TITLE: What is the ratio of Carmichael pseduo-primes to true primes for $1$ to $n$? Or is it known?
QUESTION [7 upvotes]: Let $\pi(n)$ be the prime counting function. And let $\varphi(n)$ be the count of Carmichael pseudo-primes for $1$ to $n$. Is the ratio,
$$\frac{\varphi(n)}{\pi(n)}$$
is known, as $n \to \infty$? I am asking this because of i want to know how much safe is the Fermat Primality Test (the one which only uses Fermat's Little Theorem).
I recently wrote a Python script (initially for RSA) and produced more than 10, random 1024 bit primes with the help of this test. After i used the powerful mathematical software PARI/GP to check if they are really primes or pseudo-primes. In conclusion none of them were pseudo-primes. So, if know this rate, i will know the probability that this test will encounter a pseudo-prime and fail.

REPLY [2 votes]: We have the following result (Knödel, $1953$) :-
$$\varphi(n)<n\exp\big(-k(\log n \log\log n)^\frac{1}{2}\big)$$
By strengthened version of PNT, we will have for sufficiently large $n$:
$$\pi(n)>\frac{n}{\log n}$$
Thus $\forall$ sufficiently large $n$, we have:-
$$\frac{\varphi(n)}{\pi(n)}<\log n \exp\big(-k(\log n \log\log n)^\frac{1}{2}\big)=\exp \big(\log \log n -k(\log n \log\log n)^\frac{1}{2}\big)$$
Note that:- 
$$\lim_{n \to \infty} \log \log n -k(\log n \log\log n)^\frac{1}{2}=- \infty \implies \lim_{n \to \infty} \exp \big(\log \log n -k(\log n \log\log n)^\frac{1}{2}\big)=0$$
Hence, we have:-
$$\lim_{n \to \infty}\frac{\varphi(n)}{\pi(n)}=0$$