GATE CS 2017 Set 1 QGA-5 — General Aptitude

To find the probability that a $k$-digit number does not contain the digits $0, 5,$ or $9$, we consider the choice of digits at each of the $k$ positions:

1. Total possible digits for any position: $$\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\} \quad \text{(10 possible digits)}$$
2. Excluded digits: $$\{0, 5, 9\} \quad \text{(3 digits)}$$
3. Allowed digits: $$\{1, 2, 3, 4, 6, 7, 8\} \quad \text{(7 digits)}$$

Assuming each digit position in the $k$-digit sequence is chosen independently from the $10$ possible digits:

• The probability of choosing an allowed digit for any single position is: $$P(\text{allowed}) = \frac{7}{10} = 0.7$$
• Since the choices for all $k$ positions are independent, the probability that none of the $k$ digits contain $0, 5,$ or $9$ is: $$P = \underbrace{0.7 \times 0.7 \times \dots \times 0.7}_{k \text{ times}} = 0.7^k$$

Therefore, the correct option is C.