1. Let the random variable $N$ denote the throw on which the first $2$ appears.

2. Since the die is fair:

   • Probability of getting a $2$ on any throw is $p = \frac{1}{6}$
   • Probability of not getting a $2$ is $q = \frac{5}{6}$

3. We need the probability that the first $2$ appears on an even-numbered throw: $P(N \text{ is even}) = P(N=2)+P(N=4)+P(N=6)+\cdots$

4. Using geometric distribution, $P(N=n)=q^{n-1}p$ So for even $n=2,4,6,\dots$, $P(N \text{ is even})=qp+q^3p+q^5p+\cdots$

5. Factor this as a geometric series: $P(N \text{ is even})=qp\left(1+q^2+q^4+\cdots\right)$ Since $1+q^2+q^4+\cdots = \frac{1}{1-q^2}$ we get $P(N \text{ is even})=\frac{qp}{1-q^2}$

6. Substitute $p=\frac{1}{6}$ and $q=\frac{5}{6}$: $P(N \text{ is even})=\frac{\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)}{1-\left(\frac{5}{6}\right)^2}$

7. Simplify: $\frac{\frac{5}{36}}{1-\frac{25}{36}}=\frac{\frac{5}{36}}{\frac{11}{36}}=\frac{5}{11}$

8. Therefore, the required probability is $\boxed{\frac{5}{11}}$

9. Comparing with the options:

   • A: $\frac{5}{11}$ ✅
   • B: $\frac{5}{6}$
   • C: $\frac{1}{6}$
   • D: $\frac{6}{11}$

Hence, the correct option is A.