WYZANT PROBABILITY PROBLEM
HOW MANY POSSIBLE WINNING TEAMS ARE THERE?
THERE ARE 80 POSSIBLE MEMBERS FOR THE FIRST PERSON ON THE TEAM CHOSEN AT RANDOM, AND THEN THERE ARE 79 PEOPLE THAT COULD BE CHOSEN AT RANDOM FOR THE SECOND PERSON ON THE TEAM.
THEREFORE, THERE ARE 80 * 7 = 6,320 POSSIBLE WINNING TEAMS.
NOW, HOW MANY OF THOSE TEAMS QUALIFY UNDER THE CRITERION REQUESTED?
LET'S TAKE A LOOK AT THREE POSSIBLITIES:
1. THE FIRST PERSON CHOSEN IS A MOM. THERE ARE 20 MOMS, SO THERE ARE 20 CHOICES. EACH OF THESE MOMS ONLY HAS ONE DAUGHTER, SO THERE ARE 20 * 78 = 1,560 POSSIBLE TEAMS THAT MEET THE GIVEN CRITERIA. (THE DAUGHTER MUST BE EXCLUDED.)
2. THE FIRST PERSON CHOSEN IS EITHER A DAD OR A SON. THERE ARE A TOTAL OF 40 DADS AND SONS, AND WE MUST CHOSE A MOM ON THE SECOND SELECTION TO MAKE THAT A TEAM WHICH MEETS THE GIVEN CRITERIA. THAT IS 40 * 20 = 800 POSSIBLE TEAMS WHICH MEET THE GIVEN CRITERIA.
3. FINALLY, SUPPOSE THE FIRST PERSON CHOSEN ON A TEAM IS A DAUGHTER. ONLY ONE OF THE REMAINING 20 MOM CHOICES IS HER MOTHER. SO THERE ARE 20 * 19 WHICH IS = 380 TEAMS THAT MEET THE GIVEN CRITERIA.
THIS COVERS ALL POSSIBLE OPTIONS, AND EACH OF THE THREE SCENARIOS GIVEN ABOVE ARE MUTUALLY EXCLUSIVE. SO, IF WE ADD THESE NUMBERS UP AND DIVIDE THE RESULT BY THE TOTAL POSSIBLE NUMBER OF TEAMS (6,320), WE WILL HAVE THE PROBABILITY THAT WAS REQUESTED.
1,560 + 800 + 380 = 2,740
2,740 / 6,320 = 0.4335443038 = 137/316
