This question asks us to find the probability that a number with k digits does not contain the specific digits 0, 5, or 9.

Probability Calculation for Single Digit

First, let's consider the total number of possible digits. The digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. That makes a total of 10 possible digits.

The question specifies digits that the number should NOT contain: 0, 5, and 9. There are 3 forbidden digits.

To find the digits that are allowed, we subtract the forbidden digits from the total set of digits.

Allowed digits = Total digits - Forbidden digits

Allowed digits = {1, 2, 3, 4, 6, 7, 8}

The number of allowed digits is 7.

The probability that a single randomly chosen digit is NOT 0, 5, or 9 is the ratio of the number of allowed digits to the total number of digits.

Probability (single digit is allowed) = (Number of allowed digits) / (Total number of digits)

Let $P(\text{allowed})$ denote this probability.

$$P(\text{allowed}) = \frac{7}{10} = 0.7$$

Probability Determination for k-digit Number

A k-digit number is formed by choosing k digits. Each digit position can be filled independently.

We want the probability that *every* digit in the k-digit number is from the allowed set (i.e., not 0, 5, or 9).

Since the choice for each digit is independent, the probability that all k digits are allowed is the product of the probabilities for each digit being allowed.

Probability (k-digit number does NOT contain 0, 5, or 9) = P(1st digit allowed) $\times$ P(2nd digit allowed) $\times$ ... $\times$ P(k-th digit allowed)

Using the probability calculated for a single digit:

$$P(\text{k-digit number is allowed}) = (P(\text{allowed}))^k$$

$$P(\text{k-digit number is allowed}) = \left(\frac{7}{10}\right)^k$$

$$P(\text{k-digit number is allowed}) = (0.7)^k$$

Conclusion

Therefore, the probability that a k-digit number does NOT contain the digits 0, 5, or 9 is $0.7^k$. This matches option 3.