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Math Help - [SOLVED] Probability

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    [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: \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
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by fabxx View Post
    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
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