 Hello and welcome to the session. In this session we are going to discuss the following question which says that a bar contains 30 tickets numbered from 1 to 30. 7 tickets are drawn at random and are arranged in ascending order. 1. Find the probability that the ticket contains at most 2 prime numbers. 2. Find the probability that the fifth number is 25. Probability of occurrence of any event, say A, is given by number of favorable outcomes upon total number of outcomes. If x things are to be selected from n different things then the number of ways for this selection will be ncx that is n factorial by x factorial into n minus x factorial. With this key idea we shall proceed with the solution. We are given that a bar contains 40 tickets numbered from 1 to 40 and 7 tickets are drawn at random and are arranged in ascending order. Now we need to find the probability that the ticket contains at most 2 prime numbers. There are 40 tickets out of which 7 tickets are drawn at random. From the key idea we know that if x things are to be selected from n different things then this can be done in ncx ways that is n factorial by x factorial into n minus x factorial ways. Therefore total number of outcomes when 7 tickets are to be drawn from 40 tickets is given by 40c7. Now we shall find the prime numbers lying between 1 to 40 are given by 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31 and 37. Now prime numbers between 1 and 40 are 12 and number. Therefore numbers which are not prime is given by 40 minus 12 that is 28. Therefore there are 28 numbers which are not prime. Now probability of getting at most 2 prime numbers is equal to probability of getting no prime number plus probability of getting 1 prime number plus probability of getting 2 prime numbers. Now we shall find probability of getting no prime number. Now we know that there are 28 numbers which are not prime and there are 30 tickets in all. So number of favorable outcomes is given by 28c7 as we have to select 7 tickets out of the 28 tickets which are not prime. And total number of outcomes is equal to 30c7. Therefore probability of getting no prime number is given by, we know that probability of occurrence of any event say A is given by number of favorable outcomes by total number of outcomes. Number of favorable outcomes is given by 28c7 upon total number of outcomes that is 40c7. So probability of getting no prime number is equal to 28c7 upon 40c7. Now we shall find probability of getting 1 prime number. Now one number is to be selected from 12 numbers which are prime rest 6 numbers are to be selected from remaining 28 numbers which are not prime. Therefore number of favorable outcomes is equal to 12c1 into 28c6 using fundamental principle of counting as we have to select 1 number from the 12 prime numbers and 6 numbers from the remaining 28 non-prime numbers. Therefore probability of getting 1 prime number is equal to number of favorable outcomes that is 12c1 into 28c6 upon total number of outcomes that is 30c7. Therefore probability of getting 1 prime number is equal to 12c1 into 28c6 upon 30c7. Now we shall find probability of getting 2 prime numbers where 2 numbers are to be selected from 12 numbers which are prime and the remaining 5 numbers are to be selected from the remaining 28 numbers which are not prime. Therefore number of favorable outcomes is given by 12c2 into 28c5 which are to be selected from the 12 prime numbers and 5c2 are to be selected from the remaining 28 non-prime numbers. Therefore probability of getting 2 prime numbers is equal to number of favorable outcomes that is 12c2 into 28c5 by total number of outcomes that is 30c7. Therefore probability of getting 2 prime numbers is equal to 12c2 into 28c5 upon 30c7. Now we have got probability of getting no prime number, probability of getting 1 prime number and probability of getting 2 prime numbers. And we know that probability of getting at most 2 prime numbers is given by probability of getting no prime number plus probability of getting 1 prime number plus probability of getting 2 prime numbers. Therefore probability of getting at most 2 prime numbers is equal to probability of getting 0 prime number that is 28c7 upon 40c7 plus probability of getting 1 prime number that is 12c1 into 28c6 whole upon 40c7 plus probability of getting 2 prime numbers that is 12c2 into 28c5 upon 40c7 which is equal to 28c7 plus 12c1 into 28c6 plus 12c2 into 28c5 upon 40c7. Hence the probability of getting at most 2 prime numbers is equal to 28c7 plus 12c1 into 28c6 plus 12c2 into 28c5 upon 30c7 which is the required answer. This completes our session. Hope you enjoyed this session.