Assume that the committee contains 8 men and 10 women and that three are selected at random for a subcommittee. What is the probability that the subcommittee consists of 2 men and 1 woman, given that it contains both men and women?

324/546 =  54/91

Explanation:

Using a brute force type of method in solving this problem.  There are eight possible outcomes which may be represented by

WWW   (failure)  

WWM    (failure)

WMW     (failure)

WMM     (success)   Probability =  (1/3) (10/14) (9/13)  =  90/546

MWW    (failure)

MWM    (success)    Probability  =  (2/3)(5/14) (9/13)   =  90/ 564

MMW    (success)    Probability  = (2/3) (9/14)(5/13)   =  144/ 564

MMM     (failure)

Where success is distinct as having a subcommittee with 1 woman and 2 men.   Adding up the all probabilities for the successes yields      324/546 =  54/91

answer; i believe that the correct answer is (tabula rasa);