Use hypergeometric distribution where there are two categories of identical objects/persons, each with a know size.
d=number of Democrats selected
D=total number of Democrats = 6
r=number of Republicans
R=total number of Republicans =5
Then
[tex]P(d,r)=\frac{C(D,d)C(R,r)}{C(D+R,d+r)}[/tex]
where
[tex]C(n,r)=\frac{n!}{(n!(n-r)!)}[/tex] = combination of r items selected from n,
D+R=total number of members = 6+5 =11
d+r=number of members selected = 7
[tex]P(d,r)=\frac{C(D,d)C(R,r)}{C(D+R,d+r)}[/tex]
[tex]P(4,3)=\frac{C(6,4)C(5,3)}{C(6+5,4+3)}[/tex]
[tex]=\frac{C(6,4)C(5,3)}{C(11,7)}[/tex]
[tex]=\frac{15*10}{330}[/tex]
[tex]=\frac{5}{11}[/tex]
Answer: the probability of selecting 4 Democrats and 3 Republicans is 5/11
