For a very large number, n, of trials, a binomial distribution (which this is: "success", rolling a one or a two, has probability 2/6= 1/3, "failure", rolling a three, four, five, or six, has probability 4/6= 2/3), with probability of success p, can be modeled by the normal distribution with mean np and standard deviation \(\displaystyle \sqrt{np(1- p)}\). If you were given a problem like this, surely you knew that?:wink:
