 Assalamu alaikum. Welcome to lecture number 18 of the course on statistics and probability. You will recall that in the last lecture, I discussed with you the concept of permutations and combinations. And after that, we discussed a number of important and fundamental concepts such as the random experiment, the sample space of a random experiment, the concept of events and different types of events such as the mutually exclusive, the equally likely and the exhaustive events. In today's lecture, I will discuss with you the various ways in which probability can be defined. These include the subjective approach as well as the objective approach. And then under the objective approach, we have different ways of defining probability. But students, before I begin these formal definitions of probability, I would like to revise with you briefly the concept of mutually exclusive events, exhaustive events and equally likely events. Let us begin with the concept of mutually exclusive events. As you now see on the screen, two events A and B of a single experiment are said to be mutually exclusive or disjoint if and only if they cannot both occur at the same time. In other words, they do not have any point or element in common. For example, when we toss a coin, we get either a head or a tail, but we cannot have both head and tail at the same time. Similarly, when a die is rolled, the events even number and odd number are mutually exclusive as we can get either an even number or an odd number in one particular throw and we cannot get both of them at the same time. Coming to real world situations, a student will either pass or fail in an exam. Similarly, a person will either be a teenager or he or she is not a teenager. It is not possible that both these things are happening at the same time. So, I think you will agree that it is a fairly simple concept. If we talk about three or more events at the same time, such events which are originating from the same experiment, they will be said to be mutually exclusive if they are mutually exclusive pair wise. On the contrary, if two events are such that both of them can occur at the same time, then of course, we will say that they are not mutually exclusive. For example, if we draw a card from an ordinary deck of 52 playing cards, it can be both a king and a diamond at the same time. Therefore, diamonds and diamonds are not mutually exclusive events. When we talk about playing cards, you will know that there are four types of cards in an ordinary deck of cards, hookah, cheerya, eat or paan. And in English, of course, we say spade, club, diamond and hearts. And for the three numbers 11, 12 and 13, we have pictures on the cards and so we call them face cards, the king, queen and the jack. The cards of spades and clubs are black in color and the cards of diamonds and hearts are red. And students, it is important that you keep all these points in mind because a number of books that you will study in order to study problems of probability, you will find a number of examples relating to playing cards. Alright, having discussed the concept of mutually exclusive events, I would like to revise with you the concept of exhaustive events. As you now see on the screen, events are said to be collectively exhaustive when the union of mutually exclusive events is equal to the entire sample space S. For example, in the coin tossing experiment, head and tail are collectively exhaustive events. Similarly, in the die tossing experiment, even number and odd number are collectively exhaustive events because if we take the union of the two, we get the entire sample space 1, 2, 3, 4, 5 and 6. Students, this concept of mutually exclusive events is related to the concept of mutually exclusive events. Inme, dechukiho and that is the concept of the partition of a set. As you now see on the screen, a group of mutually exclusive and exhaustive events belonging to a sample space S will be called a partition of the sample space. And with reference to any sample space S, events A and A bar, the complement of A will form a partition because they are mutually exclusive and exhaustive. This fact is clearly depicted by the Venn diagram because as you can see, the events A and A bar are mutually exclusive. They do not have any points in common and when we take the union of these two events, we obtain the entire sample space. The next concept that I would like to review with you is the concept of equally likely events. A concept that is very, very basic and important for the classical definition of probability that I will be discussing in a short while. As you now see on the screen, two events A and B are said to be equally likely when one event is as likely to occur as the other. In other words, each event should occur in equal number in an indefinite number of trials. For example, when a fair coin is tossed, the head is as likely to appear as the tail and the proportion of times each side is expected to appear in a very large number of trials is 1 by 2. Similarly, if a card is drawn out of a deck of well shuffled cards, the each card is equally likely to be drawn and the proportion of times that we would expect any card to appear in a very, very large number of draws, this proportion will be 1 by 52. Having revised the concepts of mutually exclusive, exhaustive and equally likely events, let us now begin the discussion of probability itself. As I said earlier, students probability can be approached in two ways, the subjective approach and the objective approach. As you now see on the screen, subjective or personalistic probability as it is called is a measure of the strength of a person's belief regarding the occurrence of an event A. Students probability defined in this manner is purely subjective. Subjective se matlab yeh, k different persons faced with the same evidence can arrive at different conclusions regarding the probability of a certain event. For example, suppose that a panel of three judges is hearing a trial. Aik hi evidence hai jo tino judges ke saamne pesh kia jaara hai. But it is possible that two of the judges arrive at the conclusion that the person who is accused is guilty. But one of the judges thinks that the evidence of the trial is presented is not strong enough to arrive at this conclusion. So, students you can see ke subjectivity ka jo element jo main aap se pehle baat ki, iske andar bohot zyada maujood hai. Aik hi evidence ho or aik hi situation ho and different persons would arrive at different conclusions regarding the probability of a particular event. In fact, agar hum gaur kareen, to iss tari ke se agar hum probabilistic decisions leen, to hum unhe quantify nahi kar sakte. Ham iss tara ke decisions apni rose marra zindagi mein har waqt leh rahe hote. Agar aap ghar se nikalthe waqt, chhatri leh ker nahi nikalthe, to iss ki bhaja yeh hoti hai, ke aap ke zehen me ye baat hoti hai, ke suraj sarpe nikla wa hain hai. To iss baat ki probability bohot kam hai ke paach minute ke baat baarish ho jaye. Iss tara ke decision hum lete to har waqt hi rehte hain. Lekin we are not able to quantify these probabilities the way we statisticians would like to do for our statistical work. This is why for statistical purposes the other definitions are the ones that we are more interested in and these are the ones which fall under the category of the objective approach to probability. The objective approach implies that every person faced with a particular situation will arrive at the same result or the same conclusion regarding the probability of a particular event. Under the objective approach the first definition that I would like to discuss with you is the classical definition of probability. As you now see on the screen if a random experiment can produce n mutually exclusive and equally likely outcomes and if m out of these outcomes are favorable to the occurrence of a certain event A then the probability of the event A denoted by P of A is defined as the ratio m over n. Students this definition was formulated by the French mathematician Laplace and it can be very conveniently applied in those situations where the outcomes are equally likely and where it is easily possible to count the total number of outcomes and the outcomes favorable to a particular event. Agatesh's definition is very long but in reality it is very simple and I would like to explain this concept to you with the help of an example. Suppose that a card is drawn from an ordinary deck of 52 playing cards and suppose that we are interested in finding the probability that the card is a red card. This is a very simple problem and now let us see how we approach it. The first thing is that if it is a well shuffled deck of cards you have shuffled it very well. So intuitively you can understand that when you draw a card then every card is equally likely to be drawn. And this point is the crux of the matter. If this condition is equally likely then we can apply this classical definition. If this condition is not complete then there is no way we can apply the classical definition. Here if we know that it has been shuffled very well then of course we are in a position to apply this definition. Once we are satisfied that we can apply it then we have to compute the ratio m over n. m is the number of outcomes that are favorable to what I want but n of course is the total number of possible outcomes. And as you now see on the screen in this particular example the total number of possible ways in which we can draw one card out of 52 is obviously 52. Also because we are interested in a red card and because of the fact that we already know that the cards of diamonds and hearts are red in color and there are 13 of each of them. Therefore the total number of outcomes favorable to what I want is 13 plus 13 and that is 26. Now that it is clear that m is 26 and n is 52 the probability that our card will be red is equal to 26 over 52 and that is half or in other words 50 percent. Cards case in the example may have more possibilities compute. Suppose that we are interested in this event that the card that I draw is a 10. Students it is obvious that the total number of tens in this deck of cards is 4, a 10 of diamonds, a 10 of hearts, a 10 of spades and a 10 of clubs. Lehaaza wo patte jo 10 ki occurrence ko favor karte hain unki tadaad char hain unki tadaad char hain m is 4, n the total number of ways in which I can draw a card out of this deck is as before 52 and therefore the probability that my card is going to be a 10 is equal to m over n 4 by 52. Let us consider another example. Suppose that we toss a fair coin 3 times as you now see on the screen the sample space of this experiment consists of 8 possible outcomes head, head, head, head, head, head, tail, head, tail, head, head, tail, tail, tail, head, tail, tail, tail, head and tail, tail, tail, tail, tail, tail. Suppose that we are interested in finding the probability of the event that when we throw these 3 coins or this one particular coin 3 times we get at least one head. Students abhum ushi tara approach karenge iss problem ko jistara hum ne absi thori der pehle iss se pichle example mein kiya tha. The very first thing to try to understand is ke wo josh sharth hai equally likely ki is that valid for this example. Now the point is that we said that it is a fair coin that we are tossing 3 times. If it is a fair coin agar wo iss tara se bana hua hai ki bilkul sahi hai, tera nahi hai, ek side se gisa hua nahi hai then it is obvious that we can say that the tail is as likely to occur as the head aur kha hum usko ek defa toss karen ya teen defa toss karen uski ye property fairness ki jo property hai that will be maintained. Is rationale ke through we are quite satisfied that the outcomes, the various outcomes, the two possible outcomes actually head and tail they are equally likely to occur. Once we are satisfied that this is the case, we can very quickly apply the classical definition of probability m over n, where m is the number of outcomes that are favorable to what I want and n is the total number of possible outcomes. In this example, as you noticed seven of the eight possible triplets contains an H and only one triplet tail-tail-tail is the one which is without the head. Hence the probability that my outcome, my triplet will contain at least one head is given by m over n and that is seven over eight. Abhi jo example hum ne discuss kie ye toh bhat hi zara simple example the. Lekin aapne note kiya hoga ki me ne ish baat ke upar bhat zara zor diya ke aap satisfy hon ke equally likely ki jo definition hai wo apply hoti hai ya nahi hoti. It is very easy to learn the formulas and to apply them but it is more important to realize that unless the assumptions of that particular formula are not valid, we must not apply those formulas. Let us now consider a more interesting and a little involved problem as you now see on the screen. Suppose that four items are taken at random from a box containing twelve items and the four that are drawn they are inspected for checking whether they are faulty or whether they are alright. The box is to be rejected for export purposes if more than one item is found to be faulty. Now if there are three faulty items in that box what is the probability that the box will be accepted? He is drawing these items from this box at random. Ye jo do alphaz maine isthimal kiye these are very important random kamafoom yaha par yehi hai na. It is as if it is the lottery method. Koi usme intentional selection nahi selection nahi hai or randomly you draw kar rahein. Now because of this randomness we can say that each one of those 12 items is equally likely to be drawn and the moment I say this I am in a position to apply the classical definition and that is m over n. But students in this particular problem it is not as convenient to compute m or n as it was in the earlier ones that I presented. Deke, yaha par we will be applying the rule of combinations. Aap note ki jie ke denominator ju hai humare formulae ka n that represents the total number of possible outcomes or in other words the total number of ways in which I can draw four items out of 12. But of course there is no importance of the order in which I draw these items. Order ke ahmiyat nahi hai, lehaza it is not a case of permutations but it is a case of combinations. And according to what I explained in the last lecture there are 12C4 ways of selecting four items out of 12. And as you now see on the screen 12C4 is equal to 12 factorial over 4 factorial into 12 minus 4 factorial. And when you solve this expression it comes out to be equal to 495. The computation of the numerator m is a little more involved and actually it is very interesting. Isme hum rule of combinations, rule of multiplication or rule of addition tino bayak vakt iste mal karenge. Let me explain it to you step by step. Dekhi humare box me 12 item se, out of which 3 are faulty and 9 are alright. Now I want to find the probability that my box is accepted and it is not stopped from being exported. Aur uske liye humne bahle kaha tha, ke a box is rejected if 2 or more items out of those 4 that have been selected, if 2 or more are found to be faulty. Iska matlab ye hua ke agar mere 4 selected items me se, ek faulty hai toh it is no problem. Aur agar ek bhi faulty nahi hai, then of course there is no problem at all. Iska matlab ye hua ke jo favorable outcomes hai jo iss baat ko favor karte hai, ke mera jo box hai that will not be stopped from being exported. Wo dohi hai, 1 is that situation when out of the 4 selected not even 1 is faulty and the other is that situation when out of the 4 selected only 1 is faulty. So, we will be computing the number of ways in which the first event can occur and the number of ways in which the second event that is favorable to what I want can occur and then I will add the 2 and that will give me the total number of outcomes favorable to what I want. As you now see on the screen, in order to compute the number of ways in which all 4 of my drawn items will be alright, we note that there are 3 C 0 ways of drawing 0 faulty items out of 3 faulty items and there are 9 C 4 ways of drawing 4 ok items out of the 9 which are alright. And hence the total number of ways in which the joint event no faulty item and 4 alright items can occur is given by 3 C 0 into 9 C 4 in accordance with the multiplication theorem that I discussed in the last lecture. Bilkul issi tara, we can find the number of ways in which we can have one faulty and 3 alright items and as you now see on the screen, the total number of ways in which this can be done is 3 C 1 into 9 C 3. 3 C 1 ways of drawing 1 faulty item out of 3 and 9 C 3 ways of drawing 3 alright items out of 9. Students, agla point yeh hai, ke jaisa maine pehle kaha, the box is going to be accepted if we have no faulty item or one faulty item. Yeh chur loves or maine isthimal kia, jaisa ke aapne set theory main note kia tha, or stands for plus, yani aur ka matlab hai set theory kiru se, ke aap union ki baat kar rahe. Aur jo discussion hum iss vaakth kar rahe, iss context main or stands for plus. So, as you now see on the screen, the total number of outcomes that are favourable to the fact that the box is accepted is given by 3 C 0 into 9 C 4 plus 3 C 1 into 9 C 3 and that is equal to 126 plus 252 in other words 378. Dividing the favourable number of outcomes by the total number of outcomes, the probability that the box is accepted in spite of the fact that it actually contains three faulty items is 0.76 or 76 percent. Students, as you have noticed the classical definition is quite a simple definition. Lekin dekhne ki cheese aur note karne ki baat yeh, ke this definition has a number of shortcomings, sabse peh li baat aap yeh dekhne and this is a very interesting point that this definition involves circular reasoning. Jaise ke main ne define kia tha hum ne kaha, ke if a random experiment can result in n equally likely and mutually exclusive and exhaustive outcomes and m of them are favourable to the occurrence of an event a and so on. Students, iss ke andar yeh do alfaaz main estimalki yeh that those outcomes are equally likely, iss ka matlab yeh ke main yeh kaha that the outcomes are equally probable. Ab dekhye ke probability hi ko to main define karne ki koshish kar rahe hum aur us definition ke andar hi I am using the word or using the concept of probability. This is called circular reasoning aur iss se mujhe ek latifa yaad aa gaya. Ek bachche ne ek sahab se puchhaa ke school ke dar hai aur ono ne kaha ke masjid ke saam ne to bachche ne puchhaa ke masjid ke dar hai to ono ne kaha ke school ke saam ne and this is the kind of problem that we have in this particular definition. Having said this I would like to convey to you students ke yeh jo equally likely ka concept hi yeh intuitively understandable hai aur jayaise main saari discussion jopta ki us mein explain kiya agar aapko us phenomenon ki jo physical situation hai us se yeh meh sus hota hai ke equally likely ka concept yehapa apply karta hai then you can apply this definition. Now the other problem with this particular definition is that it is very difficult to apply and it is probably not possible to apply it if the total number of outcomes cannot be counted in a convenient manner as we have done in the examples that we just did. Agar aap ka aapki sample space jo hai us me infinite number of sample points ho ya uncountable number of sample points ho then it is not going to be possible for you to apply this definition. The third problem with this definition is that it does not hold if the outcomes are not equally likely. And in my opinion this is one of the main problems with this definition. Hazar ha real life phenomena se hai ke jaha pe we cannot say that one event is as likely to occur as another. Ab yeh jo example hai ke aap coin toss kar rahe hai ya diet toss kar rahe hai aap hudi sochiye hum aam life me iss kusum ke experiment kitni mertaba kar te. Practically never except perhaps when we are playing a game as a child a game of Ludo or any other such situation. So generally speaking we will not be dealing with situations where the various possible outcomes are equally likely. And students these are exactly the situations when we will need to look for another definition of probability. The definition that I am going to present to you next is called the relative frequency definition of probability. Students iss definition ko mai in detail next lecture mai aap ke saath discuss karungi iss vaakth mai bohot muhtasir karke aap ko ye convey karna chahti hum ke according to this definition if an experiment is conducted a very large number of times and what we are interested in occurs a certain number of times then the proportion in which the event of my interest occurs that proportion is regarded as the probability of that event. And when the result arrives of course we find that a certain proportion of the students obtain the first division another proportion obtains the second division and a certain proportion also fails in the exam. Students ye jo proportions hain inhi ko hum regard karenge as the probability of obtaining a first division the probability of obtaining a second division and so on ye bilkul usi rational ke mutaabek jo mai ne aabhi aap ko convey kia. So if we find that in the metric examination of 2002 in a particular province of a particular country the proportion of first divisioners is 25 percent then we can say that the probability that a student of this particular country of in this particular kind of a situation the probability that he will obtain a first division is 25 percent. In today's lecture I discussed with you the two main approaches to probability the subjective approach and the objective approach. Under the objective approach I discussed in detail the classical definition of probability and I would like to encourage you students to practice with this concept and to attempt a number of questions involving the classical definition and involving the various counting rules such as the rule of combinations and permutations. In the next lecture I will discuss in detail the relative frequency definition that I briefly touched upon today and after that we will proceed to the axiomatic definition of probability. Until next time my best wishes to you and Allah hafiz.