 Hi and welcome to the session. Let us discuss the following question. Question says, a box contains 90 disks which are numbered from 1 to 90. If one disk is drawn at random from the box, find the probability that it bears a 2 digit number, a perfect square number, a number divisible by 5. First of all, let us understand that probability of occurrence of an event E denoted by P E is defined as number of outcomes favorable to E upon total number of possible outcomes. This is the key idea to solve the given question. Let us now start with the solution. Now we are given in the question that a box contains 90 disks which are numbered from 1 to 90. Now from these 90 disks, one disk can be chosen in 90 ways. So total number of possible outcomes is equal to 90. Now we are given that if one disk is drawn at random from these 90 disks, then we have to find the probability that it bears a 2 digit number. We know 90 disks are numbered from 1 to 90 and there are 9 numbers which are single digit number. So number of disks bearing 2 digit number is equal to 90 minus 9. Clearly we can see there are 9 single digit numbers. So if we subtract 9 from 90, we get 81. Now one out of these disks can be chosen in 81 ways. So number of outcomes favorable to a 2 digit number is equal to 81. Now we have to find the probability of getting a disk bearing a 2 digit number. It is equal to number of outcomes favorable to a 2 digit number upon total number of possible outcomes. That is it is equal to 81 upon 90. Now we will cancel common factor 9 from numerator and denominator both and we get 9 upon 10. So required answer for the first part of the given question is probability that the disk bears a 2 digit number is equal to 9 upon 10. Let us now start with the second part. Now we have to find the probability that the disk bears a perfect square number. First of all let us write all perfect square numbers from 1 to 90. So we can write perfect square numbers from 1 to 90 are 1, 4, 9, 16, 25, 36, 49, 64, 81. So these are 9 numbers. So we can write number of perfect square numbers from 1 to 90 is equal to 9. Now one number from these 9 numbers can be selected in 9 ways. So number of outcomes favorable to a perfect square number is equal to 9. Now let us find out probability of getting a disk marked with a perfect square. Or we can say we will find out probability of getting a disk marked with a perfect square number. Now it is equal to number of outcomes favorable to a perfect square number that is 9 upon total number of possible outcomes that is 90. Now we will cancel common factor 9 from numerator and denominator both and we get 1 upon 10. So required answer for the second part of the given question is probability that the disk bears a perfect square number is equal to 1 upon 10. Let us now start with the third part. Now we have to find the probability that disk bears a number divisible by 5. First of all let us write all the numbers from 1 to 90 which are divisible by 5. They are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 80, 90, 90 and so on. 85 and 90. Now clearly we can see these are 18 numbers. So there are 18 disks marked with the numbers which are divisible by 5. So we can write number of outcomes favorable to a number divisible by 5 is equal to 18. We know one number out of these 18 numbers can be chosen in 18 ways. So number of outcomes favorable to a number divisible by 5 is equal to 18. Now we will find out probability of getting a disk marked with a number divisible by 5. It is equal to number of outcomes favorable to a number divisible by 5 that is 18 upon total number of possible outcomes that is 90. Now we will cancel common factor 18 from numerator and denominator both and we get 1 upon 5. So required answer for the third part is probability that a disk bears a number divisible by 5 is equal to 1 upon 5. So this is our required answer. This completes the session. Hope you understood the solution. Take care and keep smiling.