Joint PMF for a urn ball
There are $5$ Red , $3$ Green, $2$ Blue Balls in an urn. We take 4 balls at random one after another.
Let $X$=#reds, $Y$ = #greens, calculate marginal distr. of $X,Y$ and their joint PMF.
So I listed 4 tuples (#reds,#greens) which are of interest :
$(4, 0) = (5*4*3*2)/(10*9*8*7) = 120/5040$
$(3, 1) = (4*(5*4*3*3))/5040) = 720/5040$ (all #ways to get 3 reds and 1 green)
$(2, 2) = (6*(5*4*3*2)/5040) = 720/5040$
$(1,3) =(4*(5*3*2*1)/5040) = 120/5040 $
We know that $\sum_x\sum_yP_{x,y}(x,y)=1$ but I clearly don't get $1$, if there weren't any blue balls the method above would've worked and I'm stuck now.
Let $X$=#reds, $Y$ = #greens, calculate marginal distr. of $X,Y$ and their joint PMF.
So I listed 4 tuples (#reds,#greens) which are of interest :
$(4, 0) = (5*4*3*2)/(10*9*8*7) = 120/5040$
$(3, 1) = (4*(5*4*3*3))/5040) = 720/5040$ (all #ways to get 3 reds and 1 green)
$(2, 2) = (6*(5*4*3*2)/5040) = 720/5040$
$(1,3) =(4*(5*3*2*1)/5040) = 120/5040 $
We know that $\sum_x\sum_yP_{x,y}(x,y)=1$ but I clearly don't get $1$, if there weren't any blue balls the method above would've worked and I'm stuck now.
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