 Hello and welcome to the session. In this session we discuss the following question which says a box contains 80 disks which are numbered from 1 to 80. If one disk is drawn at random from the box, find the probability that it appears a single digit number and a perfect square number. Before we move on to the solution, let's recall the formula to find the probability of an event E. This is given by the number of trials in which the event E has happened upon the total number of trials. This is the key idea to be used for this question. Now we move on to the solution. Now the question is given that the box contains 80 disks and we take out one disk from these 80 disks and these 80 disks are numbered from 1 to 80. So we say the total number of disks is equal to 80. Now we take let E1 be the event of getting a disk bearing a single digit number and we take let E2 be the event of getting a disk bearing a perfect square number. First we need to find the probability that the disk bears a single digit number that is mean to find the probability of event. Now the number of disks bearing a single digit number is equal to 9. So probability of event that is probability of getting a disk bearing a single digit number is equal to the number of disks bearing a single digit number that is 9 upon the total number of disks that is 80. So this is equal to 9 upon 80. Next we find the probability of E2 that is the probability of getting a disk bearing a perfect square number. Now the number of disks bearing a perfect square number is equal to 8. So probability of E2 is equal to the number of disks bearing a perfect square number that is 8 for the total number of disks that is 80. This is equal to 1 upon 10. So final answer is for the first part it's 9 upon 80 and for the second part it's 1 upon 10. So this completes the session. Hope you have understood the solution for this question.