Let $X \sim \text{HGeom}(w,b,n)$, what is the distribution of $n-X$? 
Let $X \sim \text{HGeom}(w,b,n)$, what is the distribution of $n-X$?

The distribution of $X$ (e.g., number of white ($w$) balls in a sample of size $n$) is hypergeometric, so 
$$P(X=x) = \frac{\binom{w}{x}\binom{b}{n-x}}{\binom{w+b}{n}}.$$
If $X$ is the number of white balls, then $n-X$ must be the number of black balls.
Let $Y=n-X$, then $P(Y=y)=P(n-X=y)=P(X=n-y)$, so plugging in, I get
$$P(Y=y) = \frac{\binom{w}{n-y}\binom{b}{y}}{\binom{w+b}{n}}$$


*

*Is this a correct proof (not only if the results is correct)?

*What kind of manipulations are actually allowed within the $P(\dots)$? I'm always tempted to do some stuff like $P(Y=y)=P(n-X=n-x)=P(X=x)$, which is obviously meaningless, but I can't figure out what is formally wrong with it. Why is it allowed to plug in $n-X$ for $Y$, but not $n-x$ for $y$?

 A: Your proof is correct, and your equivalence holds true.  Observe:
$\begin{align}
\mathsf P(Y=y) 
& = \frac{\binom{b}{y}\binom{w}{n-y}}{\binom{w+b}{n}} 
& \text{Via the hypergeometric distribution of $Y$ black balls among $n$ drawn}
\\[1ex] 
& = \frac{\binom{b}{n-x}\binom{w}{x}}{\binom{w+b}{n}} & \text{Defining $x=n-y$}
\\[1ex]
 & = \mathsf P(X=x)
 & \text{Via the hypergeometric distribution of $X$ white balls among $n$ drawn}
\end{align}$
The reason you don't substitute $x=n-y$ when asked to find the distribution of $Y$ is just that it is more sensible to express it as a function of the argument of that random variable ($y$), as requested.
Ie: When given $\mathsf P(X=x)=f(x)$ and $Y=h(X)$ then asked to find some function $g$ such that $\mathsf P(Y=y)=g(y)$, you don't look for $g(h(x))$ except, maybe, as a step along the way to the final expression: $g(y)=f(h^{-1}(y))$.
So: $\mathsf P(Y=y) = \mathsf P(X=n-y)= \frac{\binom{b}{y}\binom{w}{n-y}}{\binom{w+b}{n}}$
A: Hypergeometric distribution gives us the probability that there are exactly $k$ white balls in the $n$ balls that have been drawn from an urn containing $w$ white balls and $b$ black balls.
Given that there are exactly $k$ white balls out of the total $n$ balls, this means the rest of the $n-k$ balls are black balls. Thus $n-X$ is nothing but the distribution of black balls when $n$ balls are drawn from an urn containing $w$ white and $b$ black balls. This is nothing but the distribution of white balls when $n$ balls are drawn from an urn containing $w$ black and $b$ white balls. This is nothing but HGeom($b$, $w$, $n$). This interchanging of $w$ and $b$ is what you get in your final answer as well.
