In two papers of Nymann and Leahey from the mid-to-late 70’s, they determined the asymptotics of the probabilities of selecting $k$ relatively prime integers [1] and selecting an integer which is $k$-free [2] according to the uniform and binomial distributions. I thought it would be a fun exercise to combine the results in the obvious way, extending their results in the binomial distribution while also coming up with an alternative proof of Benkoski’s result for the uniform distribution [3].
Definition. Let $r,s$ be positive integers with $rs\geq2$. We say that $s$ integers are relatively $r$-prime when their greatest common divisor has no $r$th power factor greater than $1$.
Fix positive integers $r,s$ with $rs\geq2$, let $N_{n}=\left\{0,\ldots,n\right\}$, and let $P_{n}$ be a probability distribution on $N_{n}$. Let $S_{n}^{r,s}\subseteq N_{n}^{s}$ be the set of all relatively $r$-prime $s$-tuples of integers in $N_{n}$ and let $A_{n}\left(d\right)=\left\{j\in N_{n}:j\equiv0\;\left(\mathrm{mod}\;d\right)\right\}$. We have the following result.
Lemma 1. Let $P_{n}$ be any probability distribution on $N_{n}$. Then for $n>1$, we have $$P_{n}^{s}\left(S_{n}^{r,s}\right)=\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right)\Big( P_{n}\left(A_{n}\left(d^{r}\right)\right)^{s} – P_{n}\left(\left\{0\right\}\right)^{s} \Big)$$
Proof.
Let $p_{1} \lt \cdots \lt p_{\ell}$ be the primes less than or equal to $n^{1/r}$. Then, denoting the complement of $S_{n}^{r,s}$ in $N_{n}$ by $\tilde{S}_{n}^{r,s}$, we may write $$\tilde{S}_{n}^{r,s}=\bigcup_{i=1}^{\ell}A_{n}^{s}\left(p_{i}^{r}\right),$$ where $A_{n}^{s}\left(d\right)$ denotes the Cartesian product of $A_{n}\left(d\right)$ with itself $s$ times.
Therefore, $$P_{n}^{s}\left(S_{n}^{r,s}\right)=1-P_{n}^{s}\left(\tilde{S}_{n}^{r,s}\right)=1-P_{n}^{s}\left(\bigcup_{i=1}^{\ell}A_{n}^{s}\left(p_{i}^{r}\right)\right)$$
$$=1-\sum_{m=1}^{\ell}\sum_{\left(i_{1},\ldots,i_{m}\right)}\left(-1\right)^{m-1}P_{n}^{s}\left( A_{n}^{s}\left(p_{i_{1}}^{r}\right) \cap\cdots\cap A_{n}^{s}\left(p_{i_{m}}^{r}\right) \right),$$ where the inner sum is taken over all $m$-tuples $\left(i_{1},\ldots,i_{m}\right)$ such that $1\leq i_{1}\lt\cdots\lt i_{m}\leq\ell$.
For $\left(d_{1},d_{2}\right)=1$, we have that $A_{n}^{s}\left(d_{1}\right)\cap A_{n}^{s}\left(d_{2}\right)=A_{n}^{s}\left(d_{1}d_{2}\right)$, we so we may rewrite this as $$1+\sum_{m=1}^{\ell}\sum_{\left(i_{1},\ldots,i_{m}\right)}\left(-1\right)^{m}P_{n}^{s}\left( A_{n}^{s}\left(\left(p_{i_{1}}\cdots p_{i_{m}}\right)^{r}\right) \right).$$
Now, if $\left(p_{i_{1}}\cdots p_{i_{m}}\right)^{r}>n$, then $ A_{n}^{s}\left(\left(p_{i_{1}}\cdots p_{i_{m}}\right)^{r}\right)=\left\{0\right\}^{s}$, and so we may further write this expression as $$\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right)P_{n}^{s}\left( A_{n}^{s}\left(d^{r}\right) \right)+\sum_{m=1}^{\ell}\;\sum_{p_{i_{1}} \cdots p_{i_{m}}>n^{1/r}} \mu\left(p_{i_{1}}\cdots p_{i_{m}}\right)P_{n}^{s}\left( \left\{0\right\}^{s} \right).$$
Finally, since $$\sum_{d\mid k}\mu\left(d\right)=0$$ for all $k>1$, we have $$\sum_{m=1}^{\ell}\;\sum_{p_{i_{1}} \cdots p_{i_{m}}>n^{1/r}}\mu\left(p_{i_{1}}\cdots p_{i_{m}}\right)=-\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right),$$ and this finishes the proof.
QED
Corollary 2. If $P_{n}$ is the uniform distribution, then $$\lim_{n\to\infty} P_{n}^{s}\left( S_{n}^{r,s} \right) = 1/\zeta\left(rs\right)$$ for all positive integers $r,s$ with $rs\geq2$.
Proof.
Define $\varepsilon_{n}\left(d\right)$ by the equation $$P_{n}\left(A_{n}\left(d\right)\right) = d^{-1} + \varepsilon_{n}\left(d\right).$$
Claim. $0\leq\varepsilon_{n}\left(d\right)\lt n^{-1}$ for all positive integers $n,d$ with $n>1$.
To see this, note that $$d^{-1}=\frac{\frac{n+1}{d}}{n+1}\leq\frac{\left\lfloor n/d \right\rfloor + 1}{n+1}\leq\frac{\frac{n+1}{d}+1}{n+1}=d^{-1}+\frac{1}{n+1},$$ and $$P_{n}\left(A_{n}\left(d\right)\right) = \frac{\left\lfloor n/d \right\rfloor + 1}{n+1}.$$ Therefore, $0\leq\varepsilon_{n}\left(d\right)\leq\left(n+1\right)^{-1}\lt n^{-1}.$
By Lemma 1, we have $$P_{n}^{s}\left(S_{n}^{r,s}\right)=\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right)\Big( \left(d^{-r}+\varepsilon_{n}\left(d^{r}\right)\right)^{s}-\left(n+1\right)^{-s} \Big)$$
$$=\sum_{1\leq d\leq n^{1/r}}d^{-rs}\mu\left(d\right)+\sum_{j=1}^{s}{s \choose j}\sum_{1\leq d\leq n^{1/r}}d^{r\left(j-s\right)}\mu\left(d\right)\left(\varepsilon_{n}\left(d^{r}\right)\right)^{j}$$
$$-\left(n+1\right)^{-s}\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right).$$
Now, for positive integers $r,s$ with $rs\geq2$, the first sum is $$\sum_{d=1}^{\infty}d^{-rs}\mu\left(d\right)=1/\zeta\left(rs\right),$$ the terms $$\sum_{1\leq d\leq n^{1/r}}d^{r\left(j-s\right)}\mu\left(d\right)\left(\varepsilon_{n}\left(d^{r}\right)\right)^{j}\lt n^{-j}\sum_{1\leq d\leq n^{1/r}}d^{r\left(j-s\right)}\mu\left(d\right)$$ each go to $0$ by the claim earlier in this proof, and, finally, since $$\Big| \sum_{1\leq d\leq n^{1/r}}\mu\left(d\right)\Big|\leq n^{1/r},$$ we have that the final term $$\left(n+1\right)^{-s}\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right)$$ goes to $0$, and the desired result is proved.
QED
This result is also proved in [3] using totient functions.
Now we turn our attentions to the probability distribution in the title, the binomial distribution. Fix $0\lt p\lt 1$ and define $$\varepsilon_{n}\left(d\right)=P_{n}\left(A_{n}\left(d\right)\right)-d^{-1}=\sum_{k\equiv0\;\left(d\right)}{n \choose k}p^{k}\left(1-p\right)^{n-k} – d^{-1}$$ for integers $n,d>0$ with $d\leq n$ as in the proof of Corollary 2.
Lemma 3. $|\varepsilon_{n}\left(d\right)|\ll n^{-1/2}$ uniformly in $d$.
The proof can be found in [1].
Theorem 4. If $P_{n}$ is a binomial distribution, then $$\lim_{n\to\infty}P_{n}^{s}\left(S_{n}^{r,s}\right)=1/\zeta\left(rs\right)$$ for all positive integers $r,s$ with $rs\geq2$.
Before proving this result, note that the case $rs=2$ has already been shown in the papers of Nymann and Leahey. In this case, we have either $r=1$ and $s=2$ [1] or $r=2$ and $s=1$ [2]. Therefore, it is enough to prove the case $rs\geq3$.
Proof.
Let $r,s$ be fixed positive integers such that $rs\geq3$.
By Lemma 1, we have $$P_{n}^{s}\left(S_{n}^{r,s}\right)=\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right)\Big( P_{n}\left(A_{n}\left(d^{r}\right)\right)^{s} – P_{n}\left( \left\{0\right\} \right)^{s} \Big)$$
$$=\sum_{1\leq d\leq n^{1/r}}d^{-rs}\mu\left(d\right)+\sum_{j=1}^{s}{s \choose j}\sum_{1\leq d\leq n^{1/r}}d^{r\left(j-s\right)}\mu\left(d\right)\left(\varepsilon_{n}\left(d^{r}\right)\right)^{j}$$
$$-\left(1-p\right)^{ns}\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right).$$ As in the proof of Corollary 2, the first term approaches $1/\zeta\left(rs\right)$, and Lemma 3 shows that the terms in the second sum are $$\sum_{1\leq d\leq n^{1/r}}d^{r\left(j-s\right)}\mu\left(d\right)\left(\varepsilon_{n}\left(d^{r}\right)\right)^{j}\ll n^{-j/2}\sum_{1\leq d\leq n^{1/r}}d^{r\left(j-s\right)}\mu\left(d\right),$$ which each go to $0$ for $rs\geq3$. Finally, $$\left(1-p\right)^{ns}\sum_{1\leq d\leq n^{1/r}}\mu\left(d\right)$$ goes to $0$, and the result is proved.
QED
References
[1] J.E. Nymann and W.J. Leahey, On the probability that integers chosen according to the binomial distribution are relatively prime, Acta Arithmetica 31 (1976), 205-211.
[2] J.E. Nymann and W.J. Leahey, On the probability that an integer chosen according to the binomial distribution be $k$-free, Rocky Mountain J. Math. 7 (1977) no. 4, 769-774.
[3] Stanley J. Benkoski, The Probability that $k$ Positive Integers are Relatively $r$-Prime, J. Number Theory 8 (1976), 218-223.