<s>[INST] Urn with balls - Finding the joint distribution 
Consider an urn containing $4$ red and $6$ white balls. We draw two times with replacement. Let $X$ be the number of red balls we draw, and $Y$ the number of white balls we draw. Find the joint probability distribution of $X$ and $Y$.

Since we draw exactly two times:
$$\begin{align}P(X=0, Y= 0) &= P(X = 0, Y= 1) \\[1ex]&= P(X = 1, Y= 0) \\[1ex]&= P(X = 1, Y= 2) \\[1ex]&= P(X = 2, Y= 1) \\[1ex]&= P(X = 2, Y= 2) \\[1ex]&= 0\end{align}$$
Now we have $3$ cases left:
$P(X = 0, Y = 2) = (\dfrac{6}{10})^2 = 0.36.$
$P(X = 1, Y = 1) = \dfrac{4}{10} \cdot \dfrac{6}{10} = 0.24.$
$P(X = 2, Y = 0) = (\dfrac{4}{10})^2 = 0.16.$

These don't add up to $1$, so there must be something wrong with my solution. Where's my mistake?
 [/INST] Hint: $~~0.36+{0.24}+0.16=1-{0.24}$

It's because $\mathsf P(X{=}1,Y{=}1)= 2\cdot\tfrac{4}{10}\cdot\tfrac{6}{10}=0.48$ by reason that we are selecting 1 red and 1 white ball in any order of extraction.
</s>