PMF of HGeom & Binom Distributions (Women and Men getting promoted)

Question

Working through some book problems from "Introduction to Probability" (Blitstein)

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 1 <= t <= 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?

I believe this is a hypergeometric distribution:

X -> Hgeom(n,m,p)

= X -> Hgeom(n,m,t/(n+m))

(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.

I believe this is a binomial distribution

Women who get promoted (number of men who get promoted do not matter?):

$ X -> Binom(n,p) $

Women who do not get promoted (number of men who get promoted do not matter?):

$ X -> Binom(n,1-p) $

Total employees who get promoted:

$ X -> Binom(n+m,p) $

(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.

Let Y = t-X

P(Y=y)=P(t-X=y)=P(X=t-y)

Y->Hgeom(n,m,t-m)

$ P(X=t-y) = \frac{{n \choose {t-y}}{m \choose {t-m-(t-y)}}}{{n+m}\choose {t-m}} $

I don't think I'm right, especially on C. Can someone confirm starting logic for A and B, and point me in the right direction for C?

Answer 1

Well, $(a)$ and $(b)$ are okay.   Though perhaps you should show your working.

I believe this is a hypergeometric distribution:

Yes, though it is $\mathcal{HyperGeom}(n+m,n,t)$. Can you sa why?

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For $(c)$ take $X\sim\mathcal{Bin}(n,p)$ and $Y\sim\mathcal{Bin}(m,p)$ (independently), so $X+Y\sim\mathcal{Bin}(n+m,p)$ as you determined in (b).   You now seek $\mathsf P(X=s\mid X+Y=t)$ so use the definition for conditional probability:

$$\mathsf P(X=s\mid X+Y=t)=\dfrac{\mathsf P(X=s)}{\mathsf P(X+Y=t)}\mathbf 1_{0\leq s\leq\min( t,n)\leq m+n}$$