 Hi and welcome to the session. Let's work out the following question. The question says a die is thrown twice and the sum of numbers appearing is observed to be 8. What is the conditional probability that the number 5 has appeared at least once? Let us see the solution to this question. Here we shall consider the following events. Event A, B, the event of getting the sum of numbers. The sample space for event A will be 2, 6, 3, 5, 4, 4, 5, 3 and 6, 2. Now in each of them the first, it represents the number that we get on the first die or die for the first time and this is for the second time. Let B be the event of getting the number 5 at least once. The sample space for B will be 5, 1, 5, 2, 5, 3, 5, 4, 5, 5, 5, 6, 1, 5, 2, 5, 3, 5, 4, 5, 5. We see that the elements common in these two sets that is A intersection B is equal to the element 3, 5, 5, 3. Event A is 5 by 36. Probability of event B is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. 11 by 36 and probability of A intersection B is equal to 2 by 36. Now required probability is probability B by A that is equal to probability B intersection A divided by probability of A that is equal to 2 by 36 divided by 5 by 36 and that is equal to 2 by 5. So this is our answer to this question. I hope that you understood the solution and enjoyed the session. Have a good day.