Deck of 60 cards.
$12$ are black $b_1, \ldots, b_{12}$
$10$ are green $g_1, \ldots g_{10}$
$8$ are red $r_1, \ldots, r_8$
I draw $13$ cards and want calculate the probability to have: at least $3$ black and at least $2$ green and max $2$ red.
Can you help me calculate this somehow through the use of the hypergeometric distribution? I can calculate it without it but it will literally take me forever. If it was exact $3,2,2$ it would be very easy since I would do just this $$\Pr(X=3,Y=2,Z=2,Q=6) = \frac{\dbinom{12}{3}\dbinom{10}{2}\dbinom{8}{2} \dbinom{30}{6}}{\dbinom{60}{13}}$$
But what I want is $\Pr(X \geq 3,Y \geq 2,Z \geq 2,Q=X-3+Y-2+2-Z)$. Can I break it somehow in a product? $Q$ seems to depend on the amount of cards picked by $X,Y,Z$, so I don't think is correct to say $\Pr(X \geq 3) \Pr(Y \geq 2)\Pr(Z\leq 2)\Pr(Q=X-3+Y-2+2-Z)$
Thank you