 two events A and B are independent if and only if probability of B given A is equal to probability of B. Now probability of B given A means we are talking about the conditional probability the conditional probability of B given A that is equal to probability of B then we can say that A and B are independent that means whether A has occurred or not there is no effect of this on B so that is why they are independent or we can say that A and B are independent if probability of A given B is equal to probability of A now probability of B given A is equal to probability of B implies probability of A given B is equal to probability of A and conversely. So here we can talk about multiplicative rule what is that if in an experiment the event A and B can both occur then probability of A intersection B will be probability of A into probability of B given A. So this rule is coming from the definition of conditional probability which is already covered because we know that probability of B given A is equal to probability of A intersection B divided by probability of A provided probability of A is not equal to 0. So from this we can get probability of A intersection B is equal to probability of A into probability of B given A what is this probability of A intersection B what does it mean it means that probability what is the probability that both the events A and B will occur what is the probability that A and B both the events will occur. So this is equal to probability that A will occur into probability that B given A so probability of A what is the probability that A will occur into what is the probability that B will occur given that A has already occurred. So this is the meaning of this thus the probability that both A and B occur is equal to the probability that A occurs multiplied by the probability that B occurs given that A occurs since the events A intersection B and B intersection A are equivalent it follows from the above result that probability of A intersection B is equal to probability of B intersection A which is equal to probability of A into probability of B given A and this is same as probability of B into probability of A given B. So they are all equivalent next we consider one example suppose that we have a fuse box containing 20 fuses of which 5 are defective if 2 fuses are selected at random and removed from the box in succession without replacing the first what is the probability that both fuses are defective. So let us try this example in this way if A denotes the event that first fuse is defective and B denotes the event that second fuse is defective here we have to find the probability of A intersection B so we have to find the probability that both A and B will occur and that is we know that probability of A into probability of B given A now what is that first among 20 fuses 5 are defective. So according to the problem we know that among 20 fuses 5 fuses are defective so that is why the first fuse is defective probability of that will be 5 by 20 so probability of A is nothing but the probability that the first fuse is defective so that will be 5 by 20. Now once the first fuse is defective it is now taken out and kept aside now another fuse is randomly chosen from the box so the probability that the second fuse is defective will be that is probability that B given A so what is the probability that the second fuse is defective given that the first one is already defective. So this probability will be 4 by 90 because one defective fuse is already chosen so the number of defective fuses will be 4 and there are 19 fuses remain in the box so that is why the probability that B given A will be 4 by 19 and so the resulting probability that is probability that A intersection B will be nothing but 5 by 20 into 4 by 19 which is 1 by 19 now two events A and B are independent if and only if probability of A intersection B is equal to probability of A into probability of B so this result is coming from the fact that probability of B given A will become now probability of B so let us see here because we know that probability of A intersection B is probability of A into probability of B given A so since A and B are independent we know in that case probability of B given A will be probability of B and that is why in this case probability of A intersection B will become probability of A into probability of B so that is why we can say that two events A and B are independent if and only if probability of A intersection B is equal to probability of A into probability of B therefore to obtain the probability that two independent events will both occur we simply find the product of their individual probabilities. Let us take one example here what is the probability of getting two heads in two flips of a balanced coin now we know that flipping of a balanced coin twice so if we flip it twice they are independent so two flips of a balanced coin they are independent and as the coin is a balanced coin the required probability will be half into half that will be 1 by 4 so this is the probability that first flip will give a head into probability that the next one is head that is also half so that is why probability of two heads will be half into half and that will be 1 by 4 because they are independent. Now suppose there are three events in an experiment event A event B and event C so in an experiment three events A B C can occur so if we have to find the probability that A intersection B intersection C that is we have to find the probability that all these three events will occur A B as well as C so this will be probability of A into probability of B given A into probability of C given A intersection B so that means the probability that A will occur into the probability that B will occur given that A has occurred into probability that C will occur given that A and B both have occurred so how is it coming this very simple A intersection B intersection C now A intersection B can be taken as one event so I can write it as probability of A intersection B into probability that C given A intersection B so I am taking two events here A intersection B and C so these two events now if we consider these two events by the definition of conditional probability we can write probability of A intersection B into probability of C given that A intersection B has occurred so this can be written as probability of A into probability of B given A into probability of C given A intersection B so in this way this will come now if we have more than three events then what should we do if in an experiment the events A1 A2 to AK can occur then probability of A1 intersection A2 intersection so on intersection AK so intersection of A1 A2 to AK so that I am considering the probability of that event can be written as probability of A1 into probability of A2 given A1 into probability of A3 given that A1 intersection A2 has occurred dot dot probability of AK given A1 intersection A2 so on intersection AK – 1 so this result is coming by extending the case which we have already considered that is if there are three events so the way we have done it the same way if we extend number of events then we can get this result very easily now if the events A1 A2 to AK are independent then probability of A1 intersection A2 intersection dot dot intersection AK is equal to the product of their individual probabilities that is probability of A1 into probability of A2 dot dot probability of AK so this can be written as probability of A1 intersection A2 intersection dot dot intersection AK that will be equal to probability of A1 into probability of A2 and so on probability of AK now let us talk about total probability if the events B1 B2 to BK constitutes a partition of the sample space S such that probability of BI not equal to 0 for I equal to 1 to K then for any event A of S probability of A is equal to summation I equal to 1 to K into probability of BI intersection A and that is equal to summation I equal to 1 to K into probability of BI into probability of A given BI and this is called total probability now let us prove this result for proving this result we consider a Venn diagram partitioning the sample space from the figure it is clear that B1 B2 to BK are mutually exclusive events the event A is seen to be the union of the mutually exclusive events B1 intersection A B2 intersection A up to BK intersection A so this is so A the event A is the union of all this events B1 intersection A B2 intersection A up to BK intersection A so that is we can write the event A as B1 intersection A union B2 intersection A union not that union BK intersection A now since B1 B2 to BK are mutually exclusive events we can say that B1 intersection A B2 intersection A up to BK intersection A are also mutually exclusive and that is why probability of A can be written as probability of B1 intersection A plus probability of B2 intersection A plus dot dot plus probability of BK intersection A so this is nothing but the summation I equal to 1 to K of probability of BI intersection A and this probability can be written in this way that probability of BI into probability of A given BI so it can be written as the summation I equal to 1 to K probability of BI into probability of A given BI now let us talk about Bayes rule if the events B1 B2 to BK constitutes a partition of the sample space S where probability of BI not equal to 0 for I equal to 1 to 2 K then for any event A in S such that probability of A is not equal to 0 then probability of BR given A is equal to probability of BR intersection A divided by summation I equal to 1 to K probability of BI intersection A which is equal to probability of BR into probability of A given BR divided by the summation I equal to 1 to K probability of BI into probability of A given BI for R equal to 1 to K so this is called the Bayes rule and this Bayes rule is very very important and it has lot of applications so how to prove this rule by the definition of conditional probability probability of BR given A will be probability of BR intersection A divided by probability of A then using the above result that is the result of total probability we can find the expression in place of the denominator so probability of BR given A will become probability of BR intersection A divided by summation I equal to 1 to K probability of BI intersection A which is equal to probability of BR into probability of A given BR divided by summation I equal to 1 to K probability of BI into probability of A given BI and this completes the proof. So in this way we are getting the Bayes rule and what is this actually so if in some experiment it there are some steps so we are doing something in the first step and doing something in the second step like this so and so if it is asked in the other way that is the final event we know final event will occur that is given then what is the probability that before that it has occurred so this kind of question is asked so in that case we can use Bayes rule and we can find the probability of the conditional probability that final event has occurred then before this this event had occurred so this kind of question can be tackled by Bayes rule. So let us consider one example in this context one bag contains four white balls and three black balls and another bag contains three white balls and five black balls. One ball is drawn from the first bag and placed unseen in the second bag what is the probability that a ball now drawn from the second bag is black. Now we have chosen randomly a ball from the first bag and we have kept it in the second bag and then we are choosing a ball randomly from the second bag now we have to find the probability that if what is the probability that this ball drawn from the second bag is black. So let us try this solution in this way W denotes the event that a white ball is transferred from bag 1 B denotes the event that a black ball is transferred from bag 1. So these two events W and B they are mutually exclusive either W will occur or B will occur A denotes the event that a black ball is drawn from bag 2. Now the required probability is probability of A and what is that probability of A can be written as probability of W intersection A union B intersection A and since W and B are mutually exclusive events W intersection A and B intersection A are also mutually exclusive and that is why this can be written as probability of W intersection A plus probability of B intersection A. So this can be written as probability of W into probability of A given W plus probability of B into probability of A given B. Now let us try this in this way that we know that first bag contains four white and three black balls second bag contains three white and five black balls. Now we have to find probability of A which is probability of W into probability of A given W plus probability of B into probability of A given B. So now probability of A given W so we have to find the probability that if a white ball is transferred from bag 1 to bag 2 then what is the probability that a black ball will be chosen from bag 2. So that probability and this is the probability probability of W is nothing but the probability that a white ball is chosen from bag 1. So that will be 4 by 7 because there are four white balls in bag 1 among 7 balls. So 4 by 7 into now if a white ball is transferred from bag 1 to bag 2 in bag 2 there will be 9 balls after this transfer and number of black balls will remain same because a white ball is transferred from bag 1 to bag 2. So this probability will be 5 by 9 plus the probability that a black ball is chosen from bag 2. So that will happen with probability 5 by sorry it will be 3 by 7 because there are three black balls in bag 1 among 7 balls. So it will be 3 by 7 into and this conditional probability is nothing but that if a black ball is transferred from bag 1 to bag 2 then what is the probability that from back to a black ball will be chosen and that will become 6 by 9 because now there are 6 black balls in bag 2 and total number of balls after this transfer is 9 that is why this probability will be 6 by 9. So now we can calculate this probability that is 20 plus 18 divided by 63 so it will be 38 by 63. In a certain assembly plant 3 machines B1, B2 and B3 make 30%, 45% and 25% respectively of the products. It is known from past experience that 2%, 3% and 2% of the products made by each machine respectively are defective. Now suppose that a finished product is randomly selected what is the probability that it is defective. So let us try this solution in this way the event A represents the product is defective, B1 is the event that the product is made by machine B1, B2 the product is made by machine B2, B3 is the event that the product is made by machine B3. So we can write probability of A that is probability of B1 into probability of A given B1 plus probability of B2 into probability of A given B2 plus probability of B3 into probability of A given B3. Because B1, B2, B3 these 3 events are mutually exclusive so probability of A can be written as probability of A that is probability of B1 intersection A union B2 intersection A union B3 intersection A since B1, B2 and B3 are mutually exclusive B1 intersection A, B2 intersection A and B3 intersection A are also mutually exclusive and that is why this probability can be written as probability of B1 intersection A plus probability of B2 intersection A plus probability of B3 intersection A and this can be written as probability of B1 into probability of A given B1 plus probability of B2 into probability of A given B2 plus probability of B3 into probability of A given B3. So that is what is here this problem can be represented by this tree diagram also if we have this kind of diagram see probability of B1 is 0.3 so it has 3 branches probability of B1 probability of B2 probability of B3 these 3 possibilities are there either B1 will occur or B2 will occur or B3 will occur. So probability of B1 now from this we can go to probability of A given B1 which is 0.02 in this way if we go along this line we will get probability of B2 that is 0.45 and from this we can go to probability of A given B2 which is 0.03 and along this path we have probability of B3 which is 0.25 and from here we can go to probability of A given B3 which is 0.02. So from each of this path we are going to A so this represents the problem here so probability of A will be 0.3 into 0.02 plus 0.45 into 0.03 plus 0.25 into 0.02 and this if we calculate we will get 0.0245 now with reference to previous example if a product were chosen randomly and found to be defective what is the probability that it was made by machine B3. So now we can apply Bayes rule to find the solution using Bayes rule we can write probability of B3 given A is equal to probability of B3 into probability of A given B3 divided by probability of B1 into probability of A given B1 plus probability of B2 into probability of A given B2 plus probability of B3 into probability of A given B3. So the values are already obtained so we can substitute this values here so it will be 0.005 divided by 0.006 plus 0.0135 plus 0.005 which is the denominator is 0.0245 and the numerator is 0.005 so it will be 10 by 49 so that means if a defective product was selected the probability is very low that it was made by B3 and that is all thank you.