# Probability of choosing 2 specific cards

Let's have the typical bohemian package of 32 cards. We randomly select 2 cards from this package and consider a random vector $$(X, Y) ^ T$$ where $$X$$ gives us the number of hearts and $$Y$$ The number of kings. My original goal is to determine the probability distribution of this vector. So I created a $$3×3$$ table with columns $$0, 1, 2$$ (that is, the number of kings drawn) and the same with lines.

Now it's hard for me to fill in the table, since I need to calculate the probabilities of specific cards stolen. I have tried the following:

$$P (X = 0, Y = 0) = frac {21} {32} * frac {20} {31}$$, because I can choose 21 cards of 32 that are neither kings nor hearts and in the second selection it is the same but with one less card.

$$P (X = 0, Y = 1) = 1 – ( frac {7} {32} * frac {21} {31})$$, because I can choose 7 hearts (king of hearts not included) in the first selection and 21 no hearts and no kings in the second selection. But we need a complement for this.

$$P (X = 0, Y = 2) = frac {3} {32} * frac {2} {31}$$, because I can choose 3 not king of hearts and then 2.

$$P (X = 1, Y = 0) = frac {7} {32} * frac {21} {31}$$ , because it is a complement to the second. $$P$$.

$$P (X = 1, Y = 1) = 1 – ( frac {21} {32} * frac {20} {31})$$, because it is a complement to the first. $$P$$.

$$P (X = 2, Y = 0) = frac {7} {32} * frac {6} {32}$$, because in the first selection I can choose between 7 heart cards (king of hearts not included) and the same in the second selection but with one less card.

Other cells in the table are $$0$$.

Now I know that these calculations are not correct, since the sum of the sums of each line must be equal to 1.

Please correct me or show me the correct way. Thank you.