Geometric Probability Question Involving 'at least…'

Anne is running a rubber duck shooting stall at the local fate. She observes that her latest player, Mark, who is a keen shooter, has only 0.15 chance of missing the shot.

Anne wonders what the probability is that Mark will need to take at least 3 shots until he has a shot where he doesn't miss.

My working,

\begin{align} Pr(X \geq 3) &= 1 - Pr(X \leq 2)\\ &=1 - [Pr(X = 1) + Pr(X = 2)]\hspace{1cm} \text{(since the number of shots can only be 1 or 2)} \\&= 1 - (0.15)^0(0.85)^1 -(0.15)^1(0.85)^1 \hspace{1cm} \text{(by geometeric distribution formula)}\\& = 0.0225 \end{align}

Hence, the probability that Mark will need to take at least 3 shots until he has a one where he doesn't miss is $2.25\%$.

This answer feels correct, but I'm not sure about where I omitted $Pr(X = 0)$. Can I do this?

Random variable $X$ is the number of shots needed to get the shot where Mark doesn't miss. The distribution if $X$ is geometric. You noted that
So $Pr(X=0)=0$, as well as $Pr(X=-5)$ or $\Pr(X=-150)$. Do we need to take the last two probabilities into account?
Therefore the required probability is $0.15^2=0.0225$.