Probability of Population A population consists of $50$% men and $50$%  women of a population of $50$ people. A simple random sample (draws at random without replacement) of $4$ people is chosen. Find the chance that in the sample  
i) the fourth person is a woman?
ii) The third person is a woman, given that the first person and fourth person are both men?
My attempt:
i) $P(4\text{th Women})=\frac{\dbinom{25}4}{\dbinom{50}4}=\;$?
ii) Conditionality $P(3\text{rd person Women}\mid\text{The }1\text{st and }4\text{th are both men})$?? Way Forward?
 A: (i) You can’t simply look at sets of $4$ people, because the order in which they’re chosen matters. Still, you can start by looking at sets and then count the number of usable orderings of each set. I’ll do this below, but note that there is a much easier way to solve this particular problem: since the numbers of men and women in the pool are equal, the fourth person chosen is equally likely to be a man or a woman, and the probability that it’s a woman is therefore ... what?
To calculate the probability by brute force:


*

*There are $\binom{25}4$ sets consisting entirely of men; each of these can be drawn in $4!$ different orders, and none of these $4!\cdot\binom{25}4$ draws will give you a woman on the fourth draw.

*There are $\binom{25}3\binom{25}1$ sets consisting of $3$ men and one woman. If the woman is the fourth person drawn, the $3$ men can be drawn in any of $3!$ orders, so there are $3!\cdot\binom{25}3\binom{25}1$ ways to draw three men and then a woman.

*There are $\binom{25}2\binom{25}2$ sets consisting of $2$ men and $2$ women. There are $2$ ways to choose which woman is drawn fourth. Once you’ve done that, there are $3$ ways to decide in which of the first $3$ positions the other woman will be drawn, and then there are $2$ ways to order the $2$ men in the remaining $2$ positions. Thus, there are $2\cdot3\cdot2\cdot\binom{25}2\binom{25}2$ ways to draw $2$ men and $2$ women with a woman in the fourth slot.

*There are $\binom{25}1\binom{25}3$ sets consisting of one man and $3$ women. The man can be drawn in any of the first $3$ slots, and the $3$ women can then be drawn in $3!$ different orders, so there are $3\cdot3!\cdot\binom{25}1\binom{25}3$ ways to draw one man and $3$ women with a woman in the fourth slot.

*Finally, there are $\binom{25}4$ sets consisting entirely of women, each of which can be arranged in $4!$ ways to put a woman in the fourth slot, for a total of $4!\cdot\binom{25}4$ draws of $4$ women.
Thus, there are altogether
$$3!\cdot\binom{25}3\cdot25+12\binom{25}2^2+18\cdot25\cdot\binom{25}3+24\binom{25}4$$
draws with a woman in the fourth slot, out of a total of $4!\cdot\binom{50}4$ draws.

Now see if you can do part (ii); it’s actually quite a bit easier. Note that if the first and fourth persons drawn are men, the second and third persons are really being chosen from a pool of $25$ women and $23$ men. You can now use brute force very easily, though you can also use the idea of the very short solution to (i) to avoid even that.
A: We can split it into, the probability that the first three were all women, $2$ women, $1$ woman, or no women. So
$$P(4\text{th Woman})=\frac{\binom{25}{3}}{\binom{50}{3}}\frac{22}{47}+\binom{3}{1}\frac{\binom{25}{2}\binom{25}{1}}{\binom{50}{2}\binom{48}{1}}\frac{23}{47}+\binom{3}{2}\frac{\binom{25}{1}\binom{25}{2}}{\binom{50}{1}\binom{49}{2}}\frac{24}{47}+\frac{\binom{25}{3}}{\binom{50}{3}}\frac{25}{47}$$
For the second part:
$$\begin{align*}
P(3\text{rd Woman} | \text{$1$st and $4$th Men})&=\frac{P(\text{$3$rd Woman and $1$st and $4$th Men})}{P(\text{$1$st and $4$th Men})}\\
&=\frac{25\cdot24\cdot25\cdot23+25\cdot25\cdot24\cdot24}{25\cdot24\cdot23\cdot22+\binom{2}{1}25\cdot24\cdot25\cdot23+25\cdot25\cdot24\cdot24}
\end{align*}$$
