 Hello and how are you all today? The question says in a single throw of two dice, find the probability that neither a double nor a total of ten will appear. So let's proceed with the solution. Now here, let E be equal to the event of getting a double. And that will be having elements as 1, 1, 2, 2, 3, 3, 4, 4, 5, 5 and 6, 6. That means in all there are 1, 2, 3, 4, 5, 6 elements in it. So n of E number of elements in E are equal to 6. Similarly, let F be equal to the event of getting a total of 10. So it will includes the element that of 4, 6, 6, 4 and 5, 5. That means number of elements in this event is equal to 3. Right? And the total number of elements in this whole sample are equal to 36 as there is a throw of 2 dice over here. So we need to find number of elements in E union F first. That will be n of E plus n of F minus n of E intersection F. N of E is found out above as 6 plus n of F is 3 and there is only one element which is repeating over here. So it will be minus 1 over here that is equal to 8. So the probability of getting neither a doublet nor a total of 3, sorry, a total of 10 will be equal to 1 minus probability of getting these that was E union F. So 1 minus 8 probability of E union F will be number of elements in E union F divided by number of elements in the set. So we have 36 minus 8 upon 36 that is equal to 28 upon 36 that on simplifying will give us 7 upon 9. So 7 upon 9 is the solution to the given question. Right? So hope you understood the whole solution and enjoyed it too. Have a nice day.