 Hello and welcome to the session, I am Shashi and I am going to help you with the following question. Question says, find the probability of getting 5 exactly twice in 7th rows of a die. Let us now start with the solution. Now clearly we can see repeatedly throwing of a die 7 times are Bernoulli trials. So we can write repeated throws of a die are Bernoulli trials, we know Bernoulli trials are finite in number, they are independent, they have exactly 2 outcomes that is success or failure and the probability of success remains same in each trial. So throw of a die 7 times are Bernoulli trials. Now let us assume that x denotes number of 5 in an experiment of 7 trials. Now x has binomial distribution with n is equal to 7 and p is equal to 1 upon 6, we know probability of getting 5 in a single throw of a die is equal to 1 upon 6, we know total outcomes where we are able to 5 is equal to 1 and total number of possible outcomes is equal to 6. So probability of getting 5 in a single throw of dies is 1 upon 6 and also probability of failures that is probability of not getting 5 in a single throw of dies is equal to 1 minus 1 upon 6 which is further equal to 5 upon 6. Now we know for binomial distribution B and P probability of x successes is equal to n c x multiplied by q raise to the power n minus x multiplied by p raise to the power x where x is equal to 1 to n now we have to find the probability of getting 5 exactly twice in 7 throw of a die. So number of successes is equal to 2 so value of this x is equal to 2 now total number of trials is equal to 7 so value of n is equal to 7 value of x is equal to 2 we know value of q is equal to 5 upon 6 so we will substitute 5 upon 6 for q and here we will write 7 minus 2 we know here n is equal to 7 and x is equal to 2 now probability of getting 5 is equal to 1 upon 6 so we will substitute 1 upon 6 for p here and we will write 2 for x now this is further equal to 7 multiplied by 6 upon 2 multiplied by 1 multiplied by 5 upon 6 raise to the power 5 multiplied by 1 upon 36 we know square of 6 is 36 and square of 1 is 1 7 minus 2 is equal to 5 now simply find further we get we will cancel common factor 2 here we get 21 upon 36 multiplied by 5 upon 6 raise to the power 5 now again we will cancel common factor 3 from numerator and denominator both and we get 7 upon 12 multiplied by 5 upon 6 raise to the power 5 is equal to probability of getting 2 successes so this is our required answer this completes the session hope you understood the solution take care and have a nice day.