Conditional Probability of Y given X (both random variables) The exercise:

A box contains 5 white balls and 5 black balls. 3 balls are drawn randomly, one by one, without replacement. The following random variables are defined:
$$ \begin{align}
&\begin{aligned}
X =
 \begin{cases}
      0, & \text{if the first ball extracted is white} \\
      1, & \text{if the first ball extracted is black}
    \end{cases}
\end{aligned}
\\
&\begin{aligned}
Y = \text{Number of black balls drawn}
\end{aligned}
\end{align}$$
Find the conditional probability of $Y$ given $X$.

$\textbf{My thoughts}$ about it:
So first of all, I check the possible values both random variables can take. For $X$ that is clear, and for $Y$ its possible values are on the set $\{0,1,2,3\}$.
Now I want to use the fact that: $\text{P}(Y=x|X=x)=\frac{\text{P}(X=x,Y=y)}{\text{P}(X=x)}$, for $x=0,1$ and $y=0,1,2,3$ to calculate what I'm being asked.
So, $\text{P}(X=x)=5/10=.5$ for either $x=1$ or $x=0$. After doing this I want to figure out $\text{P}(X=x,Y=y)$ for each possible value of both $x,y$. I think my sample space looks something like this
$$\Omega = \text { \{ (w,w,w),(w,w,b),(w,b,w),(b,w,w),(b,b,b),(b,b,w),(b,w,b),(w,b,b) \} }$$ where b stands for black marble, and w for white marble and their positions tells me order in which they were extracted.
$\textbf{My issue:}$ Is how to calculate any one of those probabilities, for example $\text{P(w,w,w)}$, since trying to calculate $ \begin{align}
&\begin{aligned} 
\text{P}(Y=0,X=1)=\text{P}(\text{0 black marbles extracted and 1st extracted marble is white}) = \text{P}(\text{Extracting 3 white marbles}) = \text{P(w,w,w)}.
\end{aligned}
\end{align}
$
Since there is no replacement there is no independence and I cannot make $\text{P(w,w,w)}=\text{P(w)P(w)P(w)}$ and I doubt the answer is $\frac{1}{8}$ since if I change the initial conditions to one million white marbles essentially my model would remain the same but clearly the probability of drawing 3 white marbles would increasy drastically.
Any insights or hints are much appreciated!
Thanks for reading!
 A: Conditioned on $X=0$, there are four white balls remaining and five black balls remaining. The probability of drawing zero black balls is simply that of drawing two consecutive white balls: $$4/9\cdot3/8=1/6.$$ The probability of drawing one black ball is $$
5/9\cdot4/8 + 4/9\cdot5/8 = 5/9.
$$ The probability of drawing two black balls is that of drawing two consecutive black balls: $$
5/9\cdot4/8 = 5/18.
$$ 
Conditioned on $X=1$, there are five white balls remaining and four black balls remaining. The probability of drawing zero black balls is zero. The probability of drawing one black ball is is that of drawing two consecutive white balls: $$5/9\cdot4/8=5/18.$$ The probability of drawing two black balls is simply $$4/9\cdot5/8 + 5/9\cdot4/8 = 5/9. $$ The probability of drawing three black balls is that of drawing two consecutive black balls:
$$
4/9\cdot3/8 = 1/6
$$
So, the conditional probability of $Y$ given $X$ is
\begin{align}
\mathbb P(Y=0\mid X=0) &= 1/6\\
\mathbb P(Y=1\mid X=0) &= 5/9\\
\mathbb P(Y=2\mid X=0) &= 5/18\\
\mathbb P(Y=0\mid X=1) &= 0\\
\mathbb P(Y=1\mid X=1) &= 5/18\\
\mathbb P(Y=2\mid X=1) &= 5/9\\
\mathbb P(Y=3\mid X=1) &= 1/6.
\end{align}
A: Let $Z$ be the count of black balls among the last two balls drawn.
Since $X$ is the count of black balls among the first ball draw, then $Y=X+Z$.
$$\mathsf P(Y{=}y\mid X{=}x)=\mathsf P(Z{=}y-x\mid X{=}x)$$
In short: For any $x\in\{0,1\}$ and $y\in\{x,x+1,x+2\}$, we have $\mathsf P(Y{=}y\mid X{=}x)$ as the probability for drawing $y-x$ from $5-x$ black balls, and $2-y+x$ from $4+x$ white balls, when drawing $2$ from all $9$ balls. 
