We have a group of 15 people, 7 men and 8 women.
Randomly selecting 5 people, what's the probability to pick 3 men and 2 women? What's the probability to pick at least 1 man?
I tried solving the first question like this: ${7 \choose 3} = \frac{7!}{(7-3)!3!} = 35 $ and ${8 \choose 2} = \frac{8!}{(8-2)!2!} = 28 $,
so the probability should be $\frac 1{980}$.
But I'm stuck on the second question, how should I proceed?
You are right the number of ways to pick 3 men (m) and 2 women (w) are ${7 \choose 3}$ and ${8 \choose 2}$. And the number of ways to pick 5 people without any conditions is ${15 \choose 5}$. Therefore the probability to pick $3$ men (m) and $2$ women (w) is $\frac{{7 \choose 3}\cdot {8 \choose 2}}{{15 \choose 5}}=\frac{140}{429}\approx 32.63\%$
The formula is related to the Hypergeometric distribution.
Hint for the second question:
$P("\text{Picking at least one man}")=1-P("\text{Picking no man (5 woman)}")$