If you pick a number at random from $1 \ldots 10^k$ where $k$ is large, call this number $Z$. $M(Z)$ is defined as 0 if $Z$ is not divisible by two repeated primes. Find the probability that $M(Z) = 0$. I'm having a hard time getting started on this question, I know you use a particular identity but I'm not sure why you use it.
How do determine probability related to Möbius function
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probability
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1First of all, the definition of $M(Z)$ is unnecessary: the question is simply to find the probability that the number picked is not divisible by two repeated primes. Second, I'm not 100% sure what this phrase means: do you mean that the number is not of the form $p^2 q^2 m$ where $p$ and $q$ are primes? From the question title, I suspect you mean that the number should not be divisible by the square of a single prime. Third, are you interested in the exact probability, or in the limit of that probability as $k\to\infty$? (You should also say that the number is chosen _uniformly_ at random.) – 2011-09-21
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0I am having trouble understanding what "two repeated primes" means. Is the number $17^2$ supposed to be counted? What about $17^2 \cdot 19^2$? Finally, how about $17^2 \cdot 19$ and $17 \cdot 19$? – 2011-09-21
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The mention of "Möbius" makes me suspect that you want the probability that $\mu(Z) = 0$, i.e. that $Z$ is divisible by the square of a prime. For each nonempty set $S$ of primes whose product $Q_S \le 10^{k/2}$, let $A_S$ be the event that $Z$ is divisible by $Q_S^2$. Then $P(A_S) = \lfloor 10^k/Q_S^2 \rfloor / 10^k$. By inclusion-exclusion, $P(\mu(Z)=0) = \sum_S (-1)^{|S|-1} P(A_S)$.
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0I corrected "mobius" in the title. – 2011-09-21