 Hello and welcome to the session. In this session, first we will discuss about partition of a sample space. A set of events, E1, E2 and so on up to EN is set to represent partition of sample space S if the first condition is that EI intersection Ej is equal to phi given i is not equal to j where i, j goes from 1, 2, 3 and so on up to n. The second condition is E1 union E2 union and so on union EN is equal to S. The third condition is probability of EI is greater than 0 for all i, 1, 2 and so on up to n. Or we can also say that the events E1, E2 and so on up to EN represent a partition of sample space S if they are pairwise disjoint, they are exhaustive and they have long zero probabilities. Next we have theorem of total probability. According to this theorem we have let E1, E2 and so on up to EN be a partition of sample space S and suppose that each of the events E1, E2 and so on up to EN has long zero probability and let A be any event associated with sample space S then we have probability of A is equal to probability of E1 multiplied by probability of A upon E1 plus probability of E2 multiplied by probability of A upon E2 plus and so on up to probability of EN multiplied by probability of A upon EN. This could also be written as probability of A is equal to summation probability of Ej multiplied by probability of A upon Ej where j goes from 1 to n. Next we have Bayes theorem according to which have if E1, E2 and so on up to EN are events which constitute a partition of sample space S that is we have E1, E2 and so on up to EN are pairwise disjoint, even union E2, union so on up to union EN is equal to the sample space S and let A be any event with non zero probability then we have probability of EI upon A is equal to probability of EI multiplied by probability of A upon EI upon summation probability of Ej multiplied by probability of A upon Ej where j goes from 1 to n. Bayes theorem is also called the formula for the probability of causes. This formula gives the probability of a particular event EI given that the event A has occurred. Suppose we have 2 bags bag 1 and bag 2 in bag 1 we have 4 red balls and 4 black balls in bag 2 we have 2 red balls and 6 black balls. Let E1 be the event of choosing bag 1, E2 be the event of choosing bag 2 and A be the event of drawing a red ball. Now probability of even that is probability of choosing bag 1 is 1 upon 2 and probability of E2 that is choosing bag 2 is also 1 upon 2. Probability of A upon E1 that is probability of drawing a red ball from bag 1 is given by 4 upon 8 since we have 4 red balls in bag 1 and total 8 balls in bag 1 which is equal to 1 upon 2. Then probability of A upon E2 that is drawing a red ball from bag 2 is given by 2 upon 8 since there are 2 red balls in bag 2 and total 8 balls in bag 2 which is equal to 1 upon 4. Now we shall find the probability of drawing a ball from bag 1 given that it is red that is we have to find the probability of even upon A. From Bayes theorem this becomes equal to probability of even multiplied by probability of A upon even upon probability of even multiplied by probability of A upon even plus probability of E2 multiplied by probability of A upon E2. This is equal to 1 upon 2 multiplied by 1 upon 2 multiplied by 1 upon 2 plus 1 upon 2 multiplied by 1 upon 4. This is equal to 1 upon 4 upon 1 upon 4 plus 1 upon 8 which is equal to 2 upon 3 that is probability of even upon A that is probability of drawing a ball from bag 1 given that it is red in color is equal to 2 upon 3. So this is how we use the Bayes theorem. This completes the session. Hope you have understood the partition of a sample space, theorem of total probability and Bayes theorem.