The probability that a $k$-digit number does NOT contain the digits 0, 5, or 9 is

There are $10$ possible decimal digits: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

The digits to avoid are $0, 5,$ and $9$ (a total of $3$ digits). The allowed digits are the remaining $10 - 3 = 7$ digits.

For any single digit position, the probability that the digit chosen is not $0, 5,$ or $9$ is:

Since this condition must hold independently for each of the $k$ digits in the number, the probability is:

Hence, the correct option is (C).