There are $9$ women, who each own $4$ breweries, and $12$ men, who each own $9$ breweries. I pick $4$ people at random (without replacing).
a) Let $Y$ = the total number of breweries. Write down the pmf of $Y$ (probability of each possible number of breweries).
My work on 3a):
The probability to select $k$ shrubberies and $4-k$ shrubberies is: $$\frac{\binom{13}{k}\binom{13}{4-k}}{\binom{13}{4}}$$ *It is the permutation, with first num on top, not sure how to use correct format **
I don't think you understand the question. We pick $4$ out of the $21$ people, and count how many breweries they own. If we happen to pick $4$ men, we get $36$. If we pick $4$ women, we get $16$. The probability that all the people we choose are men is $$\frac{\binom{12}4\binom90}{\binom{21}{4}}$$ The denominator is the number of ways to choose $4$ people from the $21$ owners. The numerator is the number of ways to choose $4$ men and $0$ women. This is the only way to arrive at $36$ breweries. Therefore,
$$Y(36)=\frac{\binom{12}4\binom90}{\binom{21}{4}}$$
Can you do the other cases now?
