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 two are women.

Answer 1

Answer:

$\dfrac{14}{33}$

Step-by-step explanation:

5 men and 7 women are on a crowded elevator, 12 people in total.

At the next floor, four people get off the elevator. If two of them are women, then the remaining two are men.

Number of ways to choose 2 women from 7

$=C^7_2=\dfrac{7!}{2!(7-2)!}=\dfrac{7!}{2\cdot 5!}=\dfrac{6\cdot 7}{2}=21$

Number of ways to choose 2 men from 5

$=C^5_2=\dfrac{5!}{2!(5-2)!}=\dfrac{5!}{2\cdot 3!}=\dfrac{4\cdot 5}{2}=10$

Number of ways to choose 4 people from 12:

$=C^{12}_4=\dfrac{12!}{4!(12-4)!}=\dfrac{12!}{2\cdot 3\cdot 4\cdot 8!}=\dfrac{9\cdot 10\cdot 11\cdot 12}{24}=495$

Therefore, the probability that two are women is

$\dfrac{21\cdot 10}{495}=\dfrac{42}{99}=\dfrac{14}{33}$