Question

suppose 5 men and 7 women are on a crowded elevator. at the next floor, four people get off the elevator. find the probability that all are women?

Answer 1

The total number of arrangements for all of them to be women is:
C(7, 4) because the order does not matter; we'd still have the same arrangements.

Now, we can simplify this:
$C(7, 4) = \frac{7!}{4!(7 - 4)!}$
$= \frac{5040}{4!3!}$
$= \frac{5040}{24 \cdot 6}$
$= \frac{5040}{144} = 35$

Now, the total number of arrangements, without restriction, is simply: C(12, 4) because we don't care who we pick.

$\text{P(4 women): } \frac{35}{495} = \frac{7}{99}$