Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

Probability math problem ?

A club consists of six men and four women. Three members are selected at random to attend a conference. Find the probability that the selected group consists of two men and one woman. ?? 

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  • 4 weeks ago

    There are the following draws that would yield 2M,W.

    M.M.W => 1/6*1/5*1/4 = 0.008333333

    M.W.M => 1/6*1/4*1/5 = 0.008333333

    W.M.M => 1/4*1/6*1/5 = 0.008333333

    All the other possible draws would not. And there are three paths to success. So the total of the three is p(M,M,W) = 0.008333333 + 0.008333333 + 0.008333333 = 0.024999999 ~ .025 ANS.

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  • 4 weeks ago

    Men --> 6C2 = 15 combinations

    Women --> 4C1 = 4 combinations

    Choices are independent so we have in total:

    15 x 4 = 60 combinations

    Total choices are 10C3 = 120 

    Hence, probability 60/120 = 1/2

    :)>

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  • 4 weeks ago

    (6C2)(4C1) / (10C3) = 15(4) / 120 = 60/120 = 1/2.

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