Question

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?

Answer 1

324/546 = 54/91

Step-by-step 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