Thread: [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?

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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

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?

1. 
2. 
3. 
4. 
5. 

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