1. ## [SOLVED] Probability

Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

I did it like this: $\displaystyle \frac{3}{5}$x$\displaystyle (\frac{2}{3}+\frac{3}{6})$=$\displaystyle \frac{7}{10}$ which is obviously wrong since its not on. Can someone please explain the steps? Thanks in advance

2. Originally Posted by fabxx
Of 5 employees, 3 are to be assigned an office and 2 are to be assigned a cubicle. If 3 of the employees are men and 2 are women, and if those assigned an office are to be chosen at random, what is the probability that the offices will be assigned to 2 of the men and 1 of the women?

I did it like this: $\displaystyle \frac{3}{5}$x$\displaystyle (\frac{2}{3}+\frac{3}{6})$=$\displaystyle \frac{7}{10}$ which is obviously wrong since its not on. Can someone please explain the steps? Thanks in advance
It would have to be this: $\displaystyle \frac{\displaystyle\binom32\binom21}{\displaystyle \binom53}=\dots$

Do you see why? We want two men $\displaystyle \binom32$ and 1 women $\displaystyle \binom 21$ to occupy the three offices: $\displaystyle \binom 53$

--Chris