The probability that a \(k\)-digit number does NOT contain the digits 0,5, or…
2017
The probability that a \(k\)-digit number does NOT contain the digits 0,5, or 9 is
- A.
\(0.3^k\) - B.
\(0.6^k\) - C.
\(0.7^k\) - D.
\(0.9^k\)
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Correct answer: C
Key fact: excluding the digits 0, 5, and 9 leaves seven allowed digits: 1, 2, 3, 4, 6, 7, 8.
If each of the k positions is chosen independently from 0–9 (so leading zeros are allowed): each position has 7 allowed choices out of 10, giving probability (7/10)^k = 0.7^k.
If a "k-digit number" is taken in the usual sense (leading digit cannot be 0):
Total k-digit numbers = 9·10^(k-1).
Favourable numbers: first digit must be one of the seven digits 1,2,3,4,6,7,8 (7 choices), and each of the remaining k−1 digits also has 7 choices, so favourable count = 7^k.
Therefore the probability = 7^k / (9·10^(k-1)) = (7/10)^k · (10/9).
Conclusion: The expression 0.7^k is correct under the interpretation that digit positions are independent and leading zeros are allowed. Under the standard definition of a k-digit number (no leading zero), the correct probability is 7^k/(9·10^(k-1)). The problem is ambiguous unless the treatment of leading zeros is specified.
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