# Probability of a prime factor

We are given an arbitrary number $n$ and a sequence of primes $p_1=2$, $p_2=3$, ..., $p_k$. I am interested in the following question: Are the events "Prime $p_i$ is a factor of $n$" independent for distinct primes $p_i$?

It seems that the answer is no but I think this depends on the size of $n$. If $n$ is slightly greater than $p_k$ or lower than $p_k$, the answer is no. But what about the case where $n \geq \prod_{i=1}^k p_i$?

This is an intuition but I am not able to give a argument.

Thank you

• Isn't probability of $p_i|n$ for given $p_i$ and $n$ equal 0 or 1? – Jaroslaw Matlak Sep 22 '17 at 7:41
• If $n$ follows a uniform random distribution on any set of $N$ consecutive integers, where $N$ is a multiple of $\prod p_i$, then the answer is yes. – Hagen von Eitzen Sep 22 '17 at 7:46
• @HagenvonEitzen No. The event $p_i | n$ and $p_j | n$ are not independent. That's why prime numbers are complicated. – reuns Sep 22 '17 at 8:55
• @Adam54 When you say "an arbitrary number $n$" you meant $n$ an arbitrary integer $\le N$. The events are not independent : if $p_i | n$ and $p_i > N/2$ then $p_j \nmid n$. This is enough to prove the dependence in every other cases. Now you can create a random integer variables whose event $X_{i} = p_i| n$ are independent, simply choose $k$ primes and set $n = \prod_{i=1}^k p_i^{E_{i}}$ where each $E_i$ are independent random variables. – reuns Sep 22 '17 at 9:02
• Thanks for your comments. What about the case where the number $n$ is chosen in the interval $\left[\prod_{i=1}^k p_i , (\prod_{i=1}^k p_i)^2 \right]$? Are there results which stipulate independence under certain conditions (for instance, for $n$ an arbitrary large number that tends to $\infty$)? – Adam54 Sep 22 '17 at 10:30

The "probability" that a natural number N is divided by any prime p is $\frac{1}{p}$ as long as $p≤\sqrt{N}$. These "probabilities" are independent of each other and thus the "probabilty" that a natural number N is divided through the primes p1 and p2 is $$\frac{1}{p1}*\frac{1}{p2}$$
• Are you sure? For infinitely large $N$, maybe it's true, but when a small number can be divided by $p1$, I think that the probability that it can be divided by $p2$ decreases. – Adam54 Sep 29 '17 at 9:39