Probability that new baby born is a boy in a nursery

Before the new baby is born, there are 2 boys (Andy and Bob). The new baby (NB) is born, then we pick a baby at random and get a boy. In order for this to happen, one of the following scenarios occurred:

1. NB is a boy and we picked Andy

2. NB is a boy and we picked Bob

3. NB is a boy and we picked NB

4. NB is a girl and we picked Andy

5. NB is a girl and we picked Bob

In three of these five cases, NB is a boy, so the probability is $\frac{3}{5}$

The number of girls at the start is irrelevant, because changing that number doesn't change the relative probability of any of scenarios $1-5$ occurring

Suppose there are initially $g$ girls in the room. Let $A$ be the event that the new baby is a boy, $B$ the event that the baby picked is a boy. I presume we are supposed to assume $\mathbb P(A) = 1/2$ (although in real life that is not quite true). Then $\mathbb P(B\mid A) = 3/(g+3)$ and $\mathbb P(B\mid A^c) = 2/(g+3)$, so $$\mathbb P(A\mid B) = \frac{\mathbb P(B\mid A) \mathbb P(A)}{\mathbb P(B\mid A) \mathbb P(A) + \mathbb P(B\mid A^c) \mathbb P(A^c)} = \frac{3}{5}$$ The point is that $g$ has the same effect on both $\mathbb P(B\mid A)$ and $\mathbb P(B\mid A^c)$, a multiplicative factor of $1/(g+3)$ in both cases, which cancels out in $\mathbb P(A\mid B)$.