Correct Answer - Option 3 : 0.7
k
Consider k = 1, means 1-digit number doesn’t contain the digits 0, 5, or 9.
Remaining digits are 1, 2, 3, 4, 6, 7 ,8.
Total outcomes = 10
Favourable outcomes = 7
Probability = \(\frac{7}{{10}}\) = 0.7
Consider k = 2, means 2-digit number doesn’t contain the digits 0, 5, or 9
So, both the places can be filled by the remaining 7 digits.
Total outcomes = 10 × 10
Favourable outcomes = 7 × 7
Probability = \(\frac{7}{{10}}\) × \(\frac{7}{{10}}\) = \({0.7^2}\)
Similarly it goes for k = 3, probability = \({0.7^3}\)
It goes up to k….
The probability that a k-digit number does NOT contain the digits 0, 5, or 9 = \({0.7^k}\)
