1. Prime factorization of $2022$

We first factorize: $2022 = 2 \times 1011 = 2 \times 3 \times 337$ Since $2,3,337$ are distinct primes, $2022 = 2\cdot 3\cdot 337$

2. Interpretation of the question

We choose a number $n$ uniformly at random from $S=\{1,2,3,\dots,2022\}$ and want the probability that $\gcd(n,2022)=1.$

This is exactly the count of integers between $1$ and $2022$ that are coprime to $2022$, divided by $2022$.

So required probability is $\frac{\varphi(2022)}{2022}$ where $\varphi$ is Euler's totient function.

3. Compute $\varphi(2022)$

Using $\varphi(m)=m\prod_{p\mid m}\left(1-\frac1p\right)$ we get $\varphi(2022)=2022\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac1{337}\right).$

Now simplify step by step: $2022\left(\frac12\right)\left(\frac23\right)\left(\frac{336}{337}\right)$

Since $2022=2\cdot 3\cdot 337,$ we have $2022\cdot \frac12 = 1011,$ $1011\cdot \frac23 = 674,$ $674\cdot \frac{336}{337} = 2\cdot 336 = 672.$ So, $\varphi(2022)=672.$

4. Required probability

Hence, $P=\frac{672}{2022}.$ Reduce the fraction: $\frac{672}{2022}=\frac{336}{1011}=\frac{112}{337}.$

So the probability is $\boxed{\frac{112}{337}}.$

5. Check options

• A: $\frac{128}{1011}$ ✗
• B: $\frac{166}{1011}$ ✗
• C: $\frac{127}{337}$ ✗
• D: $\frac{112}{337}$ ✓

Therefore, the correct option is D.