 Hi, I'm Zor. Welcome to Unizor education. I would like to present another series of problems. These are the problems which I include into the chapter dedicated to theory of probabilities, which is a part of advanced mathematics course for teenagers presented on Unizor.com. I do suggest you as usually to number one, watch it from the website Unizor.com, not just from the YouTube, because the website contains comments. And number two, before you actually watch the lecture, look at the comments, look at the notes, and they contain all the problems which I'm going to present to you. And I strongly encourage you to try to solve them just by yourself. That's a good exercise. And then obviously listen to the lecture. All right, let's start. Well, the first thing which I would like to remind you is that these and many other problems in theory of probabilities are very easy represented in the graphical form. Basically, probability is a measure. And as a measure it's completely analogous to the concept of error, for instance, on the plane. So if you have something like piece of the land and then a smaller patch inside, you can compare what is the area of this corresponding to the area of the entire piece of land. And the ratio, which is obviously from zero to one, is completely analogous to the probability problem. So if you randomly drop the point inside this entire area, the probability of having this point inside this area is obviously proportional to the area of this of this little patch. So that's why basically we have this analogy. Now, let's just continue with this analogy. If this area represents, well, consider the area of an entire page, the entire piece of land is equal to one in whatever units, doesn't really matter. Then I can say that the area of this particular page is actually a concept completely equivalent to the probability. Now, so what's the probability of an opposite event? So if this page represents the event, then everything outside of this page represents an event which is not exactly this one, right? So if this represents, okay, something happens, then everything outside represents this did not happen. And the probability of this of happening, plus probability of this not happening, obviously, equals to one. Now, also, if you have two areas, for instance, this one, which is, let's say, A, and this one, which is B. So these represent two different events. So again, if the entire area of an entire piece of land has an area of one, then the area of this is the probability of the point falling into this area. And the probability of this is basically the area of this is a probability of the point randomly thrown onto this land, is to fall within the boundaries of A. Now, then I can just use a theory of sets, the set theory, to represent another event. What, for instance, representation of event when both A and B are happening? Well, obviously, it's an intersection between them, this little piece, right? Now, what's the probability of A or B? Well, that's the union of these. So, intersection is analogous to end, union is analogous to or. And the opposite, the not event I was already saying, that's outside of the event. So, I will use this to basically demonstrate the solutions to the problems, which I was talking about. Now, another thing, which is also very graphically useful to take a look at is conditional probability. If I would like to know what's the conditional probability of event, let's say, B, on the condition that A has happened. Now, what does it mean? It means that I already know that whenever I threw the point randomly onto this particular piece of land, I know beforehand that it's already inside the boundaries of A. So, what's the probability of B happening under this condition? Well, obviously, it's a ratio of this piece relative to the ratio of A. Completely equivalent to the probability of A in the first place. It's the ratio of area of A divided by ratio of an entire piece of land. Well, if it's equal to one, then it's just an area. But in any case, it's a ratio, right? So, in this case, the conditional probability is basically the probability of A and B, and I'm using the set theory intersection divided by probability of A. So, it's ratio of this area divided by this. So, this is a graphical representation of probabilities which I'm going to use. So, that's basically my preamble, preamble to all these problems. Alright, now let's go to the problems. Now, let's consider the following fact. I know that from event Y, if it happens, follows that event X is also happening. Now, first of all, before I'm asking questions, how to represent it graphically? Well, here it is. Now, this is again our piece of land and this is our X. Now, what I'm saying right now that if the point falls into the Y, it automatically means it falls inside the X. What it means? It means that the Y should be somewhere here, inside the X, right? Only in this case, from occurrence of event Y, I have automatically occurrence of event X. So, that's the graphical representation of following from Y follows X. If from Y follows X, Y should be graphically inside. Alright, now, here are the questions. True or false? Not Y, X. Is it true or false? Well, let's just think about it. What is not Y? Not Y is everything which is outside of Y, graphically, right? Now, I know that the point actually falls inside of this, not Y, which means everywhere outside of Y. Does it automatically mean that the X actually happened, that the point is inside X? No, not at all, because let's take this point, for instance. This point is outside of Y, so it belongs to this, but does it belong to X? No. So, from this, it's not following. I should have put this forward. So, this is false. Next one. From Y follows not X. Is this right? Well, let's just think about it. What is not X? Not X is everything outside X, right? It's this error. So, I know that my point falls into this error. Does it mean that it automatically falls into this error outside of X? Well, actually, absolutely not. It's just an opposite. We definitely know that it's not following. It's not falling inside of this. So, this is also false. Next. Not Y follows not X. Is this true? Let's just think about it. Not Y means that the point actually is outside of the Y area here. Not X means it's outside of X, which is this area. But let's consider this point, which is in X, but not in Y. Well, since it's not in Y, it's actually corresponding to this particular condition. Not Y happens. So, the point is in the not Y. Does it mean that it's automatically in the not X? Not X is outside of this, and the point is inside X. So, that's not true. So, that's also not true. That's false. Next. From X follows Y. Okay. Is this the true statement? So, if my point falls inside the X, does it mean that it automatically falls inside Y? Obviously not, because the point can be again in this position. It falls inside X, but it does not fall inside Y. So, X happens and Y does not happen. So, this is not the correct implication. False. Next. Not X, Y. So, again, if I randomly drop the point inside this square, I know that not X happened, which means the point is somewhere here, outside of X. Does it mean that it's inside Y? Absolutely not. It's just an opposite. It's not. So, this is false too. Next. From X follows not Y. Is this true? So, X happens, which means I'm inside this area. Question is, does it automatically means that I am, that not Y happens? Well, not Y is here, outside of Y. All I know is that the point is inside X. But now, if the point is here, it's inside X, but it's inside Y, not outside of Y. So, again, this is not true. And the last but not least, not X follows not Y. Is this true? From not X. Okay. Not X means everything outside this bigger area, right? This one. Now. So, I know that I am inside of this area. This is not X. Now, my question is, does it mean that I'm automatically outside of the Y? Not Y. Which means I'm in this area. Yes, absolutely. Because every point which is inside not X is also inside the bigger area. So, not Y is bigger than not X, which means whatever, if I know that not X happens, then not Y definitely happens automatically. So, this is actually true. And that's the only a true condition which is related to this one. If this is given, then this is also true. If this is true, this is true. Or others are false. So, that's my first problem. And as you see, my graphical representation is very useful in these cases. Basically, the whole language of theory of probabilities, I have transferred to the language of the set theory, like included, union, intersection, excluded and stuff like this. The problem number two. I know that both are true. So, I have events X, Y and Z. And I know that if Y happens, X happens. If Z happens, X happens. So, X follows both from both, from Y and from Z. Again, before I'm presenting the problem, let me graphically explain how it works. Again, I have this big piece of land. I have somewhere X. And since this X actually follows from Y and follows from Z, both Y and Z are supposed to be inside. Then I know that if my point randomly dropped onto this square, if it falls inside Y, it also falls inside X. Which means if Y happens, X happens. If it falls inside Z, it also falls inside X. So, again, from Z, I have an implication to X. X follows from Z. So, that's my condition. Now the questions again, similarly. Now the first. Y and Z, from this, follows X. Is X following from Y and Z? Well, let's just think about what is Y and Z. Y and Z means that both events happened. So, my point falls inside Y and inside Z. It means it's inside this intersection between these two. So, if this is true, if the point falls inside the intersection, does it mean it's inside? And it's inside X, which means X happens? Yes, absolutely. So, this is true. Next is OR Z. Is this true? Well, what is OR? Graphically, OR is represented as a union. So, I know that Y or Z happened. It means my point randomly dropped on this square falls inside Y or inside Z, which means it's inside the union of these two. OR is equivalent of the union. And is equivalent of intersection OR is equivalent of the union, right? Now, if my point is inside this union. Now, I know that Y is inside and Z is inside X, both. Well, union is obviously as well inside, which means that if my point is within the boundaries of this union of Y and Z, it's also inside X, which means X happens and this is also a true condition. Next. Oh, this is complicated. NOT X OR NOT Z. And my question is whether NOT Y follows from this. NOT X OR NOT Z. Okay, let's just think about it. What is NOT X? NOT X is everything outside X, which means this area. NOT Z is outside of Z, which means this area. Connecting them with OR, which means all the points with this type of shading and this type, they are OR, which means we are unionizing them all together. But now let's just think about it. NOT Z is definitely bigger than NOT X, right? Because Z is inside, so whatever is outside of Z would be definitely including in itself whatever is outside of X. So basically the union of these is exactly the same as NOT Z. And now my question is whether from NOT Z follows NOT Y. Well, NOT Z is everything outside of Z. NOT Y is everything outside of Y. Now, whether from one follows another. Well, obviously, no. Because if the point is somewhere here, which is outside of Z, which means it belongs to NOT Z, it's inside Y. So it definitely inside means it's not outside, right? So NOT Y would not be true. So this is an example of the point which which is inside of this area, but not inside that area, which means from this I cannot say that this one follows. So this is false. This is false. Next, I think I have to redraw it because I have too much drawn already on it. So this is my X. This is my Y. And this is my Z. All right, next. NOT X OR Z. NOT X OR Z. And I'm asking if NOT Y follows from it. X OR Z. X OR Z. Well, since Z is inside X, X OR Z, which means all the points which belong to X or Z are actually exactly the same as X. So this is actually the same as X. So NOT X, NOT X, and this is NOT Y. NOT X means here. NOT Y is outside of Y. Now, as you see, everything outside of Y is bigger than everything which is outside of X. Outside of X is only this area. Outside of Y is this area. So if I know that the point is here, which means outside of here, it's definitely inside of NOT Y. So this actually does follow. This is false. Again, I have to redraw this. X. Okay, next. NOT X, NOT Y and Z. Okay. Let me just wipe up. NOT X, NOT Y and Z. Y and... All right, let's just think about it. Y and Z is intersection. NOT Y and Z is a rather big area, everything outside of this shaded intersection. Now, NOT X is everything outside of X. Definitely smaller than everything outside of this, because outside of this includes everything here, right? So this is smaller. This is part of this, and since this is part of this, if points falls into this area, it definitely falls into this area. So from this condition, this does follow. NOT X, NOT Y or Z. Well, this is also kind of the same thing, because Y or Z is this union, right? Now, everything outside of this union, which is this, is bigger than everything outside of X, which is just this. So again, if my point falls into this area, which is outside of X, it definitely falls into this area, which is outside of this area, which is much bigger. So this is also true. That's the second problem. Next. X and Y are mutually exclusive. Okay, again, first of all, let's just present it in a graphical format. This is X. Mutually exclusive, it means there are no common elements, no common elementary events. So in this case, that's the way how it's presented. There are no common elements. Intersection is empty, so to speak. Okay? So that's what mutual exclusivity is. Now, question is, are they independent? Now, back to my definition of conditional probability. You remember conditional probability actually was B conditioning of A is equal to NB divided by probability of A. Now, what is independence? Independence between them means that this is exactly the same as probability of B. If B is independent of A, then the probability of B occurring if A occurs should be the same as the probability of B occurring by itself. That's what independence actually, by definition, is. So all we have to do right now, we have to check if this is true. Well, in this case, it's not X, A and B, it's X and Y. So let's just check this particular identity if it's true. So the probability of X intersection Y should be equal to probability of X times probability of Y, right? I just multiply both things by probability of A. So that's what we have to check if we want to talk about independence. That's what independence means. Independence of simultaneous occurrence of two events is exactly the same as the product of their probabilities. Okay? Alright, so let's just think about it. Now, obviously, the probability of X and probability of Y, which in this particular geometrical model are just areas of these two. If the area of entire square is equal to one, then these are areas which are really probabilities. Now, what's the probability of their intersection? Zero. Right? There is no intersection, so the probability is zero. There are no elementary events coming. So this is non-zero and this is zero, so definitely there is no independence. It's not independent. If they are mutually exclusive events, they are not independent. Because if something happens, then we definitely know that it affects the results of this. We cannot say that if they are mutually exclusive, then the chances of this completely independent outcomes are completely independent of the outcome of this. Because as soon as this is happening, we already cut out certain number of elementary events which have happened and they could not happen for this guy, since they are mutually exclusive. And that's what makes the whole, that's what fails the independence test. Next, how about X and not Y? Okay, X and not Y, probability of X and not Y. Again, it should be equal to the product of their probabilities. Is this true? No, it's not. Because again, this is definitely not equal to zero. Probability of X is area of this. Probability of Y is area which is outside of Y. Which is also not zero. And X and not Y are X and not Y. This is equal to, let's just think about what's the intersection between X and not Y. Now this is X, not Y is everything outside of Y. So their intersection is everything which is X actually. So this is P of X. So if this, can this be equal to this? No, only if this is equal to one and Y is definitely. I mean in certain cases, yes, that's true. If Y is an entire sample space but we are talking about general case. In general, since probability of not Y is not equal to one, that identity is not true. So this is also not true as well as this one. Next, not X and not Y. Okay, let's do it again. Not X and not Y. Probability of not X and not Y. Now if they are independent, it should be equal to the probability of not X multiplied by the probability of not Y. Is it true or not? Let's just think about it. Now not X is outside of X, right? Not Y is outside of Y. So if I'm talking about end condition, that's actually the area outside both of them, right? So whatever is right now shaded, that's the area which is this. Now this is not X. It's a probability of area outside of X and this is the probability of area outside of Y. Now both of them are bigger than this one. So it's not really obvious whether the product of their probabilities is equal to this one. I think to approach this problem, it's probably better to use some concrete example. Now what can be a concrete example? Let me just think about it. What if let's say my area, now what if my area is, this is X and the area is one-eighths for instance. And this is Y and the area is one-quarter. Just let me try to use this as an example. Now what is not X and not Y? That's basically this area which is what? One minus one-quarter and one minus and minus one-eighth which is what? It's eight minus four is four minus one-three-three-eighths. Now what's the probability of not X? If X is one-eighth then not X is seven-eighths. What is the probability of not Y? Y is one-quarter so not Y is three-quarters. Is this product of these equal to three-eighths? Well obviously no, I mean in this case you see it's three-eighths times seven-fourth. So this is just an example of the case when they are not equal. So if I found just one example it means that this is not really a true implication. So this does not follow from this in general case. Okay so that was my third problem and let me go to the number four. Number four is about standard deck of cards. So we have 52 cards. We have ranks from two to ten, Jack, Queen, King and Ace. And we have four different suits. We have spades. We have hearts. We have diamonds. And we have clubs. Now my experiment is pulling one card out of this deck of 52 cards. So I'm talking about certain number of events. Event number one, Queen is pulled. Event number two, Space pulled. Number three, Spades or Hearts are pulled. And number four, Queen of Spades is pulled. Now I have questions. Whether event E1 is independent from E2 or E3 or E4. Let's just think about it. Okay, let's just do the arithmetic. The same thing as in the previous case. What's the probability of E1 by itself? Well, there are four different Queens in the deck of cards. So we have the probability equals to 452, which is one-thirtieth. Probability of E2 equals two Spades. Now how many Spades I have? I have four different suits. So I have 13 of each. So the probability is 1352, which is one-quarter. Probability of event E3, Spades or Hearts. So it's 13 Spades, 13 Hearts, so it's 26. 2652, which is one-half. And finally, what's the probability of having a Queen of Spades? Well, that's only one in an entire deck, so the probability is 152. Okay, now let's talk about combination of these events. E1 and E2. So what's the probability of picking a Queen and at the same time of Spades? Well, obviously this is E1 and E2 equals to 152, right? The Queen of Spades is only one card. Now let me compare it with the product of their probabilities. One-thirteenths and one-fourths, which is actually product would be 152. So this E is E1 and E2 are independent. Now let's talk about E1 and E3. Probability of E1 and E3. So how many cases when both are happening? Spades or Hearts, and it's a Queen. So there are only two cards, Queen of Spades and Queen of Hearts. There are no others. So the probability is equal to 252. Now if I will multiply probability of E1 and probability of E3, it's 113 times one-half. It's 126. Now 252 is 126 actually, right? So that corresponds as well. So the product of probabilities is equal to the probability of their intersection. So this is also independent. Okay, now about E4 and E1. So this says Queen is pulled and this says Queen of Spades is pulled. The probability is actually 152. So I know it's Queen and I know it's a Queen of Spades. So there is only one card out of 52, right? So it's 152. But if you will compare the probabilities, the P of E4 is 152 already and P of E1 is 113. So if you multiply them together, obviously it would be some very small number. One over 13 times 52. Definitely not equal. So this is not independent. E1 and E4. And for a good reason, because in the condition of this problem, Queen is in both cases. You see, here we are talking about different characteristics of the cards. One is a rank, another is a suit. And this is also rank and suit. But this is rank and this is a mixture of rank and suit. And that's what disturbs the independence. It's no longer independent. So that was my fourth problem. Thanks very much. I do suggest you to go again through these problems on theunisor.com and try again to solve just by yourself. And then if you cannot, just go back to the lecture. That will be it for today. Thanks a lot and good luck.