Five cards are randomly chosen from a standard deck, one at a time with replacement. Let $X$, $Y$, $Z$ be the numbers of chosen queens, kings, and other cards.
(a) Find the joint PMF of X, Y, Z.
(b) Find the joint PMF of X and Y .
a) The joint PMF of $X$, $Y$ and $Z$ $$ P(X=x,Y=y,Z=z) = \left(\frac{1}{13}\right)^{x+y} \left(\frac{11}{13}\right)^z, $$ where the support is $\{(x,y,z) \in \mathbb{Z} \mid x+y+z = 5 \text{ and } x,y,z \geq 0 \}$.
b) To find the joint PMF of $X$ and $Y$ alone, I marginalize out $Z$.
$$ P(X=x,Y=y) = \left(\frac{1}{13}\right)^{x+y} \sum_{z=0}^{5-x-y} \left(\frac{11}{13}\right)^z = \left(\frac{1}{13}\right)^{x+y} \frac{1-\left(\frac{11}{13}\right)^{6-x-y}}{1-\frac{11}{13}} $$
Is this correct?
You are dealing here with multinomial distribution and 'forgot' the coefficients.
Under $x+y+z=5$ and $x,y,z\in\{0,1,\dots\}$:
$$P\left(X=x,Y=y,Z=z\right)=\frac{5!}{x!y!z!}\left(\frac{1}{13}\right)^{x}\left(\frac{1}{13}\right)^{y}\left(\frac{11}{13}\right)^{z}=\frac{5!}{x!y!z!}\frac{11^{z}}{13^{5}}$$
Under $x+y\leq5$ and $x,y\in\{0,1,\dots\}$:
$$P\left(X=x,Y=y\right)=P\left(X=x,Y=y,Z=5-x-y\right)=\frac{5!}{x!y!\left(5-x-y\right)!}\frac{11^{5-x-y}}{13^{5}}$$