A hand of six cards is dealt from a standard poker deck. Let X denote the number of aces, Y the number of kings, and Z the number of queens.
a) write a formula for p_(XYZ) (x,y,z).
b) Find p_(XY)(x,y) .
Solution for a) : A hand of six cards can be selected in 52_C_6 ways.
Since there are 52 cards, there are 4 aces, 4 kings, and 4 queens.
Let x represent aces, y kings and z queens.
Then X,Y,Z form a hypergeometric distribution. Then x + y + z = 6. then let x denote the number of aces cards selected, since there are 4 , 0 =< x =< 4. Similarly for y and z.
Then p_(XYZ) (x,y,z) = [ 4_C_x * 4_C_y * 4_ C_z]/(52_C_6).
solution for b):
p_(XY)(x,y) = [ 4_C_x * 4_C_y]/(52_C_6)
Is this right? Can someone please help me? I am not sure. Any feedback would be appreciated.
Best Answer
The numerator should be $$\binom{4}{x}\binom{4}{y}\binom{4}{z}\binom{40}{6-x-y-z},$$ for any $x,y,z$ with $x+y+z\le 6$, with the obvious restrictions on $x,y,z$.