Question
A company with n women and m men as employees is deciding which employees to promote.
(a) Suppose for this part that the company decides to promote t employees, where t belongs to [1, n + m], by choosing t random employees (with equal probabilities for each set of t employees). What is the distribution of the number of women who get promoted?
(b) Now suppose that instead of having a predetermined number of promotions to give, the company decides independently for each employee, promoting the employee with probability p. Find the distributions of the number of women who are promoted, the number of women who are not promoted, and the number of employees who are promoted.
(c) In the set-up from (b), find the conditional distribution of the number of women who are promoted, given that exactly t employees are promoted.
Attempt
a) As mentioned that each set of employees is equally likely, I now understand that X ~ HGeom(n, m, t). There is another doubt as to the answer to part c comes the same as part a, is that true?
b) I understand that the total no of employees is a Bin(m+n, p) however, how do I determine the distribution of no of women employees promoted P(X=k) = $\ sum_{j=0}^k (_nC_j) (P^j) (1-P)^{n-j} (_mC_{k-j}) P^{k-j} (1-p)^{m-(k-j)}$ Is this correct, if X=k is the r.v. representing the no of women employees promoted?
Thanks for going through it all, Kindly let me know if my approach is correct or not for each part as it has been a major confusion in multiple question for me.
a) $P(X=k)=\frac{\binom{n}{k}\binom{m}{t-k}}{\binom{n+m}{t}} $
b) $P(X=k)= \binom{n}{k}p^k(1-p)^{n-k} $ where $k=$ number of women selected.
c) $P(X=k|t \text { selected})= \frac {\binom{n}{k}\binom{m}{t-k}p^t(1-p)^{n+m-t} }{\sum_{i=0}^t\binom{n}{i}\binom{m}{t-i}p^t(1-p)^{n+m-t} } = \frac{\binom{n}{k}\binom{m}{t-k}}{\binom{n+m}{t}} $