A brute force method of analysis can be used for this problem. Think of choosing the subcommittee by three successive selections from the remaining pool at each stage. The eight possible outcomes can 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 defined as have a subcommittee with 1 woman and 2 men. Adding up the probabilities for the successes yields 324/546 = 54/91
Richard P.
tutor
If the second part of the problem statement is interpreted as: "There is additional a priori information that the result of the selection process resulted in either a group of 2 men and 1 woman OR a group of 1 man and 2 women" ; then a similar
analysis of the probability of a 1 man , 2 women group must be carried out. The result of this is 120/546.
The final answer is then (324/526) /[ 324/526 + 120/546 ] which reduces to 54/74
Report
04/03/15
Katie C.
04/01/15