1) This is a binomial distribution with n = 50 and p = 0.12. Now,
P(X >= 3) = 1 - P(X <= 2) = 1 - binomcdf(50, 0.12, 2) = 0.9487
2) If it contains four digits, then there are (10)(10)(10)(10) = 10^4 = 10000 different combinations.
If it contains five digits, then there are (10)(10)(10)(10)(10) = 10^5 = 100000 different combinations.
If it contains six digits, then there are (10)(10)(10)(10)(10)(10) = 10^6 = 1000000 different combinations.
Therefore, the total number of combinations is 10000 + 100000 + 1000000 = 1110000.