Total K positions, and to make it a K-digit number the 1st position will not have 0 (so 9 choices to fill 1st position i.e. 1 to 9) while the remaining K-1 positions can be filled in 10 ways (0 to 9). So, total K-digit numbers that we can have = 9*10^(k-1).
#cases in which number does NOT contain the digits 0,5 or 9 = 7
so, probability = 7^k/9*10^(k-1) = 1.11(0.7)^k