 Hello, I am welcome to the session, I am Deepika and I am going to help you to solve the calling question. The question says two numbers are selected at random without replacement from the first six positive integers. Let X denote the larger of the two numbers obtained, point EX. Now, we know that the mean of a random variable X which is denoted by mu is equal to summation of xi Pi where i varying from 1 to n. Now, again the mean of a random variable X is also called the expectation of X that is the mean or expectation of a random variable X is the sum of the products of all possible values of X by their respective probabilities. So, this is the key idea behind our question, we will take the help of this key idea to solve the above question. So, let's start the solution. Now, according to the given question, two numbers are selected at random from the first six positive integers and X denote the larger of the two numbers obtained. So, the sample space of the given experiment consists of 36 elements in the form of ordered pairs xi, yi where xi is equal to 1, 2, 3, 4, 5, 6. And yi is also equal to 1, 2, 3, 4, 5, 6. But if we have to select larger of two numbers, we should ignore pairs where the two numbers are same. Thus, we should ignore the pair with numbers 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6. That is if we have to select larger of two numbers obtained, then we should ignore the pairs 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6. That is we should ignore pairs where the two numbers are same. Hence, now our sample space contains 36 minus 6 which is equal to 30 elements. So, the random variable that is the larger of two numbers obtained takes the values 2, 3, 4, 5 or 6. So, the probability distribution of x is for x is equal to 2, that is when the larger of two numbers is 2, when out of 30 pairs we have only two pairs where the larger of two numbers is 2. So, when x is equal to 2, px is equal to 2 over 30. Similarly, when the larger of two numbers is 3, so px is equal to 4 over 30. Again, we have six pairs where the larger of two numbers is 4. So, when x is equal to 4, px is equal to 6 over 30. Now, we will find the number of pairs where the larger of two numbers is 5. So, there are eight pairs. So, when x is equal to 5, px is equal to 8 over 30. Now, we will find the number of pairs where the larger of two numbers is 6. Now, here we have two pairs where the larger of two numbers is 6. So, when x is equal to 6, px is equal to 10 over 30. So, according to our key idea, ex is equal to sigma xi pi i varying from 1 to n. So, this is equal to 2 into 2 over 30 plus 3 into 4 over 30 plus 4 into 6 over 30 plus 5 into 8 over 30 plus 6 into 10 over 30. This is equal to 2 over 15 plus 6 over 15 plus 12 over 15 plus 20 over 15 plus 30 over 15. So, this is equal to 17 over 15 and this is again equal to 14 over 3. Hence, the answer for the above question is 14 over 3. So, this completes our session. I hope the solution is clear to you. Bye and have a nice day.