, exactly the same shortcut formula that we discussed in an earlier lecture. Now, this can also be written as square root of sigma X square F over sigma F minus sigma X F over sigma F whole square. But in case of a probability distribution, we will be replacing the frequencies by the probabilities and our formula now becomes square root of sigma X square into P of X over sigma P of X minus sigma X P X over sigma P X whole square. But because of the fact that the sum of the probabilities of a discrete probability distribution is 1, the values in the denominator become equal to 1 and our formula reduces to the square root of sigma X square P of X minus sigma X P X whole square. In order to apply this formula for the example of the number of petals on the flowers of a particular species or for any other such example, of course, we should multiply the X column with the column of probabilities in order to obtain X into P of X just as we did in the case of the mean and we will create another column and that is the column of X square into P of X which will be obtained by multiplying the column of X by the column of X into P of X. So, as you see on the screen in this case multiplying 3 by 0.05 gives us 0.15 and multiplying 0.15 by 3 we get 0.45. The sums of these columns are 5.925 and 36.925 and substituting them in the formula we obtain the standard deviation of X equal to 1.3. Students, the notation for the mean of a probability distribution is mu and the notation for the standard deviation is sigma. The sigma that I am talking about now is the small sigma of the Greek alphabet as you know capital sigma stands for summation. Now that we have computed the mean and the standard deviation of our probability distribution students we can represent it on the graph of our distribution exactly in the same manner as we did in case of a discrete frequency distribution. So, as you now see on the screen the mean of this particular probability distribution is 5.925 and the standard deviation is represented by a horizontal line segment which is equal to 1.3 units. Now that we have computed both the standard deviation and the mean of our distribution students is it not obvious that we can also compute the coefficient of variation as you remember in case of a frequency distribution the coefficient of variation is defined as S over X bar into 100. So, in the case of a probability distribution it will be defined as sigma over mu into 100. And as you now see on the slide the coefficient of variation for this particular example comes out to be 21.9 percent. Students, abhi abhi amne jo example ki aap ne note kiya ke usme jo probabilities hai we arrived at those by the relative frequency definition of probability. Lekin aap jante hi hai ke there are certain situations where probabilities are arrived at according to the classical definition. So, let us consider another example in which we obtain the probabilities using the classical definition of probability. As you now see on the screen if we are interested in finding the probability distribution of the sum of the dots when two fair dyes are thrown we are going to apply the classical definition of probability. In this problem we are also interested in utilizing this probability distribution in order to find the probabilities of obtaining number 1 a sum that is greater than 8 and number 2 a sum that is greater than 5, but less than or equal to 10. In order to solve this problem students the first thing to realize is that if the dice are fair this is going to be the case of equally likely outcomes and we are going to be able to apply the classical definition of probability and we do make this assumption we will assume that the dice are properly made and they are fair and so we apply this definition and according to this definition we have to compute probabilities of the form m over n where n is the total number of possible outcomes whereas, m is the number of outcomes favoring what we want. So, now in this lesson we proceed step by step. First of all the sample space of this experiment as you remember the total number of possible outcomes is 36 which ones are they 11, 12, 13 and so on. After that we have to note that the random variable we are interested in that is the sum of the dots on the two dice and students you will realize immediately that it is a discrete random variable why because 1 plus 1 is 2, 1 plus 2 is 3 and so on or up to sum 2.7 or 4.9. So, in any situation may be may be they are all going to be whole numbers and they range from 2 to 12, 1 plus 1 as I said is 2 but the last outcome 6 and 6 gives us a sum of 12. Now, when the sample space is determined now the probabilities are talk about the table which we have to construct in order to construct the probability distribution. So, as you see on the screen we will first create the column of x which goes from 2 to 12 and then we create the column of probabilities p of x which is to be filled out in such a way that the sum of the column will come out to be 1. Now, in order to fill out this column of course we should count the number of outcomes that are favoring a sum of 2, a sum of 3, a sum of 4 and so on or as you see on the screen the number of outcomes that favor a sum of 2 is only 1 and that is 1, 1. So, the probability that x is equal to 2 where x represents the sum of the dots on the two dice is 1 by 36. Similarly, in order to compute the probability that the sum is equal to 3 we notice that there are two outcomes that favor this particular sum and they are 1, 2 and 2, 1. Hence, the probability that x is equal to 3 is 2 by 36. Similarly, we can compute all the probabilities or if you look at that sample space then you will notice that these are very interesting things that you will keep going diagonally and you will get the sum of the particular sum of those diagonals. So, as you now see on the screen the probability distribution for this particular example is x is equal to 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 and the probabilities are 1 by 36, 2 by 36 and so on. They keep on rising until they attain the value 6 by 36 against a sum of 7 and then they decline and we have 5, 4, 3, 2 and 1. As you have noted, the numerators of these probabilities are increasing in a steady fashion and then they decline in the same manner in a mirror image way and hence I hope you realize that if we draw the line chart of this particular distribution it will be absolutely symmetric. But in this problem, our next question is that we should compute the probability of the event that our sum is greater than 8. How do we proceed? Obviously, greater than 8 means either 8 or 9 or 10, 11, 12 or your love's or is the malhua that reminds us of the addition theorem and the special case of the addition theorem probability of A or B if A and B are mutually exclusive is equal to probability of A plus the probability of B. So, extending this theorem to the case of more than two events, we obtain in this example as you see on the screen the probability that x is greater than 8 is equal to the probability that x is equal to 9 or 10 or 11 or 12 and that is equal to the probability that x is equal to 9 plus the probability that x is equal to 10 plus the probability that x equal to 11 plus the probability that x is equal to 12. Applying the values of these probabilities in this formula, the probability that we obtain a sum greater than 8 comes out to be 4 plus 3 plus 2 plus 1 by 36 and that is 10 by 36. Similarly, the other part of the question was find the probability that the sum is greater than 5 but less than or equal to 10 and this is equal to the probability of x equal to 6 or 7 or 8 or 9 or 10 and adding those probabilities this singular probability comes out to be 23 by 36. Now that we have discussed the concept of the discrete probability distribution in considerable detail, we have seen its basic format. Just two columns, a column of x and a column of probabilities such that the sum is 1, the next concept that I would like to discuss with you is the concept of the distribution function. As you now see on the screen, the distribution function of a random variable x denoted by capital F of x is defined as the probability that capital X is less than or equal to small x. Students, you have noted that capital F of x and capital F bracket is a small x and I said that the probability that capital X our random variable capital X is less than or equal to small x. So, what is the difference between capital X and small x students? Capital X represents the random variable that we are talking about and small x represents some particular value of the random variable. In other words, if I am interested in determining the probability that the sum of the dots on the two dice is less than or equal to 5, I am talking about capital F of 5. As you see on the screen, capital F of 5 is equal to the probability that capital X is less than or equal to 5 and this is equal to the probability that x is equal to 2 or 3 or 4 or 5 and because of the fact that all these events are mutually exclusive and they cannot occur together, this probability is equal to the probability that x is equal to 2 plus the probability that x is equal to 3 plus the probability that x is equal to 4 plus the probability that x is equal to 5. Substituting the values of these probabilities in the formula, we obtain 1 by 36 plus 2 by 36 plus 3 by 36 plus 4 by 36 and evidently this is equal to 10 by 36. In other words, students, the distribution function of a discrete probability distribution is simply the cumulation of the probabilities from the beginning up to that particular value that we are talking about. In this case, 5. You have seen that this concept is exactly the same as the concept of the cumulative frequency distribution. As you see on the screen, if we construct the column of cumulative probabilities for the same example that we just discussed, we obtain 1 by 36, 3 by 36, 6 by 36, 10 by 36 and so on and the last one, the 1 against x equal to 12 comes out to be 36 by 36 exactly equal to 1. These cumulative probabilities represent capital F of x and as you can see the probability that x is less than or equal to 5 that is 10 by 36, it is seen at a glance if we have a look at the column of the cumulative probabilities. So, you have seen that this concept, this is also a very useful concept, whenever you have a particular value or less than a particular value, you can think of the distribution function capital F of x. Students, next time we will take this concept a bit further and after that I will discuss with you in a formal mathematical manner, the concept of mathematical expectation. I wish you the very best in your study of this new topic that we have started today and until next time, Allah Hafiz.


Let us turn now to the conduct of proceedings at the ICJ and see a little bit more concretely what are the various steps and also the incidents that may occur along the way. That is, between the moment that the court is seized of a dispute and the moment it delivers its judgment. The court is seized of a dispute, either through the notification of a special agreement, as you know, or through the filing of an application, if the basis of jurisdiction invoked by the claimant state is a compromise reclose, is two optional closers or an expected consent in a case of foreign progratum. From that moment onwards, the judicial machinery is set in motion. And a first procedural step is taken by the registry of the court, under various provisions of the statute and the rules, the details of which are rather unimportant here. Under those provisions, the registry notifies all UN, the Secretary General of the UN and all the UN member states and other states entitled to appear before the court. A special notification is also addressed to states that are parties to multilateral treaties whose interpretation is in question in the case. And furthermore, if that treaty is the constituent element of an international organisation or if it has been adopted within the framework of that organisation, such organisation will also be notified and it will receive copies of the written pleadings. Notified states may also request to receive copies of the written pleadings of the parties. All these may sound very administrative and purely procedural, but the publicity given in this way to the case and the automatic character of such publicity stands in sharp contrast to arbitral proceedings. When a dispute is brought before the ICJ, the case is public and it resonates within the world community and as we shall see later, it is because a certain publicity is given to a case that various types of interventions in the proceedings exist. Then after having a certain devues of the party, the court issues a first procedural order by which time limits are fixed for the filing of the party's written pleadings. And there are usually two rounds of written pleadings. The claimant writes a memorial exposing all its arguments concerning the jurisdiction of the court and the merits of its own claims and providing of course all evidence in support of its case. The respondent is then offered the same amount of time to write in response what is called a counter memorial. And then usually a second round of written pleading takes place, with the claimant filing a reply and the respondent filing a rejoinder in response to the reply. When the parties have filed all their written pleadings, oral pleadings take place and there again two rounds of oral pleadings are usually organized. At the end of the oral pleadings, the agents, that is the official state representatives having usually the rank of an ambassador, the agents of each disputing state reads out their respective final submissions. And then the court begins its deliberation and the judgment is delivered a few months later. I shall come back on the internal deliberative process leading to the actual drafting of judgments in a separate video. If the respondent state does not appear in court or if it fails to defend its case, the claimant may call upon the court to decide in favor of its claim. But before doing so, and as article 53 of the statute states, the court must however satisfy itself not only that it has jurisdiction but also that the claim is well founded in fact and law end of quote. Well, this is the procedure in its streamlined form when no procedural incidents occur. But incidental proceedings do happen most of the time and they will increase the complexity of the entire process. Let us turn now to those incidental proceedings in the next videos and readings.


So, there's multiple layers of stupidity to this story that we're about to talk about, so I will try to make it as comprehensible as possible to people who aren't on Twitter and or terminally online like myself. So, it starts with self-proclaimed theocratic fascist Matt Walsh reacting to last week's Nashville mass shooting carried out by a trans suspect by using that opportunity to claim all trans people are evil saying, quote, I came to the conclusion years ago that the trans movement is the greatest evil our country faces, I only become more and more sure of this fact with each passing day and more and more determined to oppose it until my last breath.  So now, the right is claiming that AOC is the one who actually wished death on Matt Walsh because that account is actually AOC's burner account according to them, and because of that claim, well, that response to Matt Walsh went viral with a lot of right-wingers thinking that she actually said this to Matt Walsh. For example, Miles Ian Chong tweeted, what did Alexandria Ocasio-Cortez mean by this? Matt Walsh also called her out saying, hi AOC, this isn't you wishing death on me, is it? Many are speculating that this is your burner account. You need to come out and denounce this kind of rhetoric, or we can only assume that you support it.  That is a disturbing name. Excited to see an unhinged trans murderer, a member of the conservative press, Matt Walsh? Yes, that is demonic so the name fits.  Well, it's because of a now viral tweet from leftist turned right-wing propagandist, Nico House.  Now after Tulsi's campaign was over, since leftists no longer took him seriously, he predictably started grifting towards the right almost immediately, just like his mommy Tulsi. And now the quintessential Tulsel himself here claims that he has definitive proof that Zaza Smoka on Twitter is AOC's burner account. And that's because of this. Here's his now viral claim, AOC has burned accounts. I think he means burner accounts. I responded to one of her posts reminding her that she sends money to Nazis and some random account responded to me speaking in first person as if I attacked them. Then that same person deleted the tweets. All right, so let's check it out. He responded to AOC saying, but you vote to send money to Nazis and to fund Israeli apartheid. But hey, at least you stood up to a TikTok star. Zaza Smoka responded to that saying, LOL. And what makes you think that I did anything to support Nazis? Your delusional seek help. So there you have it. AOC responded to Nico House in the first person while not realizing that she was still logged into her burner account. Got her. He's got her dead to rights, right? Not so much, because as people quickly pointed out, when Nico originally responded to AOC, he tagged both AOC and Zaza Smoka, who also responded to AOC, meaning that Zaza Smoka got a notification after Nico House responded to them and seemingly thought that Nico House was directing his Nazi accusation to them. Hence the confusion in Zaza's reply. And as Ravenna put it, this bozo replied to an account that retweeted AOC and didn't untag that account before he did it, which I think is a lot more believable than AOC having an alt named Zaza Smoka with an anime avatar. But hang on a second. If Zaza Smoka is not AOC, then why delete the response to Nico House?  I've done it. Most people on Twitter have done this. So it's not that suspicious. And it's just a little bit ironic of Nico House of all people to call somebody else a Nazi defender or supporter of Nazis considering he's friends with a very popular white nationalist. Now the claim that Zaza Smoka is AOC's burner account gets even more dumb. Given that as the surface points out, the account liked Nazi for work posts on Twitter, including risque photographs of other women. So if we're going to assume that this is AOC's burner account, we have to assume that she didn't just create it to respond to shit posters on Twitter, but she created it also to presumably, I guess inconspicuously like thirst traps of other women. I mean, I guess it's not impossible, right? But you also have to assume that she's either a closeted lesbian or at least bisexual at a minimum if this is indeed her account. And listen, I want to be very clear here. I wish this were true. I think that AOC using a burner account to covertly wish death on Matt Walsh is objectively hilarious. So I want it to be true as we all do. But I'm just saying that there's not enough evidence to support that claim. But regardless, that was apparently enough evidence for right wingers and even a sitting member of Congress, Marjorie Taylor Greene decided to promote this claim. Now this shouldn't be too surprising that they're running with this given that Marjorie's proof that all Democrats are pedophiles is their support of bottom surgeries for trans kids, which is something that she claimed was happening with zero evidence in an interview with Leslie Stahl of 60 Minutes. So their standard for evidence, it's obviously not very high. Nonetheless, I mean, why not run with this? Why let evidence or a lack thereof get in the way of a good narrative, right? But regardless, the right will continue to push this false narrative because that's just kind of what they do. So with that being said, it's incumbent on all of us normal people who base our worldviews on empirical reality to push back and debunk false claims wherever we see them. And this is one of those instances where we need to push back and stop this from spreading like wildfire, even though it may be a little bit too late. And I've got to say that it's a little bit worrying that so many people took this claim and ran with it because we're entering this new era where AI is getting increasingly convincing and deep fakes are looking a lot like reality. So if people already don't really have that high of a standard for evidence, how bad is it going to get when we enter this era where it's going to be even more difficult to disprove things? So this is why we all need to just hope that people are a little bit more rigorous in things that they're willing to accept. Because I mean, if we're just willing to assume that anyone has a burner account because they respond to a reply in the first person, then that's bad for all of us on Twitter because what's stopping somebody else from just creating an account and going under a public figure, responding to the reply in the first person and getting other people to think that that's them. I mean, do we really want this? This isn't good for Nico House. This isn't good for Miles Ian Chong or Matt Walsh or Marjorie Taylor Green or myself. It's not good for any of us. So the best thing that we can all do is just ask for a higher standard of evidence before falling for something, hook, line and sinker. That being said, though, I agree that it would be hilarious if this was indeed AOC. But unfortunately, I'm not buying it. This isn't the Matt Bender Lib Tardo one moment that they were hoping for. But I understand why they would want this to be true, because that is in fact very incriminating and simultaneously hilarious. But you just need more evidence to push this claim. Sorry.


So you want rake may is the next quarterback of the New England Patriots May has the prototypical build you look for in a QB He's athletic with arm strength and accuracy His size and skill set are similar to Chargers quarterback Justin Herbert for a fraction of the cost Where he gets into trouble is his penchant for hero ball instead of living to play another down plus He's on the young side if the commanders fall in love with Jane Daniels at number two May could fall into the Patriots lap in number three and all it would cost them is a little patience Committing to drafting a quarterback allows the Patriots to spend in free agency This receiver class boasts T Higgins, Mike Evans, Michael Pittman Signing any of those players and drafting a tackle with their early second round picks starts turning the ship in the right Direction if Drake may falls to you at number three Now all you have to do is hope that he can play quarterback in the NFL hit on free agents and draft Well at other positions piece of cake You


Hello and welcome to the session. In this session we will discuss the change of base in log rhythms. Now we have to prove that log N to the base A is equal to log N to the base B into log B to the base A and here N, A and B are all positive numbers. Now let us start with its proof. Now let log N to the base A is equal to X and log N to the base B is equal to Y and log B to the base A is equal to Z. Now using the definition of logarithm that is if three numbers A, X and Y are written as A is super X is equal to Y and in logarithmic form we can write them as log Y to the base A is equal to X. So by using the definition of logarithm in these three equations for the first equation we will have A raised to power X is equal to N for the second equation we will have B raised to power Y is equal to N and for this equation we will have A raised to power Z is equal to B that is A raised to power X is equal to N and here N is equal to B raised to power Y and also B is equal to A raised to power Z. So this is equal to A raised to power Z whole raised to power Y which is equal to A raised to power YZ. This implies A raised to power X is equal to A raised to power YZ. Now here the bases are same so we can compare the parts. So this implies X is equal to YZ. Now using the values of X, Y and Z this equation will be log N to the base A is equal to log N to the base B into log B to the base A. So this formula changes the base from A to B. Now in the same way we can prove that log N to the base B is equal to log N to the base A into log A to the base B that is here we are also changing the base by using this formula also from this formula now let it be equation number one. So from one N to the base B is equal to log N to the base A over log B to the base A. Now let it be equation number two. So from two we have log A to the base B is equal to log N to the base B over log N to the base A and also log B to the base A is equal to log N to the base A over log N to the base B. Now this is the equation number one so for N is equal to A in the formula which is given by equation number one log A to the base A is equal to log A to the base B into log B to the base A. Now log A to the base A is equal to 1 so this implies 1 is equal to log A to the base B into log B to the base a. Now this is a very important result which is to be remembered. Now let us discuss some corollaries. Now the corollary 1 is log y to the base s is equal to log a to the base x over log a to the base y. That is by using this result which we have discussed earlier. Now by using this result which we have discussed earlier, we are getting the results that is log a to the base x is equal to 1 over log x to the base a whole upon log a to the base y is equal to 1 upon log y to the base a. So this is equal to log y to the base a whole upon log to the base a. Therefore log y to the base x is equal to log y over log x. So we can write this as log y over log x as both are having the same bases and also we have y, x and a are greater than 0 and also all these three are not equal to 1. Now let us discuss one example. Here log 18 to the base 2 can be written as. Now by using this result it can be written as log 18 to the base 10 log 2 to the base 10. Now log 18 to the base 10 is equal to 1.2553 over log 2 to the base 10 is equal to 0.3010 which is equal to 4.17043. This is how by using this result we can solve log 18 to the base 2. Now let us discuss the second problem is log a to the base b into log b to the base c into log c to the base a is equal to. Now by using this result which we have discussed earlier. Now log a to the base b can be written as log a to the base m over log b into. Now log b to the base c can be written as log b to the base m over log c to the base m into this can be written as log c to the base m over log 2 turning this will give 1. Therefore log a to the base b into log b into the base c into log c to the base a is equal to 1. Now let us discuss this problem which is log y raise to power m whole to the base x raise to power n is equal to log y raise to power m whole upon log x raise to power m to the base a. Now using the third law of logarithm log m raise to power n to the base a is equal to m log m to the base a. So this is equal to log y to the base a whole upon m log x to the base a. Now again using this result. So here log y to the base a over log x to the base a will become log y to the base x. So this is equal to m over n into log y to the base x. Now let us discuss one example. Now log 125 to the base 4 can be written as log 5 raise to power 3 to the base 2 raise to power 2. Now this is of the form log y raise to power m whole to the base x raise to power n. So here m is 3, m is 2, y is 5 and x is 2. So by applying this formula this will be equal to 3 by 2 into log y to the base x that is log 5 to the base 2. Now let us discuss the next problem which is x raise to power log y to the base a is equal to x raise to power log log y to the base x into log to the base a which is further equal to raise to power log y to the base x whole raise to power log x to the base a which is further equal to y raise to power log x to the base a. Now here we are using the change of base formula and here x raise to power log y to the base x is equal to y because x are positive real numbers and also a is not equal to 1 then a raise to power log x to the base a is equal to x. So in this case also x and y are positive real numbers and also x is not equal to 1 this implies x raise to power log y to the base x is equal to y. So these were the various corollies and in this session you have learnt about the change of base formula in logarithms. So this completes our session hope you all have enjoyed this session.


Hello ladies and gentlemen fellow traders, this is Thao Stow from Cybertrain University Just want to do another round of teaching and educating all of you regarding how to spot iceberg orders so what we do here at Cybertrain University and we are showcasing and showing you again the book viewer which is right here on the left-hand side we like to call it the level 3 screen Which is where you get to see where are the institutions all the dominating Market makers you know hiding and putting their orders and something called an ECN If you subscribe to our channel You'll see a couple of alerts and I talk about what our ECNs, but I want to kind of showcase What we're looking at right here and trying to educate Everybody to show how we use these tools and and why this is so important and why a lot of you are trading blind and I want to point out microsoft because Microsoft you could see we're looking at two over here on the right We have two charts. We have a one-minute chart and we have a one-day chart and what really Stands out here if you look at the intraday chart appear up on top You'll notice that there is a big big support levels Right around here the stock has been hovering here from 10 o'clock Up until 10 30 up until you know 10 44 in the morning and then even right now Up until 3 30 and you're trying to figure out like why is this stock having of tough time going past 164 It had every time it hits it it goes back to 165 comes back to 164 goes back to 165 Then it went all the way up to 166 and then it came back down But why 164? Well, it's very simple ladies and gentlemen buyers What makes support levels on a chart Are buyers unfortunately you can't see the buyers on the chart That is what happened in the past the only way to see And that is what's indicative of the future Is seeing the orders that are out there and that's why we're looking at the nasdaq book viewer and if you look over here Remember as the stock is trending down You're looking for buyers and the buyers are right here on the buy side over here on the left And we've got three columns. You've got orders chairs and bids There is a order at every single penny 100 chairs are going to be bought at 164 39 1,400 as you're working way down But you'll notice out of all those orders you got to work all the way down to Finding where the next predominant order is and if you look right here You don't see anything unless I scroll down And then voila You have 297 orders out there that make up 21 596 shares at 164 Now listen if you look over here going back you got orders hundred chairs you got One order three orders four orders, you know two orders, but when you work your way down, this is like almost like black and white Well that ladies and gentlemen fellow traders Explains why the stock is having a tough time breaking 164 so to answer your question that when you're looking at a important resistance levels and you're trying to figure out Is it going to go past it? Will it break it or will it be a support? Well, ask yourself.  Because when I was trained being a market maker, I was always taught that it didn't matter what fowl still thought It matters what they thought Okay, you're not trading 21,000 shares of 164 dollar stock Making up over 299 different people out there that are making up that order So the thing is if you want to know how to trade the market You first got to learn how to follow the market. You want to learn how to follow the market? You follow the money So to find more videos like this ladies and gentlemen Just click on the like on the bottom if you like this video and just click subscribe And I'll be sending more videos and more alerts to see more What we call here at Cybertree University iceberg orders, which are big block orders So you can identify these orders so you can help you in today's market conditions Thanks for watching and look forward to seeing you on my upcoming videos