 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says three coins are tossed describe two events which are mutually exclusive but not exhaustive. So let us start with the solution to this question. First of all let us write down the sample space for the event that three coins are tossed. So sample space will be getting head, head and head in three coins, then getting head, head and tail in first, second and third coin respectively, getting head, tails and head in three coins respectively and so on. We have HTT, THH, THT, TTH and TTT. So this is the sample space for the event when three coins are tossed. Now let A be the event of getting exactly one head. So the sample space for this event A will be all the elements where we have just one head that is HTT, THT and TTH. Here we have exactly one head. Now let B be the event of getting exactly two heads. So sample space for B will be HHT, HTH and THH. Now we have to show that they are mutually exclusive. That means we see that A intersection B is equal to 5 because no element is common to both these sample spaces. Therefore we can say that A and B are mutually exclusive and we know that two sets are mutually exclusive only if their intersection equals to 5. That means if they do not have any element in common, now we consider A union B. We see that A union B does not give us the entire sample space since A union B is not equal to the sample space. Therefore A and B are not mutually exhaustive because they would have been mutually exhaustive if their union that means all the elements of A and B together if they would have given the sample space this then they would have been mutually exhaustive but since A union B is not equal to the sample space therefore they are not mutually exclusive exhaustive. So this is our answer to the question. I hope that you understood the question and enjoyed the session. Have a good day.