 Assalamu alaikum. Welcome to lecture number 21 of the course on statistics and probability. Students you will recall that in the last lecture I discussed with you the addition theorem and the multiplication theorem. And in the discussion of the multiplication theorem we utilized the concept of conditional probability. In today's lecture I will present to you the special case of the multiplication theorem for the case of independent events. And towards the end of the lecture I will discuss with you an important theorem called the Bayes theorem. Before we begin the discussion of independent events students let us revise the concepts that we did in the last lecture by way of a very interesting example. All you need is a methodological approach. As you now see on the screen suppose that a bag contains 10 white and 3 black balls. Also another bag contains 3 white and 5 black balls. Suppose that 2 balls are transferred from the first bag and placed into the second one and then one ball is taken from the second bag. What is the probability that the ball drawn from the second bag is a white ball? In order to answer this question let us first consider the situation that we have in the beginning of the experiment. In the beginning of the experiment we have 10 white and 3 black balls in the first bag and 3 white and 5 black balls in the second bag. Now let a represent the event that 2 balls are drawn from the first bag and transferred to the second bag. Then a can occur in the following three mutually exclusive ways. A1 the first way is that 2 white balls are transferred from the first bag to the second one. A2 the second way is that 1 white ball and 1 black ball are transferred from the first bag to the second one. And A3 the third and last way of doing this of transferring 2 balls is that 2 black balls are transferred from the first bag to the second one. Students, why did I say that these three ways are mutually exclusive? You will remember that mutually exclusive means the events which exclude each other. Those events which cannot be done at the right time. If one is happening then the other is not happening. And it is obvious that in this problem these three things cannot be done at the same time. Either both white balls will be transferred or either 1 white and 1 black will be transferred or either both of them will be transferred black. So at the same time one of these three can be done at the same time. And therefore we say that these three ways of transferring 2 balls are mutually exclusive. So as you now see on the screen, the total number of ways in which 2 white balls can be drawn out of 10 white balls is 10 C2 because as indicated in the last lecture the rule of combinations is to be applied in this kind of a situation. Also the total number of ways in which 2 balls can be drawn out of a total of 13 balls is 13 C2. And dividing 10 C2 by 13 C2 we obtain the probability of A1 that is the probability that 2 white balls are transferred from the first bag into the second one. Solving this expression the probability of A1 comes out to be 45 by 78. In a similar way we can compute the probability of the event A2 that is that 1 white ball and 1 black ball are transferred from the first bag into the second one. Now in this case the total number of ways of drawing 2 balls out of 13 is 13 C2 just as before and this expression occurs in the denominator of the formula of probability. But the numerator is obviously different and in this case it is 10 C1 into 3 C1 students. So 10 C1 ways of drawing 1 white ball out of 10 white balls and 3 C1 ways of drawing 1 black ball out of 3 black balls and chunke in dono baton ke darmean and kalavs isthimal ki ajarai. Lehaaza we will multiply that first expression with the second one. After all do you not remember that example that we did in an earlier lecture that for example if you have 3 ways of ordering a soup and 2 ways of ordering a sandwich then the total number of ways of ordering your lunch is 3 into 2. Divide 10 C1 into 3 C1 by 13 C2 the probability of A2 comes out to be 30 by 78 in a similar way the probability of the event A3 that is 2 black balls transferred from the first bag into the second is given by 3 C2 divided by 13 C2 and that is 3 by 78. I hope that you are realizing that the definition that I have applied in computing these probabilities is the classical definition of probability. Yaad hain hain ke jab bhi aap ke various outcomes equally lightly hoon then you are in a position to apply the classical definition. Yeha pe wo jo bag hain usmise hum jo ball draw kar rahe hain we are doing it without looking into the bag. We are doing it in a random manner aur jab kabhi aap random sampling karthe hain astari kise then of course the various elements of your population are equally likely to be drawn. Now after having transferred 2 balls from the first bag what happens to the second bag? This is a very very important question and we have to analyze it one by one. In the first situation if 2 white balls are transferred students then in our second bag we now have 3 plus 2 equal to 5 white balls and as before 5 black balls and now the total number of balls in that bag is not 8 but 10. Under this situation the probability that a white ball is drawn out of the second bag is given by 5 over 10 because there are 5 white balls in the bag at this time. This probability is denoted by P of W given A1 and as you can see it is a conditional probability. Yeh sliye ke yeh jo answer hum ne obtain kia hai 5 by 10. This is under the condition that 2 white balls have been transferred into the second bag. In other words it is under the condition that the event A1 has already occurred. The second situation is that 1 white and 1 black ball is transferred from the first bag into the second one and if this is the case then the second bag contains 3 plus 1 equal to 4 white balls and 5 plus 1 equal to 4 black balls. Also the total number of balls in the bag is 8 plus 2 equal to 10. Hence the probability of obtaining a white ball from the bag number 2 in this particular situation is 4 by 10 and once again this is a conditional probability because the probability of getting a white ball in this situation is being computed under the condition that 1 white and 1 black ball have been transferred into this bag. In other words this probability is being computed under the condition that the event A2 has already occurred. In a similar way we can compute the probability of W given A3 that is that we draw a white ball from the second bag under the condition that 2 black balls had been transferred into the second bag. In this situation we have 3 white and 5 plus 2 equal to 7 black balls in the second bag and the total number of balls now is 10. Hence the probability of getting a white ball from the second bag given that 2 black balls had been transferred into that bag is equal to m over n favorable over the total number of outcomes and that is 3 by 10. Students, do you remember what the question was? The question was what is the probability that after this transferring has happened if we draw one ball from the second bag it is going to be white? So, we will be applying the special case of the addition theorem. You remember that special case of the addition theorem was that if A and B are mutually exclusive events then the probability of A union B is equal to the probability of A plus the probability of B and if there are more than 2 events then we will extend this theorem. Probability of A union B union C is equal to probability of A plus the probability of B plus the probability of C. So, in this problem A, B and C which are mutually exclusive events? So, the probability of obtaining a white ball is equal to the probability of this first situation plus the probability of the second situation plus the probability of the third situation. And as you now see on the screen, mathematically speaking the probability of a white ball is equal to probability of A 1 intersection W plus the probability of A 2 intersection W plus the probability of A 3 intersection W. Probability of A 1 intersection W means the probability that the event A 1 occurs and a white ball is drawn from the second bag. And according to the general multiplication theorem of probability, probability of A 1 intersection W is equal to probability of A 1 into probability of W given A 1. Substituting the values 45 over 78 and 5 over 10 in this formula, probability of A 1 intersection W comes out to be 15 over 52. In a similar way, probability of A 2 intersection W comes out to be 8 by 52 and the probability of A 3 intersection W comes out to be 3 by 260. Hence, the required probability is probability of W equal to 15 over 52 plus 8 over 52 plus 3 by 260 and that is equal to 0.45. In other words, the chances are 45 percent that if two balls are transferred from the first bag into the second one and then a ball is drawn from the second bag, this ball is white. So, this is the methodological way in which you attempt probability problems. Let us now proceed to the concept of independent events. As you now see on the screen, two events A and B in the same sample space S are defined to be independent or statistically independent if the probability that one event occurs is not affected by whether or not the other event has occurred. In other words, if probability of A given B is equal to probability of A and probability of B given A is equal to probability of B, then we say that the events A and B are statistically independent. Now, I will explain this point that why we said that if probability of B given A is equal to the probability of B, then A and B are independent. Students, probability of B given A is equal to probability that B occurs given that A has already occurred or probability of B is equal to probability that B occurs or it simply means the probability that B occurs regardless of whether or not A has occurred. So, this statement that if P of B given A is equal to P of B, then A and B are independent. We can state this way that if the probability that B occurs given that A has already occurred is the same as the probability that B occurs regardless of whether or not A has occurred then A and B are independent. So, if you pay attention to this, then this is intuitively acceptable. If you take the concept of independence from the common life, then it is the same that it has nothing to do with the other thing. So, when we say that the probability of B given A is equal to probability of B that is the same that whether A occurs or not, the probability of B is equal to probability of A or not, then there is no effect on probability of B. And this is exactly the concept of statistical independence. I think you are getting confused. There is no such thing as students. Let me explain it to you with the help of a very simple example. Suppose that I have a die in my right hand and a coin in my left hand. Die is perfect or coin is perfect. So, the classical definition applies. All 6 outcomes on the die are equally likely to occur and both the outcomes of the coin are also equally likely to occur. Now, if I toss the two, I toss the die from the right hand and the coin from the left hand, then won't you agree that the event on the die has nothing to do with the coin type? Then either it has an even number or odd number. I have an even number of head and tail, so there is nothing to do with the head. This head cannot affect this even number and this even number cannot affect the outcome of the coin. So, a coin has an odd number of head and eye and this not affect the probability of getting an even number on the die. words the probability of B given A is equal to the probability of B, regardless of whether or not A has occurred. Students, because of this fact that I have just conveyed to you, the multiplication theorem takes a special form as you now see on the screen. If A and B are independent, then probability of A intersection B is equal to probability of A into probability of B. The reason is very simple. If we substitute P of B given A equal to P of B in the original multiplication theorem, we obtain this simple form that I just mentioned. Here, a very interesting thing is coming up. It is often seen that students are mutually exclusive or independent. They mix these two concepts. Although, if you pay attention, they are very different. In fact, two events which are mutually exclusive cannot be independent. You see, mutually exclusive means that if one event occurs, the other cannot occur. So, they have a connection. The connection is that if one is happening, then the other cannot happen. Whereas, independent means that there is no connection. So, the two concepts must not be mixed. As I said, mutually exclusive means that they cannot occur together. They are repelling each other. Here, I remember that beautiful lion. If it is necessary, then we should understand it. If it is necessary, then we should understand it. If it is not necessary, then what is cheating? Independent means that there is no connection. So, students, the two concepts must never be mixed. The equation that I presented to you, probability of A intersection B is equal to probability of A into the probability of B. If A and B are independent, let us apply this to an example. As you now see on the screen, two fair dyes, one red and one green are tossed. Let A denote the event that the red dye shows an even number and let B denote the event that the green dye shows a 5 or a 6. Show that the events A and B are independent. In order to solve this question, students, how may three quantities compute? Probability of A, probability of B and probability of the joint event A intersection B. Or, agar probability of A intersection B, probability of A into probability of B ke baabar aajai that is the left hand side comes out to be equal to the right hand side, then of course, we can say that A and B are independent. So, in order to compute these three probabilities, how do we proceed? A is the event that the red dye shows an even number. So, even number ke probability kya ho sattie? Obviously, 3 by 6, because there are 6 possible outcomes 1, 2, 3, 4, 5, 6 and there are 3 even numbers 2, 4 and 6. Isitara B is the event that the green dye shows a 5 or a 6 and what is the probability of B? Obviously, 2 by 6. Ye toh aasaan hai, but how do we compute the probability of the joint event A intersection B? The event that the red dye shows an even number and a green dye shows a 5 or a 6 students for that we will have to consider the sample space of the joint situation. And as you now see on the screen, the sample space consists of the 36 possible outcomes 1, 1, 1, 2, 1, 3, 1, 4 and so on up to 6, 6. Now it is obvious that the total number of possible outcomes for the joint experiment is 36. But how do we determine the number of outcomes that are favoring the joint event A intersection B? If you look carefully, you find that there are 6 outcomes which favor this joint event and they are 2, 5, 4, 5, 6, 5, 2, 6, 4, 6 and 6, 6. The reason is that the first number represents the outcome on the red dye whereas the second number represents the outcome on the green dye. Dividing the number of outcomes favoring this joint event by the total number of possible outcomes, we obtain probability of A intersection B is equal to 6 by 36 and that is 1 over 6. Multiplication theorem equation, probability of A intersection B is equal to probability of A into the probability of B in the case of independent events. So, this equation left hand side, we have determined its numerical value 6 by 36, yeah 1 by 6. Now what about the right hand side? We have to compute probability of A into the probability of B and it comes out to be 3 by 6 into 2 by 6 that is 6 by 36 or 1 by 6. Hence the equation is validated and we say that the events A and B are independent. Let us now extend the concept of independent events to the case of more than two events. Students, three events A, B and C all defined on the same sample space are said to be mutually independent if they satisfy the following conditions. Number 1, they should be pair wise independent that is probability of A intersection B is equal to probability of A into probability of B, probability of A intersection C is equal to probability of A into probability of C and a similar equation for B and C. The second condition is that A, B and C should be mutually independent that is the probability of A intersection B intersection C should be equal to probability of A into probability of B into probability of C. Let me explain this to you with the help of an example. Suppose that a coin, a red dye and a blue dye are tossed together. Let A denote the event that the coin shows head, let B denote the event that the red dye shows an even number and let C denote the event that the blue dye shows a six. We would like to show that the events A, B and C are statistically independent. Now students, in order to solve this problem first of all let us consider the three events A, B and C separately. Assuming that the coin and the two dyes are fair, we have probability of A is equal to 1 by 2, probability of B is equal to 3 by 6 and probability of C is equal to 1 by 6. This is so because A represents a head, B represents an even number and C represents a 6 on the dye. Next let us consider the joint events A intersection B, A intersection C and B intersection C. Students when the coin and the red dye are tossed together, our sample space will consist of twelve outcomes that is h 1, h 2, h 3 and so on up to t 6. Out of these three ordered pairs favor the event A intersection B that is the event that we obtain a head on the coin and an even number on the red dye or h 2, h 4 and h 6. Hence the probability of A intersection B is equal to 3 by 12. Similarly, when the coin and the blue dye are tossed together, our sample space once again consists of twelve outcomes h 1, h 2 and so on. Now out of these one ordered pair favors the event A intersection C that is the event that we obtain a head on the coin and a 6 on the red dye or is equal to h 6. Hence students the probability of A intersection C is 1 over 12 because we have one outcome favoring this particular event out of twelve possible outcomes. Students when the red dye and the blue dye are tossed together, our sample space consists of 36 ordered pairs which are 11, 12, 13 and so on. Now out of these three ordered pairs favor the event B intersection C that is the event that we obtain an even number on the red dye and a 6 on the blue dye or h 3 ordered pair 2, 6, 4, 6 and 6, 6. Hence the probability of B intersection C is 3 by 36. Thirdly students let us consider the tossing of the coin and the two dyes together. Now if they are being tossed together then according to the rule of multiplication our sample space consists of 2 into 6 into 6 that is 72 possible outcomes and these are 72 ordered triplets of the form h 11, h 12, h 13 and so on and students the second last one will read t 6 5 and the last one t 6 6. Out of these 72 possible outcomes only three favor the event head and even number and these three are h 2 6, h 4 6 and h 6 6. Hence the probability of the joint event head and even number and 6 that is the probability of A intersection B intersection C is given by 3 by 72. Students in order to verify that the events A, B and C are mutually independent we will be substituting all the above computed probabilities in the equations that were presented earlier and we will be noticing that the left hand side will be coming out to be equal to the right hand side. First of all in order to determine whether or not A, B and C are pair wise independent we would like to check whether or not the equation probability of A intersection B is equal to probability of A into probability of B is verified numerically. Students as you can see both sides of this equation are numerically equal to 1 over 4 and hence this particular equation is verified. Similarly, probability of A intersection C is 1 over 12 and probability of A into probability of C also comes out to be 1 over 12 and hence this particular equation is also verified. And as you now see on the screen a similar situation holds for B and C. Hence we conclude that the events A, B and C are pair wise independent. Now in order to verify the second condition we note that probability of A into probability of B into probability of C comes out to be 3 over 72 exactly the same as what we obtained for probability of A intersection B intersection C. Hence this particular equation is also verified and students since both the conditions of independence are fulfilled. Therefore, we can conclude that in this particular experiment the events A, B and C are statistically independent. In general k events A 1, A 2 so on up to A k will be defined to be mutually independent if and only if the probability of the intersection of any 2, 3 so on up to k of them equals the product of their respective probabilities. So, this is the concept of independence. Technically speaking once again the 2 events A and B are statistically independent if the probability of A intersection B is equal to the probability of A into the probability of B. Students let us go back to that very interesting example that we considered in the last lecture the births that are occurring in various regions of England and Wales and as you remember we were talking about the sex of the baby and whether it was life born or still born. As you now see on the screen the data was that out of a total of 7,16,740 births in 1956 in that region of the world there were 3,59,881 life born male babies, 8,609 still born male babies and so on and the corresponding proportions were male life born 0.5021, male still born 0.0120, female life born 0.470 and female still born 0.0109. As you will remember the total number of births in this data set is large enough for us to regard these proportions as probabilities according to the relative frequency definition of probability. So, if in this problem we are interested in determining whether the sex of the baby and the nature of birth are they independent or are they dependent. So, how we proceed students we have two options either we look at the equation probability of B given A is equal to the probability of B or we look at the equation probability of A intersection B is equal to probability of A into probability of B. This is the second equation that we have discussed in detail. In this problem let me look at the first equation. When we defined independent events, we said that if the conditional probability is the same as the unconditional probability then the two are independent. So, as you now see on the screen the total number of male births is 368,490 and the babies who were still born among these 368,490 were 8609. So, the probability of still birth given that it is a male baby is equal to 8609 over 368,490 and that is equal to 0.0234. On the other hand the total number of female births is 348250 and the number of still births among the female babies is 7,796. So, the probability of still birth given that it is a female baby is equal to 7,796 over 348250 and that is 0.0224. This is conditional probabilities and what is the unconditional probability of still birth? In all there are 7,16,740 births out of which 16,405 are still births. So, the probability of still birth without any condition of the sex of the baby is 16405 divided by 716740 and that is 0.0229. Students, come and analyze this situation. You have seen that the unconditional probability that is 0.0229. But the conditional probability for the male babies that if that is a male baby, then still birth probability you remember it was 0.0234 and that is slightly higher than 0.0229. Similarly, the conditional probability for the female babies that was 0.0224 which is slightly lower than 0.0229. So, then we can draw conclusion that the probability of still birth in males is slightly higher than the overall unconditional probability of still birth. Whereas, the probability of still birth in females is slightly lower. Hence, we cannot say that still birth and sex are statistically independent. If the probability of still birth given male was exactly equal to the unconditional probability of still birth and the probability of still birth given female also exactly equal to the unconditional probability of still birth. So, the gist of the whole discussion is that in this problem, we cannot say that the gender of the baby and the nature of birth are statistically independent. In fact, we should realize that among the babies who are still to be born, the male baby is at a slightly higher risk of being still born as compared with the female baby. Students, the table that you saw in this example leads to another important concept in probability theory and that is the concept of marginal probabilities. As you now see on the screen, we have the bivariate table and we have the four joint events male life born, male still born, female life born and female still born. But the probabilities that I am pointing to now are the ones that occur in the margins of this table. I am referring to the last column of the table as well as the last row of the table. Considering the column, the first marginal probability is 0.5141 and the second one 0.4859 and what do these represent? The first one is the probability that the baby is a male whereas the second one is the probability that the baby is a female. If we look at the last row of the table, the first probability 0.9771 is the probability that a baby is life born whereas the second marginal probability 0.0229 is the probability that a baby is still born. The point to realize is that when we talk about marginal probabilities, we are only talking about one of the two variables that we have in front of us. When we say that the probability of a male birth is 0.5141, we are saying this regardless of whether it is a life birth or a still birth. The event male birth in this problem can happen in two ways. Either it is a male birth and a life birth or it is a male birth but a still birth. So, the probability of a male birth is equal to the probability of male intersection life birth or male intersection still birth and that is equal to the probability of male intersection life birth, union male intersection still birth and this is equal to the probability of male intersection life birth plus the probability of male intersection still birth. This is equal to 0.5021 plus 0.0120 and that is 0.5141. Hence, we realize the very important relation between the joint probabilities and the marginal probabilities and that is that in any row or any column the sum of the joint probabilities is equal to the corresponding marginal probability. Another very important equation is that conditional probability is equal to joint probability over marginal probability. In this example, the probability that it is a still birth given that it is a male birth is equal to the probability of the joint event that it is the male birth and the still birth divided by the probability of male birth and this is totally in accordance with the formula of conditional probability that I convey to you in the last lecture. Probability of b given a is equal to probability of a intersection b divided by the probability of a. Substituting the numerical values of these probabilities, we obtain probability of still birth given male birth equal to 0.0233 and in a similar way, we can compute other conditional probabilities. In today's lecture, we discussed the concept of independent events and we studied the special case of the multiplication theorem in the case of independent events. Also, we considered the concept of marginal probability or in the first example, we did an interesting example in which two balls are transferred from the first bag into the second and then a ball is drawn from the second. That example was intended to be done in such detail that you develop a confidence that if you approach any probability problem in a methodological way, then you can tackle it. You can simply start with that probability which is the required probability and then see what are the different ways in which that particular event can happen. I would like to encourage you students to shed off your fear of probability theory and to try to attempt as many questions as you can, the more you practice, the better off you will be. I wish you the very best in your pursuit of knowledge in this area and until next time, Allah Hafiz.