 Hello and welcome to this session. In this session we are going to discuss the following question which says that the first 10 prime numbers 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 are written in any order. What is the probability that there are 3 prime numbers between 2 and 3? 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 things then this can be done in Cx number of ways that is given by n factorial by x factorial into n minus x factorial and if n things are to be arranged among themselves then this can be done in n factorial number of ways. With this key idea we shall proceed with the solution as there are 10 prime numbers in all let the positions be denoted by 1, 2, 3, 4 up to 10. Now when 2 is kept as the first position then 3 must occupy the first place to have 3 prime numbers in between that is when we place 2 at the first position then 3 must occupy the fifth position in order to have 3 prime numbers in between then 2 is kept at the second position then 3 must occupy the sixth position and so on till at the sixth position and 3 is kept at the tenth position to have 3 prime numbers in between also 2 and 3 can interchange their positions we know that if n things are to be arranged among themselves then this can be done in n factorial number of ways so 2 and 3 can interchange their positions in 2 factorial number of ways thus n3 can take the required position in 2 factorial into 6 number of ways or 12 ways. Now out of remaining 8 prime numbers 3 prime numbers can be selected in c3 ways that is when the key idea that if x things are to be selected from n things then this can be done in ncx number of ways. Now these 3 numbers 5 others can be arranged in c factorial into 5 factorial number of ways therefore the number of arrangements there are 3 prime numbers between 2 and 3 is given by 12 into 8c3 into 3 factorial into 5 factorial and the total number of arrangements 10 prime numbers can be put is given by 10 factorial therefore the required probability is given by number of favorable outcomes that is the number of arrangements in which there are 3 prime numbers between 2 and 3 that is 12 into 8c3 into 3 factorial into 5 factorial upon total number of outcomes that is total number of arrangements in which 10 prime numbers can be put that is 10 factorial which can be written as 12 into 8c3 is given by 8 factorial by 3 factorial into 8 minus 3 factorial into 3 factorial into 5 factorial whole upon 10 factorial which is given by 12 into 8 factorial by 3 factorial into 5 factorial into 3 factorial into 5 factorial by 10 factorial which is equal to 12 into 8 factorial by 10 into 9 into 8 factorial which is equal to 12 upon 10 into 9 that is 2 by 15 therefore the probability in which there are 3 prime numbers between 2 and 3 is given by 2 by 15 which is the required answer. This completes our session. Hope you enjoyed this session.