# Why is the probability that a prime p is a factor of a number n equal to 1/p

I'm learning some number theory and I can't seem to understand why this is the case.

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However, what we can say is that if you take a large interval of natural numbers, and take a uniform distribution on them, the probability that a given number is a multiple of $m$ is roughly $1/m$. To see this, simply observe that it is equivalent to it being $0\mod m$, and there are $m$ things that you can be mod $m$, so any distribution in which they will be equally likely will give $1/m$ as the probability of any one.
The probability that $p$ is a factor of $n$, $1\leq n \leq kp$, is $$P_k = \frac {\lvert \{p, 2p, \dotsc, kp \} \rvert} {\lvert \{1, 2, \dotsc, kp \} \rvert} = \frac k {kp} = \frac 1 p$$ So the probability that $p$ is a factor of $n\in \mathbb N$ is $$\lim_k P_k = \frac 1 p$$