 Hello and welcome to this session. In this session we discuss independence of two events. Now in one of our session, we have discussed about the compound events and probabilities of dependent and independent events. In this session, we shall learn if probability P of event A into section B is equal to probability P of event A into probability P of event B, then the two events independent otherwise dependent. Now let us recall two events are independent events. If the occurrence of one event does not affect the likelihood that the other event will occur, that is when the outcome of one event does not affect the outcome of certain event, then the two events are independent events. Now let us direct formula for finding probability of independent events. Now if A and B are two independent events, then probability of both events occurring the product of the probability of event A and probability of event B, that is probability P of event A and B, this also means probability of both events occurring is equal to probability of event A into probability of event B or we can write it as probability P of event A intersection B is equal to probability P of event A into probability P of event B. Now here dot represents multiplication. The physical description of two events A and B is not sufficient to describe whether the two events are independent or not. A and B are declared to be independent events if probability of events A intersection B is equal to probability of event A into probability of event B. Now let us see its verification. Now if probability of two events A and B occurring together, that is probability of event A intersection B is not equal to the product of two probabilities, that is probability of event A into probability of event B, then the two events not independent. So here B and B are dependent events. Now let us just give you an example for this. Here a pair of dice is thrown and A is the event of numbers are obtained. We have to show that A and B are dependent events. Now let us start with the solution. First of all we have to find the sample space of this experiment. Now we know that whenever we throw a dice, then total number of outcomes are six, that is getting a number one, number two, three, four, five and six. Now when we roll two dice simultaneously or we can say when a pair of dice is thrown, then the outcomes will be getting a number one on the first die and getting a number one on the second die, then getting a number one on the first die and getting a number two on the second die, then getting a number one and getting and getting number 3 on the second die then getting a number 1 on first die and getting a number 4 on second die and continuing this way we get all these outcomes. Now sample space of this experiment is a set containing all these ordered pairs and when we count these ordered pairs they are 36 in number it means number of elements is equal to 36. Now here a is the event sum is 8 it means here we have to check for those ordered pairs whose sum of first and second components is 8. Now here you can see ordered pair, ordered pair 35 then ordered pair 53, ordered pair 6, ordered pair 44 are the required ordered pairs and here we can see that 2 plus 6 is 8, 3 plus 5 is 8, 4 plus 4 is 8, 5 plus 3 is 8 and here 6 plus 2 is 8. So event A is a set containing the ordered pairs 2, 6, 3, 5, 4, 4, 5, 3 for event B we will write those are both dies the outcomes are the ordered pairs 1, 1, 1, 3, 1, 5, 3, 1, 3, 3, 3, 3, 5, 5, 1, 5, 3, event B is a set containing all these ordered pairs. Now let us find event A intersection B and this is equal to a set containing elements which are common to both event A and event B. So it is a set containing the ordered pairs 35 and 53 as the ordered pairs 35 and 53 are common to both these events. Now let us calculate probabilities. First of all let us find probability P of event A which is equal to number of elements favorable to event A upon total number of elements in sample space S. Now number of elements in event A is equal to 1, 2, 3, 4 and 5 and number of elements in sample space S is equal to 36. So probability P of event A is equal to 5 upon 36. Now let us find probability of event B and this is equal to number of outcomes or number of elements favorable to event B. Now here it is 1, 2, 3, 4, 5, 6, 7, 8 and 9th upon total number of elements in sample space S which is 36. So probability of event B is equal to 9 upon 36 and this is equal to 1 upon 4. Now let us find probability of event A intersection B. Now number of elements favorable to event A intersection B is 2. So probability of event A intersection B is equal to 2 upon total number of elements in sample space S that is 36 and this is equal to 1 upon 18. Now let us find the product probability of event A into probability of event B and this is equal to 5 upon 36 into 1 upon 4 which is equal to 5 upon 144. So from here you can see that probability of event A intersection B is not equal to probability of event A into probability of event B. Therefore A and B are not independent events thus they are dependent events. So here we have proved that A and B are dependent events. So in this session we have learnt how to determine the independence of two events and this completes our session. Hope you all have enjoyed the session.