3.2.11 problem 30

(a)

The distribution is hypergeometric. We select a sample of $t$ employees and count the number of women in the sample.

$$P(X=k)=\frac{\left(\frac{n}{k}\right)\left(\frac{m}{t-k}\right)}{\left(\frac{n+m}{t}\right)}$$

(b)

Decisions to be promoted or not are independent from employee to employee. Thus, we are dealing with Binomial distributions.

Let $X$ be the number of women who are promoted. Then, $P(X=k)=\left(\frac{n}{k}\right)p^k{(1-p)}^{n-k}$. The number of women who are not promoted is $Y=n-X$ and so is also Binomial.

Distribution of the number of employees who are promoted is also Binomial, since each employee is equally likely to be promoted and promotions are independent of each other.

(c)

Once the total number of promotions is fixed, they are no longer independent. For instance, if the first $t$ people are promoted, the probability of the $t+1$-st person being promoted is $0$.

The story fits that of the hypergeometric distribution. $t$ promoted employees are picked and we count the number of women among them.

$$P(X=k|T=t)=\frac{\left(\frac{n}{k}\right)p^k{(1-p)}^{n-k}\left(\frac{m}{t-k}\right)p^{t-k}{(1-p)}^{m-t+k}}{\left(\frac{n+m}{t}\right)p^t{(1-p)}^{n+m-t}}=\frac{\left(\frac{n}{k}\right)\left(\frac{m}{t-k}\right)}{\left(\frac{n+m}{t}\right)}$$