 Hello and welcome to the session. My name is Mansi and I am going to help you with the following question. The question says, a die is thrown, describe the following events. Post is A, that is a number less than 7. Second is B, a number greater than 7. C is a multiple of 3. Event D is a number less than 4. E is an even number greater than 4. Event F is a number not less than 3. Also find A union B, A intersection B, E union F, D intersection E, A minus C, D minus C, F dash and E intersection F dash. Now we see that A union B, this is event A or B. A intersection B is event A and B. Similarly for these two, F dash means event not F and so on. So now we can start with the solution to this question. First of all, we have to describe the events. The first part is A, that is a number less than 7 when a die is thrown. So we see that the sample space for event A will be 1, 2, 3, 4, 5, 6. So where this A die is thrown, so number that can turn up is 1, 2, 3, 4, 5 or 6. We have to tell an event where number is less than 7. So this is the sample space for event A. Now second, we have to tell an event where we get a number greater than 7. Since we know that in a die, we can only get 1, 2, 3, 4, 5 or 6. So event B will be 5. Now third event is C such that the event is a multiple of 3. So event C will be 3 or 6. So this will be the sample space for event C. Now fourth is D that is a number less than 4. So the sample space for event D will be 1, 2 or 3 because these are the numbers less than 4 that we can get by throwing a die. Fifth is E that is an even number greater than 4 and the only even number greater than 4 here is 6. So event E is getting a 6. This is the sample space for event E and sixth is F that is a number not less than 3 that means equal to or greater than 3. So event F will be 3, 4, 5, 6 getting a 3 or 4 or 5 or 6. This will be the sample space for event F. The next thing that we have to find is A union B. So we see that A union B is the event either A or B or both. So that will be the set containing the elements 1, 2, 3, 4, 5, 6 because that is 1, 2, 3, 4, 5, 6 union 5. So this is same as this. So this is our answer to the first thing that is this is A union B. Now we have to find A intersection B. Now we see that A intersection B denotes the events A and B that will be equal to 5 because we see that the elements in this set are not there in this set. So their intersection that is the common elements is 5 that is nothing. Now we have to find B union C. B union C we can similarly find it by the way we find A union B will be equal to the set containing the elements 3 and 6. Next thing that we have to find is E intersection F. We see that elements common to the set E and F are just the element 6. So this is E intersection F. Now the next thing we have to find is D intersection E. We see that this is D and this is E. Now we see that there is no point common to both of them. So we can say that D intersection E is equal to 5. Next we have to find A minus C. Now what we do here is we subtract the elements of C from the elements of A. So we get 1, 2, 3, 4, 5. We just get 1, 2, 4, 5 because 3 and 6 they are there in the event C so they get subtracted from this. So we have A minus C else the set containing elements 1, 2, 4, 5. Next we have to find D minus E again we do the same thing we subtract the elements of E from the elements of D. But we see that no element is common to both of them so we can say that D minus E is the set containing the elements 1, 2, 3. Next we have to find F dash that is event not F. We see that on throwing a die the sample space would be 1, 2, 3, 4, 5, 6. Now these are the possibilities that we can get. These are the numbers that we can get on a die. But we see that event F is getting 3, 4, 5, or 6. So F dash would be event not F that is we have 1 and 2. These two would be the remaining elements if we subtract this from the main sample space. So even not F would be the set containing 1 and 2. So F dash that is event not F would be 1, 2. Next we have to find E intersection F dash. Now this is F dash this is event E. So we see that we do not have any element common to these two. So we can say that E intersection F dash is equal to 5. So this was our answer to the question. I hope that you understood the question and enjoyed the session. Have a good day.