thp510 wrote:Can anyone show me how to solve this problem WITHOUT using combinatorics and instead, just using probability/fractions?
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
OA: [spoiler]3/7[/spoiler]
The probability that an event will occur exactly
n times = P(one way) * total number of possible ways.
One way to choose exactly 2 women is to select first 2 women and then 2 men:
P(WWMM) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14.
The result above needs to be multiplied by the total number of ways to choose exactly 2 women:
WWMM
WMWM
WMMW
MWWM
MWMW
MMWW
Since there are 6 ways to choose exactly 2 women:
6 * 1/14 = 3/7.
We can determine the answer more efficiently if we use combinatorics to count the number of ways 2 women can be chosen. The list above represents the total number of ways to arrange the letters WWMM: 4!/2!2! = 6.
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