Picking chips distribution

4 red and 4 white chipsI will set up the distribution.
P[nr] is probability of choosing n red chips
mCn= combination of m things n at a time

P[4]=4C4 4C0 / 8C4
P[4]=1*1/70

p[3]=4C3 * 4C1 /8C4
P[3]=16/70

P[2]=4C2*4C2 / 8C4
P[2]=36/70

P[1]=4C1*4C3 /8C4
P[1]=16/70

P[0]=4C0*4C4 / 8C4
P[0]=1/70

This should help.
Arthur
 
Unless I missed something in the question, wouldn't we need P(X=0,1,2,3) since we are only picking three chips? Therefore P(4) is unnecssary. With similar logic, I'd assume this is how we'd do P(3):

P(3)=(4c3)(4c0)/(8c3)

and so on for the rest?
 
Assuming that is correct, would P.m.f. just be the function for each probability P(X=0,1,2,3) or is there some general equation that can be made for this? I know that P(X=0)=P(X=3) and the other two are equal to each other as well.

Also, for the Y=X-2, would I simply subtract 2 from the mean of X for part b?

Thanks.
 
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