Basic probability and combination theory So, I ran into a question based on probability, however, I came to a point where I questioned some of the reasoning and can't quite understand the why factor of it. The question was based around the probability of picking exactly one male out of a group of 5 people who consisted of MMFFF where 3 were chosen at random one by one.
I started by saying that the pick order could be MFF, giving the calculation of $\frac{2}{5} * \frac{3}{4} * \frac{2}{3}$ to give $0.20$, however I realized I must multiply this by 3 as there are different combinations that exist, but what stumped me was why? How does this idea of combinations affect the probability in real life situations, I am struggling to visualize/understand it.
 A: If you're studying this kind of problem in a course you'll learn techniques like the one in @AlonYariv 's correct answer.
If you just "ran into it" you have to think it through clearly from scratch.
Since you'll be checking for "one male" after you have all three selected, the selecting "one by one" is irrelevant. So is the order. All that matters is the number of men in the final group of three.
There are $5\times4 \times 3$ ways to pick three people in order. But that counts each of the groups of three people six times (because they could come in that many orders). So there are $60/6 = 10$ ways to pick the three. To count the number with exactly one man, you pick the man $2$ ways and the women $3$ ways (because you just pick the one to leave out). That's $6$ ways all together. So the chance that you get exactly one man is $6/10$ or $60\%$.
A: Acording to my understanding that is a classical case of Hypergeometric distribution.
In this specific case with population of size 5 with 2 sucsses in the population with 3 draws looking for 1 succses among the draws:
$$\frac{{2\choose 1} \times{3\choose 2} } {5\choose3}$$ 
