Asix-sided die, in which each side is equally likely to appear, is repeatedly rolled until the total of all rolls exceeds 400. approximate the probability that this will require more than 140 rolls.

Let S = X1 + X2 + ... + X140

Get E[X] and Var[X]

E[X] = (6 + 1)/2 = 7/2
Var [X] = (6^2 - 1)/12 = 35/12

So E[S] = 140 (7/2) = 490 while Var [S] = 140 (35/12) = 1225/3

Use the central limit theorem with continuity correction in finding the probability.

Pr {S ≤ 400}
Pr {S < 400.5 - 490 / √(1225/3)}

The answer should be 0.000004