 Hello and welcome to the session. In this session we will discuss a question which says that tickets are numbered from 1 to 75. One ticket is drawn. Find the probability that number of tickets is either divisible by 3 or 5. Now before starting the solution of this question, we should know a result. And that is if A and B are two events such that they are not mutually exclusive, then probability P of event A or B is equal to probability of event A plus probability of event B minus probability of event A intersection B. Now this result will work out as a key idea for solving out the given question. Now let us start with the solution of the given question. Now we have given that tickets are numbered from 1 to 75 and one ticket is drawn from the given tickets. You have to find the probability that the number of the tickets drawn is either divisible by 3 or 5. First of all, let us write sample space for this question. Now sample space will consist of numbers of the tickets from 1 to 75. That is, sample space S is equal to search container elements 1, 2, 3, 4 and so on up to 75. So here you can see that sample space S consists of all the numbers of the tickets from 1 to 75. So sample space is a set containing all these elements which are 75 in number. So the number of outcomes is equal to 75. Now let us define the events. Event is equal to number divisible by 3 and event B is equal to number divisible by... Now let us write elements of event A elements will be the numbers from this sample space that are divisible by 3. So event A is a search containing elements 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 63, 64, 72 and 75. You can see that all these numbers are divisible by 3. So number of outcomes for event A is equal to... Now when we count these elements we are 25 in number. So number of variable outcomes for event A is 25. Now here total number of outcomes is equal to 75. Number of favorable outcomes for event A is 25. So probability P of event A is equal to number of favorable outcomes for event A that is 25 upon total number of outcomes that is 75. Now 25 into 3 is 75. So this is equal to 1 upon 3. So probability of event A is equal to 1 upon 3. Now let us write elements of event B where we find the numbers from this sample space that are divisible by 5. Event B is a search containing elements 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70 and 75. Now here you can see all these numbers are divisible by 5. Number of favorable outcomes for event B. Now when we count set B they are 15 in number. So number of favorable outcomes for event B is equal to... So probability of event B is equal to number of favorable outcomes for event B that is 15 upon total number of outcomes that is 75. Now 15 into 5 is 75. So this is equal to 1 upon 5. Now this is the event B. Now A and B have some that they are not mutually exclusive events. Now we have event A intersection B which is a set to event A and event B. So this is a search containing elements 15, 30, 35, 75. So number of favorable outcomes event A intersection B is equal to 5 of event A intersection B which is equal to number of favorable outcomes for event A intersection B that is 5 upon total number of outcomes that is 75. Now 5 into 15 is 75. So this is equal to 1 upon... So we have to find probability P of event A of B because we will use this result which is given to us in the key idea as A and B are not mutually exclusive events. So probability P of event A of B is equal to probability of event A plus probability of event B minus probability of event A intersection B. Now this is probability of event A and this is probability of event B and this is probability of event A intersection B. Now putting values here 1 upon 3 plus 1 upon 5 minus 1 upon 15. Now each variable of 3, 5 and 15 is 15. In the denominator we have 15. 3 into 5 is 15 and 5 into 1 is 5 plus 5 into 3 is 15 and 3 into 1 is 3 minus 15 into 1 is 15 and 1 into 1 is 1. So this is equal to each minus 1 whole upon 15 which is equal to 7 upon 15. So probability P of event A of B is equal to 7 upon 15. So this is the required answer and this concludes our session. Hope you all have enjoyed the session.