 Hello and welcome to the session. The given question says two dices are rolled once. Find the probability that first the numbers on the two dice are different and second is the total of numbers on the two dice is at least four. Let's start with the solution and we are given that two dices are rolled. So when two dices are rolled once then the sample space let us denote it by s is equal to on the first die if one appears and on the second die 1, 2, 3, 4, 5, 6 can appear. Similarly if on the first die two appears then again on the second die 1 to 6 numbers can appear since a die has six faces having numbers 1, 2, 6, 2, 3, 2, 4, 2, 6. Similarly when three appears on the first dice so we have 3, 1, 3, 2, 3, 3, 3, 4, 3, 5 and 3, 6 and similarly doing it for 4, 5 and 6 which appears on the first die we have 1 to 6 numbers on the second die. So this is the sample space when two dices are rolled and here the total number of possible outcomes is equal to 36. Now let us find the probability that the numbers on the two dice are different. So first let us find the probability that the numbers on the two dice are same. Let us start with the first one. Let us denote a by the event of getting same number both the dice. So here on observing we find that 1, 1, 2, 2, 3, 3, 4, 4, 5, 5 and 6, 6 are the events of getting the same number on both the dice. So a it is equal to the sample space having 1, 1, 2, 2, 3, 3, 4, 4, 5, 5 and 6, 6. Number of elements in a is equal to 6. Now we know that probability of an event is equal to the number of favorable outcomes to the event a divided by the total number of possible outcomes. Therefore here we have probability of a is equal to number of outcomes favorable all divided by the total number of possible outcomes. So here probability of a is equal to the number of possible outcomes favorable to a are 6 and the total number of possible outcomes are 36. So this is equal to 1 divided by 6. Now according to the equation we have to find the probability that the numbers on the two dice are different. That is we have to find the probability of a complement. That is we have to find the probability that the event of getting three numbers on both the dice and this is equal to 1 minus probability of a which is further equal to 1 minus 1 divided by 6 which is equal to 5 by 6. Hence answer to the first part is 5 divided by 6. Now let us proceed on to the second part. Find the probability that the total of numbers on the two dice is at least 4. So in the second part let us denote the event b by the event of getting total equal to 4 on both the dice. Since we have to find the probability that the total numbers on the two dice is at least 4. So the minimum is 4 and maximum it can go up to 12 since on both the dice the numbers which can appear last are 6 and 6 which gets the total of 12 which is the maximum total. So here we have to find the probability of b. So first let us write down the sample space for b. 1 appears on the first die then the smallest number which can appear on the second die such that sum of two numbers is 4 or greater than 4 is 3. Since 1 plus 3 is 4 then 4 can also appear on the second die. Since 1 plus 4 is 5 and we can get a total greater than or equal to 4 then we have 1 5 and 1 6. Then with 2 we can have 2 on the second die if 2 appears on the first die then we can have 2 3, 2 4, 2 5 and with 3 we can have 3 1, 3 2, 3, 3 4, 3 5 and 3 6 then with 4 we can have 4 1, 4 3, 4 4, 4 5, 4 6 but 5 also we can have all the probabilities 1, 2, 6 on the second die. So we have 5 4, 5 5 and 5 6 and on the last we can get 6 on the first die then again we have the probabilities of getting 1 to 6 on the second die. Observing we find that the number of elements in set b is equal to 33. So the probability of b is equal to the number of possible outcomes to the event b which are 33 divided by the total number of possible outcomes and simplifying this we get 11 divided by 12. Hence the answer to the second part is 11 divided by 12. So this completes the question buy and take care.