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Find the probability that two primes $p_1$ and $p_2$ divide a positive integer $x$

Solution given in my reference is $\dfrac{1}{p_1p_2}$. But I don't seem to find any logical clue to obtain the required probablity from the given question.

Is it the right soltuion in my reference ?

Or is the given question incomplete ?

Attempt in My Reference

Since $p_1$ is prime in a given set of $p_1$ consecutive $+$ve integers, $p_1$ will divide exactly $1$. $$ \text{P($x$ is divisible by )}p_1=\frac{1}{p_1} $$ Similarly for $p_2$. $$ \text{Req. Probability} =\frac{1}{p_1p_2} $$

SOORAJ SOMAN
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    What sort of reference is this that talks about "probabilities" of primes dividing integers without explaining what it means by that? There are ways to make this question precise, and if the reference specified such a way, that would also give you an indication how to prove it. In this form, it's just a vague, ill-defined statement that can't be proved. – joriki Apr 19 '20 at 10:13
  • @joriki This is usually always defined in terms of natural density. – Luke Collins Apr 19 '20 at 10:14
  • @joriki thanx for the comment. My reference is not a credible one. Thats why I am having reservations about the question itself. – SOORAJ SOMAN Apr 19 '20 at 10:16
  • @ss1729 The "probability" in this sense is always defined as a limiting one, i.e., the limit of the probability as $N\to\infty$ when picking from ${1,\dots,N}$. – Luke Collins Apr 19 '20 at 10:19
  • @LukeCollins, joriki I have edited OP to include the attempt to solve this problem given in my reference. – SOORAJ SOMAN Apr 19 '20 at 10:26
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    @ss1729 That is precisely the reasoning I explained in my answer, I don't see why joriki is being so pedantic here. I added the note on natural density which is how you'd want to formalise these things properly, but basically I think it is intuitive that since multiples of 3 occur every 3 positive integers, then the probability that a random number is divisible by 3 is $\frac 13$ (for instance). – Luke Collins Apr 19 '20 at 10:28
  • @LukeCollins: It may seem pedantic to you because you know the underlying concepts well and how to avoid associated pitfalls, but we get people asking about "uniform distributions over the natural numbers" and the like all the time, and I don't think it's a good idea to contribute to this confusion by talking loosely about these things to people who don't have as firm a grip of them as you do. – joriki Apr 19 '20 at 10:38
  • @joriki Fair enough, I assumed it was obvious that speaking about probability when "picking from $\mathbb N$" is a subtle notion to formalise. – Luke Collins Apr 19 '20 at 10:40

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Edit: Usually, these sorts of probabilities are always defined in terms of natural density. So when I say below that "the probability that $x$ is divisible by $p$ is $1/p$", this means that, if we pick a random $x$ in $\{1,\dots,N\}$, and ask about the probability that $x$ is divisible by $p$, and then let $N\to\infty$, we arrive at the intuitive fact that the probability is $1/p$.

Multiples of $p$ occur every $p$ integers, so the probability that $p$ divides $x$ is $1/p$. For distinct primes, the events are independent, so you have probability $$\frac1{p_1}\cdot\frac1{p_2}=\frac1{p_1p_2}.$$

Luke Collins
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  • I don't think it's a good idea to reinforce the idea that talk about about such "probabilities" makes sense without specifying what is meant by it. – joriki Apr 19 '20 at 10:14
  • @joriki I edited my answer. In number theory it's very common to talk about probabilities this way. For instance, the fact that the probability of $x$ being divisible by $p$ is $1/p$ leads to the fact that $\phi(x) = x\prod_{p\mid x}(1-\frac1p)$. Here is Ram Murty giving this example. – Luke Collins Apr 19 '20 at 10:23
  • @joriki If you downvoted my answer, I think it is fair that you revoke your downvote given my edit clarifying what is meant by the probability. – Luke Collins Apr 19 '20 at 10:40
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    You're right, I was going to answer you first, but now I've upvoted first :-) Answer to follow... – joriki Apr 19 '20 at 10:40
  • I'm aware that probabilistic considerations play an important role in number theory. I've written many answers on this site about probabilities of primes dividing numbers (e.g. here). But I precede them with something like "on the random model of the primes" or a definition of natural density (as you've now done) or the like; I don't write exercises for beginners asking about "probabilities" without defining what I mean. Ram Murty in that video is talking about the probability of random number $\le n$ being coprime to $n$. – joriki Apr 19 '20 at 10:45
  • @joriki Fair enough, I did not realise these questions were so common! A bit later he deduces that $\phi(n)/n = \prod_{p\mid n}(1-\frac1p)$ because the probability that any given $n$ is divisible by $p$ is $1/p$. – Luke Collins Apr 19 '20 at 10:46
  • Yes, I watched up to that point, but that's still the continuation of the equation that starts "$\operatorname{Prob}(j\le n\ldots$"; there's an equal sign between that and this equation. I agree he sometimes sounds as if he's talking about general random natural numbers, but it's all in a context where he started with a well-defined probability and could if necessary refer everything to that probability. That's quite different from asking the class, without ever saying $j\le n$, and without introducing natural densities: "What's the probability of a random number being divisible by $p$"? – joriki Apr 19 '20 at 10:51
  • @joriki You're right. – Luke Collins Apr 19 '20 at 10:53
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    Thanks – I don't want to belabour the point, just to emphasize that I'm not at all generally about talking about probablities in number theory: here's an answer I gave a long time ago where I argued in favour of probability talk (and provided a philosophical reference to support it) where others were criticizing it :-) – joriki Apr 19 '20 at 10:55