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?
---- My (wrong) solution follows: ---
The probability to choose 1 woman out of the group: 5/(3+5) = 5/8
Then, the probability to choose another woman after the first one was chosen: 4/(3+4) = 4/7
Then, the probability to choose one man after two woman were chose: 3/(3+3) = 3/6
Then, the probability to choose another man is: 2/(2+3) = 2/5
Multiplies those probabilites gives (5/8)*(4/7)*(3/6)*(2/5) = 1/14
I was pleased to see it was one of the options and not so pleased to see it is wrong. I will appreciate an explanation as to what is wrong with my solution.