This is a binomial probability
P( no 3's) = (5/6)^10
P( one 3) = C(10, 1) * ( 1/6) (5/6)^9
P( two 3's) = C(10, 2) *(1/6)^2 (5/6)^8
P( three 3's) = C(10, 3) *(1/6)^3 (5/6)^7
P(four 3's) = C(10, 4) * (1/6)^4 (5/6)^6
So....the sum of these ≈ .9845 ≈ 98.45 %
