2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Have you experimented with small values of $m \text { and } n $? What were your results? Say for $1 \text { and } 2 $, $1 \text { and } 3 $, $2 \text { and } 3 $, $3 \text { and } 3 $? $\endgroup$ – Stephen Meskin Dec 10 '17 at 7:29
  • $\begingroup$ @StephenMeskin I experimented with those combinations and received some intuition on part a, however I'm still stuck with b i.e. I highly doubt my solution for part b as it is equivalent to Bin(m+n, p). Kindly help me with that and whether part a and part c are similar question or not? $\endgroup$ – shubham kumar Dec 10 '17 at 8:18
  • $\begingroup$ You say you understand the distribution of the total number of employees promoted. Why doesn't the same logic apply to the distribution of the number of women promoted? $\endgroup$ – Stephen Meskin Dec 10 '17 at 15:49
  • $\begingroup$ @StephenMeskin $$P(X=k) = \ sum_{j=0}^k (_nC_j) (P^j) (1-P)^{n-j} (_mC_{k-j}) P^{k-j} (1-p)^{m-(k-j)}$$, I have put different values of m, n, k and j. All i understand is that for a particular value of k, different values of j give the distribution of women promoted. Though I'm pretty sure now of the result, it would be great if you could give a nod. $\endgroup$ – shubham kumar Dec 11 '17 at 5:46
1
$\begingroup$

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}} $

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

For part b, one can also think as follows

Let $R_j$ be the event that j people are promoted

and $W_i$ be the event that i women are promoted

Then

$P(W_i)=\sum_{j=0}^{n+m}P(W_i|R_j)P(R_j)$

where

$P(R_j)=p^j(1-p)^{n+m-j} \binom{n+m}{j}$

$P(W_i|R_j)=\frac{\binom{n}{i}\binom{m}{j-i}}{\binom{n+m}{j}}$

The sum simplifies to the above answer

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.