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I was studying probability theory using the textbook Introduction to Probability (2e) - Blitzstein & Hwang. I came across an example problem from a chapter regarding join distributions and had a question regarding a step in the solution. Here's the specific example question that I'm referring to:

Suppoe a chicken lays a random number of eggs, $N$ where $N$ follows Pois($lambda$). Each egg independently hatches with probability $p$. Let $X$ be the number of eggs that hatch and $Y$ be the number of eggs that fail to hatch. What is the joint PMF of $X$ and $Y$?

Textbook Solution (paraphrased):

Because the number of eggs depends on $N$, the total number of eggs laid, we must condition on $N$.

$$P(X=i, Y=j) = sum_{n = 0}^infty P(X=i, Y=j | N=n)P(N=n)$$

Among the possible values of $n$, the only one that is relevant is when $n=i + j$ due to the fact that any other case would not be consistent with the number of eggs hatched and the number of eggs that failed to hatch, since $X+Y$ must equal $N$. Therefore, we can drop all of the other terms in the summation, which leaves us with,

$$P(X=i, Y=j) = P(X=i, Y=j | N=i + j)P(N=i+j)$$

Conditional on $N=i+j$, the events $X=i$ and $Y=j$ are exactly the same event. We can therefore drop either $X$ or $Y$ and find the joint PMF with,

$$P(X=i, Y=j) = P(X=i, | N=i+j)P(N=i+j)$$

I omitted the remaining calculations as they are irrelevant to my question. The sentence that I put in bold print is the particular sentence that is confusing me.

How are $X$ and $Y$ - the number of eggs that hatch and don't hatch, respectively - the same events?

My initial understanding would be that due to the strong dependent relationship they have (i.e. knowing the value of one random variable automatically gives us the value of the other), we are able to reach this conclusion.

Is my assumption correct? Perhaps the way that the sentence was worded is confusing me.

Any feedback is appreciated. Thank you.

The given statement means that $P(X=i)=P(Y=j)$ provided that $i+j=N$ is a constant. Now from the description of $X$ and $Y$, $Xsimoperatorname{binom}(N,p)$ and $Ysimoperatorname{binom}(N,1-p)$, so
$$P(X=i)=binom{i+j}ip^i(1-p)^j$$
and
$$P(Y=j)=binom{i+j}j(1-p)^jp^i$$
It is well-known that $binom{i+j}i=binom{i+j}j$. The equality follows.