Question

Statistics Question:

Three cards are drawn without replacement from the 12 face cards (J,Q,K) of the ordinary deck of 52 playing cards. Let X be the number of Kings selected and let Y be the number of Jacks. Find the joint probability distribution of X and Y.

Answer 1

I have found that for the situation f(1,2) I get 6/55 by using:
\[\frac{\left(\begin{matrix}4 \\ 1\end{matrix}\right)\left(\begin{matrix}4 \\ 2\end{matrix}\right)}{\left(\begin{matrix}12 \\ 3\end{matrix}\right)}\]
but this does not work for the case of f(0,0) which should be 1/55, or for some of the other cases. This is not graded homework, and I have a solution manual, but it is not making any sense to me. Can you help? I have a quiz on this tomorrow and am SO confused now.

Answer 2

I know my cases are f(0,0), f(0,1), f(0,2), f(0,3), f(1,0), f(1,1), f(1,2), f(1,3), f(2,0), f(2,1), f(2,2), f(2,3), f(3,0), f(3,1), f(3,2), f(3,3) and that \[0 \le x+y \le 3\] since there are 3 cards being drawn we cannot have more than 3 jacks or 3 kings or combination of 3 of the two, or 0 jacks and 0 kings.

Answer 3

\[P(X=x,Y=y)=\left\{\begin{matrix}\frac{{4\choose x}{4\choose y}{4\choose 3-x-y}}{{12\choose 3}}& 0\le x+y\le3\\0& \text{o.w.}\end{matrix}\right.\]

Answer 4

Okay, so what is with the 3rd piece. Prof mentioned it, but said we will go into more detail in 2 weeks on it. Does it always use these 3 pieces? I guess I am asking is this formula generic for say if I had P(x,y,z)? Thanks for the quick response.

Answer 5

first is for kings, second is for jacks, and the third part is for queens

Answer 6

for example if you pick 0 kings and 0 jacks then you must have picked 3 queens

Answer 7

I see. :) That helps a lot! Thanks so very much!