and considering it $(\frac{7}{10})^{k}$

15 votes

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

- $0.3^{k}$
- $0.6^{k}$
- $0.7^{k}$
- $0.9^{k}$

24 votes

Best answer

6 votes

Probability of digits containing (0, 3 or 5) if we consider a 'k' digit number is of type base 10 = (3/10) ^k

Hence, p(not containing 0, 3 or 5)=(7/10)^k

Hence, p(not containing 0, 3 or 5)=(7/10)^k