What is the probability for which :$\gcd ((\phi(n),n-1)=1 ) $ for integers?

Question

let $n$ be an integer and $\phi$ is The Euler's Totient function.

I want to know the probability for which $$\gcd ((\phi(n),n-1)=1 ) $$

I only know if $n$ is a prime number then the probability is 0 because we have $\phi(p)=p-1$ then $$\gcd(\phi(p),p-1) =p-1=\phi(p)$$ , then my question is :What is the probability for which :$$\gcd ((\phi(n),n-1)=1 ) $$ for integers ?

Answer 1

This is an unsolved problem. For details see https://en.wikipedia.org/wiki/Lehmer%27s_totient_problem