 Hello and welcome to the session. In this session we will discuss a question which says that a coin is tossed three times and all possible outcomes are assumed to be equally likely. E is the event. Both head and tails have occurred and F is the event at most one thing has occurred. Show that events E and F are independent. Now before starting the solution of this question we should know a result. And that is two events A and B are independent of event A intersection B is equal to probability of an event A into probability of an event B. Now this result will work out as a key idea for solving out the given question. Now let us start with the solution of the given question. First of all let us write the sample space of the given experiment. Now here we have tossed a coin three times. Now remember that a coin is tossed M times where number of elements in sample space is 2 raised to power 3. Now here the coin is tossed 3 times. So number of elements in sample space is equal to 2 raised to power 3 which is equal to 8. Now when we toss a coin three times then we get these outcomes that is getting all heads. Now here H denote head and T denote then getting exactly two heads. Now here we have the outcome H H T it means we can get exactly two heads and one tail. Now here we also have head in the first toss in the second toss we can have tail. In the third toss we can have head again then the other outcome is T H H. The next outcome can be T T H then the next outcome T H T then the next outcome which is H P T then next outcome. Now this T T T means getting all tails it means here the total number of elements in sample space 1, 2, 3, 4, 5, 6, 7 and 8. So the sample space S can be written as a set containing the elements. Now we have event E that is both heads and tails have occurred. Now in the sample space the outcomes H H H and T T T have only heads or tails. Remaining elements have both heads so leaving these two outcomes rest belong to event E. So event E is a set containing elements H H T, H T H, T H H, E T H, T H T and H T T. So number of outcomes or we can say number of elements in event E is equal to 1, 2, 3, 4, 5 and 6. Now probability P of an event E is equal to number of elements favorable to event E upon total number of elements in sample space S. Total number of elements in sample space is equal to 8 and number of elements in event E is 6. So this is equal to 6 upon 8 which is equal to 3 upon 4. Thus probability P of an event E is equal to 3 upon 4. Now event F is at most one tail has occurred. It means maximum there should be one tail or less than one tail. It means they have outcomes having no tail outcomes having one tail only. Now H H H has no tail so it belongs to event F then outcomes H H T, T H H have one tail only. So they also belong to event F. So number of elements in event F is equal to 1, 2, 3 and 4. Thus probability P of event F is equal to number of elements favorable to event F that is 4 upon total number of elements in sample space S that is 8 and this is equal to 1 upon 2. Now we want to show that events E and F are independent. Now from the key idea we need that two events A and B are independent. A probability of event A intersection B is equal to probability of event A into probability of event B. Now here we have to prove that E and F are independent events. It means here we have to show that probability P of event E intersection F is equal to probability of event E into probability of event F. Now first of all let us find event E intersection F it will include both elements which are in event E as well as in event F. So from the event E and event F you see D H H H H H T are common to both events thus event E intersection F is a set containing element H H T H T H and T H H. So here we can see that number of elements in event E intersection F is equal to 3. So probability P of event E intersection F is equal to number of elements favorable to event E intersection F that is 3 upon total number of elements in sample space S that is 8. So probability of event E intersection F is 3 upon 8. Now let us find probability of event E into probability of event F which is equal to probability of event E is 3 by 4 and probability of event F is 1 by 2. So this is equal to 3 upon 4 into 1 upon 2 which is equal to 3 upon 4 into 2 that is 8. Thus probability of event E intersection F is equal to probability of event E into probability of event F. Therefore events E and F are independent events. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed this session.