 Hello and welcome to the session. In this session we discussed the following question which says, what is wrong with the following case? The mean of a binomial distribution is 8 and its variance is 12. We know that for a binomial distribution, mean is given as np and variance is given as npq where this n is the number of trials, p is the probability of getting success, q is the probability of getting failure. This is the key idea that we use for this question. Now we move on to the solution. We take let x be a binomial variant with mean equal to 8 and variance equal to 12. Now as we have mean is equal to 8 so this means np is equal to 8 and as the variance is equal to 12 so this means npq is equal to 12. Let this be equation 1 and this be equation 2. By using equation 1 and equation 2 we get 8 into q is equal to 12 so from here we get q is equal to 12 upon 8. Now 4, 2 times is 8 and 4 3 times is 12 therefore we get q is equal to 3 upon 2 which is greater than 1. That is we get q is greater than 1 and we know that q is the probability of getting the failure. Also we know that probability cannot be greater than 1 therefore the given statement that is the mean of a binomial distribution is 8 and variance is 12 is wrong. This completes the session. Hope you have understood the solution of this question.