 Hello and welcome to the session. I am Deepika and I am going to help you to solve the following question. The question says, find the binomial distribution for which the mean is 4 and variance 3. Now, the mean of the binomial distribution is given by mp where n is the number of trials and p is the probability of success of each trial and variance of binomial distribution is equal to mpq where q is the probability of failure. So, this is the key idea behind our question. We will take the help of this key idea to solve the above question. So, let's start the solution. Now, we have to find the binomial distribution for which the mean is 4 and variance 3. So, according to our key idea, we have mean of binomial distribution is equal to mp. So, we have mp is equal to 4 and variance of binomial distribution that is mpq is equal to 3. So, this implies q is equal to 3 over 4 because mpq over mp is equal to 3 over 4 that is q is equal to 3 over 4. Now, we know that q is equal to 1 minus p or p is equal to 1 minus q. So, p is equal to 1 minus 3 over 4 which is again equal to 1 over 4. Now, mp is equal to 4 implies n into 1 over 4 is equal to 4 and this implies n is equal to 4 into 4 which is equal to 16. So, the binomial distribution with n is equal to 16, p is equal to 1 over 4 and q is equal to 3 over 4 is given by probability of x successes is equal to 16 Cx into q raise to power n minus x that is 3 over 4 raise to power 16 minus x into p raise to power x where x is from 0 to 16. So, the answer for the above question is probability of x is equal to 16 Cx into 3 over 4 raise to power 16 minus x into 1 over 4 raise to power x where x is from 0 to 16. So, this is the answer for the above question. This completes our session. I hope the solution is clear to you. Bye and have a nice day.