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

  1. $0.3^k$
  2. $0.6^k$
  3. $0.7^k$
  4. $0.9^k$

I'm confused since the question has asked the number should not contain 0,5 OR 9. So, should the answer be $0.9^k$ or something else?

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    Out of ten digits, three are forbidden. Note: the problem allows the $k$-digit number to begin with a zero. – 2017-02-14
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    So, the answer is $0.7^k$ ? – 2017-02-14
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    Yes. As for the language used, it is unfortunate how sometimes in english the words "or" and "and" seem to be used interchangably in some contexts. Here, the usual interpretation of how the problem is written is "... *does not contain the digit 0 AND does not contain the digit 5 AND does not contain the digit 9*." The answer to the question of "*does not contain the digit 0 OR does not contain the digit 5 OR does not contain the digit 9*" one should approach via inclusion-exclusion principle and the answer will appear a bit messier than the given answers. – 2017-02-14
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    As for the issue of whether or not leading zeroes count, almost always we treat the number $00053$ as a two digit number (*as it can be rewritten as $53$*) but as a five digit **string**. To correct the phrasing of the question to have $0.7^k$ actually be the right answer, it should instead be written "The probability that a number **with at most $k$ digits** happens to be a $k$-digit number without any of the digits $\{0,5,9\}$ appearing..." or written "The probability that a **length $k$ numerical string**..." to avoid that issue. Otherwise correct answer is $\frac{7^k}{9\cdot 10^{k-1}}$ – 2017-02-14
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    @JMoravitz How did you get $9.10^{k-1}$ – 2017-02-14
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    What is the probability the first digit doesn't violate the condition? The second digit? The third digit? etc... As each of these are independent events, we multiply. So, it is $\frac{7}{9}\cdot \frac{7}{10}\cdot \frac{7}{10}\cdots \frac{7}{10}$, my point being that the correct probability for the first digit to not be any of 0,5,9 is going to be $\frac{7}{9}$, not $\frac{7}{10}$ as the given answer key suggests since there exist no $k$-digit **numbers** with $0$ as the first digit under the usual definition of $k$-digit numbers. – 2017-02-14
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    Alternatively worded, there are $7^k$ valid $k$-digit numbers which contain none of the digits $0,5,9$. On the other hand, there are $9\cdot 10^{k-1}$ actual $k$-digit numbers regardless (*seen by multiplication principle, again noting that the leading digit is not allowed to be zero and has only $9$ available choices, not ten like the rest of the digits*). Taking the ratio yields the result. – 2017-02-14
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    Thank you @JMoravitz That really cleared my doubt. – 2017-02-14
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    It is confusing, but a quick sanity check can be obtained by examining the question for $k = 1$. – 2017-02-14

1 Answers 1

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Case I

Assumptions:

  • The left most digit can be $0$.
  • The questioner wants $0$, $5$ and $9$ to not occur simultaneously.(Here, or & and can be interchangeably used in this case.)

Solution:

Total no. of k digit numbers $= 10^k$

If $0, 5$ and $9$ can't be used, total no. of k digit numbers $= 7^k$

Hence, $$Probability=(7^k/10^k)=0.7^k$$

Case II

Assumptions:

  • The left most digit can be $0$.
  • The questioner wants one of $0$, $5$ and $9$ to not occur.

Solution:

Total no. of k digit numbers $= 10^k$

If one of $0, 5$ and $9$ can't be used, total no. of k digit numbers = $9^k$

Hence, $$Probability=(9^k/10^k)=0.9^k$$

Case III

Assumptions:

  • The left most digit cannot be $0$.
  • The questioner wants $0$, $5$ and $9$ to not occur simultaneously.

Solution:

Total no. of k digit numbers=$9 \times 10^{(k-1)}$

If one of $0, 5$ and $9$ can't be used, total no. of k digit numbers=$7^k$

Hence, $$Probability=\frac{7^k}{9 \times 10^{k-1}}$$

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    It would be really nice if someone can help me in editing the last line. – 2017-02-14
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    use `{ }`'s to make everything go in superscript. `$a^{everything} b^JusttheJ$` produces $a^{everything} b^JusttheJ$. [More MathJax and $\LaTeX$ tips here](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference). – 2017-02-14
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    @YashSwaraj: Please check that my edit accurately captures your intent. – 2017-02-14
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    @BrianTung, thanks for your edit. – 2017-02-14