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

  1. $0.3^{k}$
  2. $0.6^{k}$
  3. $0.7^{k}$
  4. $0.9^{k}$
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8 Answers

Best answer
31 votes
31 votes

Total possibilities $=9 \times 10^{k-1},$ because most significant digit has $9$ options (excluding $0)$ and every other digit has $10$ options from $0$ to $9.$

Possibility of not containing any of the digits $0, 5, 9 = 7^{k},$ as now every digit has $7$ options.

Required probability $=\dfrac{7^{k}}{9 \times10^{k-1}} \approx  (0.7)^k$

So C is the answer.

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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
Answer:

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