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Let $a,b \in \mathbb{N}-\{0\}$. Denote the set of prime numbers by $P$. Here it is mentioned that the probability that $\gcd(a,b)=1$ tends to $\frac{6}{\pi^2}=0.60792 \ldots$

What is the probability that $\gcd(a,b) \in P$ or $\gcd(a,b) \in 2P$?

Thank you very much!

Edit: It was mentioned in the comments that my question before editing (= probability of $\gcd(a,b) \in P$) has already been asked and answered here, and the answer is $0.274933 \ldots$.

I still have several (probably trivial) questions concerning the above probabilities:

(1) Are the above calculations for $a,b \in \mathbb{Z}$? What happens for $a,b \in \mathbb{N}$ and $a,b \in \mathbb{N}-\{0\}$? (If, for example, $a=0$, then $\gcd(a,b)=b$).

(2) If the above calculations are for $a,b \in \mathbb{Z}$, is it true then that the probability that $\gcd(a,b) \in \{1\} \cup P$ is $0.60792 \ldots + 0.274933 \ldots= 0.881 \ldots$? (I think yes).

(3) I have now added to my question "or twice a prime", namely, $\gcd(a,b) \in 2P$. Edit 2: I now see that this was also answered in the above mentioned MSE question (please notice the last two comments there).

user237522
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    This has been answered already here. – Dietrich Burde Oct 16 '18 at 20:28
  • @DietrichBurde But having a prime factor in common isn't the same as having a gcd of 1. – Théophile Oct 16 '18 at 20:31
  • @DietrichBurde, thank you very much! Indeed, it is proved there that the probability that $\gcd(a,b) \in P$ is 0.274933... – user237522 Oct 16 '18 at 20:41
  • $\mathbb Z$ vs $\mathbb N$ doesn't matter, as $\gcd(a,b) = \gcd(|a|,|b|)$. – eyeballfrog Oct 16 '18 at 21:19
  • @eyeballfrog, thank you! What about $\mathbb{N}-{0}$? Are the two probabilities (for $\gcd(a,b)=1$ and for $\gcd(a,b) \in P$) the same as for $\mathbb{N}$? The first probability seems to me the same, since we only exclude $(a,b) \in {(0,1),(1,0), (0,0) }$. I am not sure about the second probability (we exclude $(a,b) \in (0,P) \cup (P,0)$). – user237522 Oct 16 '18 at 21:57
  • @user237522 Including 0 makes no difference in the probabilities; the two boundary lines make up avanishingly small percentage of the plane. (More formally, the probability is generally defined as the limit for $N\to\infty$ of the probability for $a,b\leq N$, and the portion of that set with either $a$ or $b$ equal to $0$ is proportional to $1/N$.) – Steven Stadnicki Oct 16 '18 at 23:26
  • @StevenStadnicki, thank you for your explanation (https://math.stackexchange.com/questions/37520/on-the-probability-that-two-positive-integers-are-relatively-prime?rq=1 is somewhat relevant). – user237522 Oct 16 '18 at 23:35

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