Two players shooting a target alternately

Two players A and B shoot a target alternately, until somebody hits it and wins the game. Player A takes the first shot. Each time they take a shot, their probabilities of hitting the target are $$p$$ and $$q$$ respectively. Which is the probability $$\mathbb{P}$$ that A wins?

Is the following solution correct ?

A wins if :

he wins with the $$1$$st shot or

they both fail to hit it during the first $$2$$ shots, and A hits it with the $$3$$rd shot or

they both fail to hit it during the first $$4$$ shots, and A hits it with the $$5$$rd shot or

...

So,

$$\mathbb{P} = p + (1-p)(1-q)p + (1-p)^2(1-q)^{2}p + ... = p\sum_{n=0}^{\infty}[(1-p)(1-q)]^n = \cfrac{p}{p+q-pq}$$

Alternatively, you can condition on the first outcome. If $$A$$ doesn't win in the first outcome, the first two trials must be failure and then it is a renewal process.
$$\mathbb{P}=p + (1-p)(1-q)\mathbb{P}$$ $$(1-(1-p)(1-q))\mathbb{P}=p$$ $$\mathbb{P}=\frac{p}{1-(1-p)(1-q)}$$