In a survey, we asked $7$ men and $5$ women. Is randomly selected without replacement persons one by one until a man. Let $X$ be a random variable of the number of prints required.
- Determine the values of $X$ and its probability distribution.
The original French text version:
My thoughts: for example we can have this outcomes - M , WM , WWM , WWWM , WWWWM and WWWWWM
$\Pr(X=1)=7/12,\ \Pr(X=2)=(5.7)/(12.11),\ \Pr(X=3)=(5.4.7)/(12.11.10),$ $ \Pr(X=4)=(5.4.3.7)/(12.11.10.9),\ \Pr(X=5)=(5.4.3.2.7)/(12.11.10.9.8).$
$\Pr(X=6)=(5.4.3.2.1.7)/(12.11.10.9.8.7).$
$\mathbb{E}[X] = \sum_{k=1}^6 k \, P(X=k),$
$\operatorname{Var}(X)= \operatorname{E}\left[X^2 \right] - (\operatorname{E}[X])^2$
Is my proof correct?
- Is my translation of French text correct?
- Is there any probability distributions from that list can apply it to solve the exercise?
I see that you have already figured it out from the comments, but I think that an alternative approach can be helpful in gaining a better understanding of this kind of problem.
If you think of all the ways in which you can pick all the individuals randomly without replacement, it equals the number of ways in which you can pick the 5 positions for the women times possible orderings. That's
$$ {12 \choose 5} 5! 7! $$
Now, if you consider the possibilities in which the first man comes up at position $k$, for some $k$ such that $1 \leq k \leq 6$, that amounts to fixing the genders at the first $k$ positions and considering all possibilities for the remaining $12-k$, which again multiplied by the possible reorderings is
$$ {12-k \choose 5-k+1 } 5!7! $$
Therefore, the probability that the first man will appear at position $k$ is simply
$$ P(X=k) = \frac{{12-k \choose 5-k+1 }}{{12 \choose 5 }} $$
for all $k$ such that $1 \leq k \leq 6$.