You are going to throw a 6-sided die. What is the probability that you will throw at least three "one's" in 4 tries?
P ( three 1's) = C(4/3) ( 1/6)^3 (5/6) = 5/324 = 20/1296
P(four 1's) = (1/6)^4 = 1/1296
So...the total probability = (20/1296) + (1/1296) = 21 / 1296 = 7 / 432