Question about primes in square-free numbers For any prime, what percentage of the square-free numbers has that prime as a prime factor?

Let $A(n)=\{\mathrm{squarefree~numbers~\le n}\}$ and $B_p(n)=\{x\in A(n); p\mid x\}$.
Then the asymptotic density of $B_p$ in $A$ is $b_p = \lim_{n\rightarrow \infty} |B_p(n)|/|A(n)|$. (It seems from the comments that this is not what @RudyToody is looking for, but I thought it's worth writing up anyway.) Let the density of $A$ in $\mathbb{N}$ be $a = \lim_{n\rightarrow \infty} |A|/n$.
Observe $B_p(pn) = \{px; x\in A(n),p\nmid x\}$, so for $N$ large $b_p$ must satisfy
$$
\begin{align}
b_p a (pN) & \simeq (1-b_p)aN \\
b_p &= \frac{1}{p+1}
\end{align}
$$
as @joriki already noted.
To illustrate, here are some counts for squarefree numbers $<10^7$. $|A(10^7)|=6079291$.
$$
\begin{array}{c|c|c|c}
p & |B_p(10^7)| & |A(10^7)|/(p+1) & \Delta=(p+1)|B|/|A|-1 \\
\hline \\
2 & 2026416 & 2026430.3 & -7\times 10^{-6} \\
3 & 1519813 & 1519822.8 & -6\times 10^{-6} \\
5 & 1013234 & 1013215.2 & 1.9\times 10^{-5} \\
7 & 759911 & 759911.4 & -5\times 10^{-7} \\
71 & 84438 & 84434.6 & 4\times 10^{-5} \\
173 & 34938 & 34938.5 & -1.3 \times 10^{-5}
\end{array}
$$
