 Hi and welcome to the session. Let us discuss the following question. Question says, A dice is thrown once find the probability of getting a prime number, a number lying between 2 and 6 and odd number. First of all, let us understand that probability of occurrence of an event E denoted by P E is defined as number of outcomes favourable 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 know in a single throw of a dice we can get any one of the 6 numbers that is 1, 2, 3, 4, 5, 6. We know a dice has 6 phases and these 6 numbers are present on its phases. So clearly we can see one number can appear in 6 phase. We know on throwing a dice once we can get any of these 6 numbers. So total number of possible outcomes is equal to 6. Now in the first part of the given question we have to find the probability of getting a prime number in single throw of a dice. Now we know there are 3 prime numbers marked on the dice. They are 2, 3, 5. So we can write prime numbers marked on the dice are 2, 3 and 5. Now clearly we can see a prime number can be selected in 3 ways. So number of outcomes favourable to a prime number is equal to 3. From key idea we know probability of an event E is equal to number of outcomes favourable to E upon total number of possible outcomes. So here the event is getting a prime number in a single throw of dice. So probability of getting a prime number in a single throw of dice is equal to number of outcomes favourable to a prime number upon total number of possible outcomes. Now we know number of outcomes favourable to a prime number is equal to 3 and total number of possible outcomes is equal to 6. So we will cancel common factor 3 from numerator and denominator both and we get 1 upon 2 is equal to probability of getting a prime number on the dice in a single throw. Now this completes the first part of the given question. Now let us start with the second part. Now we have to find the probability of getting a number lying between 2 and 6. Now clearly we can see there are 3 numbers 3, 4 and 5 lying between 2 and 6. So we can write numbers lying between 2 and 6 are 3, 4 and 5. Clearly we can see a number lying between 2 and 6 can be selected in 3 ways. So number of outcomes favourable to a number lying between 2 and 6 is equal to 3. Now let us find out probability of a number lying between 2 and 6 in a single throw of dice. Now this is equal to number of outcomes favourable to a number lying between 2 and 6. That is 3 upon total number of possible outcomes we know total number of possible outcomes is equal to 6. Now we will cancel common factor 3 from numerator and denominator both and we get 1 upon 2 is equal to probability of getting a number lying between 2 and 6. Now this completes the second part of the given question. Now let us start with the third part. Now we have to find the probability of getting an odd number in a single throw of dice. Now clearly we can see 1, 3 and 5 are 3 odd numbers marked on the dice. So we can write 1, 3 and 5 are 3 odd numbers marked on the dice. Now clearly we can see an odd number can appear in 3 ways in a single throw of dice. So number of outcomes favourable to an odd number is equal to 3. Now let us find out probability of getting an odd number. Probability of an odd number is equal to number of outcomes favourable to an odd number. That is 3 upon total number of possible outcomes and total number of possible outcomes is equal to 6. So required probability is equal to 1 upon 2. So we get probability of getting an odd number is equal to 1 upon 2. Now this completes the third part of the given question. So our required answer for the first part is 1 upon 2. For second part is 1 upon 2 and for third part also it is 1 upon 2. This completes the session. Hope you understood the solution. Take care and keep smiling.