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So I first thought to approach this as the complement of an inclusion exclusion problem, $P(A_{1}\cap A_{2}\cap A_{3})=1-P(A_{1}\cup A_{2}\cup A_{3})$ Where $A_{i}$ is the event that digit i appears in the decimal string. I know how to evaluate this using inclusion exclusion. But I thought that this solution would also work:

$P(A_{1}\cap A_{2}\cap A_{3})={m\choose 3}\frac{1}{10^3}$. The number of ways of picking 3 out of m events weighted by the probability that these occur.

However, plugging in m=3 my answer is off by a factor of 3! Where am I going wrong?

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  • $\begingroup$ The formula you used should be with the chosen digit $i$ absent $\endgroup$ – true blue anil May 1 '16 at 2:44
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The clue is that you are off by 3! which is the number of ways to arrange 3 objects...

To flesh this out: Consider when $m=3$, you get $ 3 \choose 3$$ (1/10^3)$ =$1/1000$ but you know that there are 6 ways - namely every possible combination of 123 (123, 132, 213, 231, 321, 312).

What your formula is giving is the probability that 1,2 and 3 appear in one particular order, but since the digits 1,2,3 can appear in any order, you have to multiply by 3!. I'd write this as $ \frac{m!}{(m-3)!10^m}$ or $ \frac{m(m-1)(m-2)}{10^m}$ but that's because I don't like the choose function.

Note that the last form gives an indication as to the how many choices you have left...

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  • $\begingroup$ would it be correct to say that we don't divide by $3!$ since these objects are different and order matters? $\endgroup$ – qbert May 1 '16 at 1:24
  • $\begingroup$ Precisely - it counts 1230 and 1320 as the same thing, when clearly they are not. $\endgroup$ – Dan Barry May 1 '16 at 1:29
  • $\begingroup$ got it. serves me right for using the choose function without thinking about what it means. $\endgroup$ – qbert May 1 '16 at 1:31
  • $\begingroup$ With this all being said, I'm not convinced by the correctness of the formula - I feel like mine give probability that's too low, as if I'm forgetting about another factor. $\endgroup$ – Dan Barry May 1 '16 at 1:56
  • $\begingroup$ I'm afraid it will work only for a $3$ digit number. $\endgroup$ – true blue anil May 1 '16 at 3:19
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You will have to apply inclusion-exclusion in this manner:

All strings - strings with

at least one from $1,2,3$ absent + at least two of $1,2,3$ absent - all from $1,2,3$ absent

Allowing zeroes at start, it would work out as:

$10^m - \binom31 9^m + \binom32 8^m - 7^m$

Added

Oh, you wanted the probability, so , of course, you'd have to divide by $10^m$

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    $\begingroup$ The other formula worked for $m=3$ because the only options were to include all the $3$ chosen numbers, the rest of the digits didn't figure at all. $\endgroup$ – true blue anil May 1 '16 at 2:57

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