[SOLVED] Probability

**fabxx** · Sep 26th 2008, 08:55 PM

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

**Chris L T521** · Sep 26th 2008, 09:15 PM

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

It would have to be this: $\frac{\displaystyle\binom32\binom21}{\displaystyle \binom53}=\dots$

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

--Chris

Thread: [SOLVED] Probability

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