Question 128922: Manager of Muddlepool FC has a squad of 16 players including 2 goakeepelrs, 5 defenders, 5 midfeilders and 4 forwards. He plays a 4 4 2 formation. So must always have 1 goalkeeper, 4 defenders, 4 midfeilders, and 2 forward. All players can be chosen but they need to play a fair amount of games.
How many different teams coud be chosen?
In the squad are 4 brothers. 1 is a goalkeeper, 1 is a defender, 1 is a midfeilder and 1 is a forward. The brothers parents come only to a game with at least two of the brothers. The team thinks the parents will only watch half of the games but the brothers think it is more than half.
In a season of 36 games how much more than half of the games are they likely to watch?
How would it differ using a 4 3 3 formation?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Manager of Muddlepool FC has a squad of 16 players including 2 goalkeepers, 5 defenders, 5 midfeilders and 4 forwards. He plays a 4 4 2 formation. So must always have 1 goalkeeper, 4 defenders, 4 midfeilders, and 2 forward. All players can be chosen but they need to play a fair amount of games.
How many different teams coud be chosen?
# of ways to pick a goal keeper: 2
# of ways to pick 4 defenders:5C4 = 5
# of ways to pick 4 midfielders: 5C4 = 5
# of ways to pick 2 forwards: 4C2 = 6
Total # of teams = 2*5*5*6 = 300
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In the squad are 4 brothers. 1 is a goalkeeper, 1 is a defender, 1 is a midfielder and 1 is a forward. The brothers parents come only to a game with at least two of the brothers. The team thinks the parents will only watch half of the games but the brothers think it is more than half.
In a season of 36 games how much more than half of the games are they likely to watch?
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Find # of teams possible with no brothers:
Ways to pick goalkeeper: 1
ways to pick defenders: 4C4 =1
Ways to pick midfielders: 4C4=1
ways to pick 2 forwards: 3C2 = 3
# of teams possible with no brothers = 1*1*1*3 = 3
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Find # of teams possible with only one of the brothers:
If goalkeeper is a brother # ways is : 1
If defender cannot be brother # ways is 4C4=1
If midfielder cannot be brother #ways is 4C4 = 1
If forwards must not be one of the brothers 3C2 = 3
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Do the same making a defender one of the brothers and get 3 teams possible
Do the same for midfielders and get 3 teams possible
Do the same for forwards and get 1 team possible.
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So there are a total of 3+3+3+3+1 = 13 teams possible with no brothers or
one brother.
Probability of a team having two or more of the brothers is 287/300
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In 36 games what is the probability 2 of the brothers are picked
more than 18 times. That is a binomial problem with n=36,p=287/300,18
Use 1-binomcdf(36,287/300,18) = practically one
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Conclusion: the probability is practically one that the parents will
attend more than half the games
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How would it differ using a 4 3 3 formation?
Available.........2.goalie.....5.defend.......5.middie......4.forward
Needed............1............4..............3.............3......
Brothers..........1............1..............1.............1......
Not Bros..........1............4..............4.............3......
# ways if no Bro..1...........4C4............4C3............3C3...= 3 teams
# ways Goalie/Bro.ONE BRO.....4C4............4C3............3C3...= 4 teams
# ways Defend/Bro.1...........ONE BRO........4C3............3C3...= 4 teams
# ways Middie/Bro.1...........4C4............ONE BRO........3C3...= 1 team
# ways Forwad/Bro 1...........4C4............4C3..........ONE BRO.= 4 teams
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There are 16 teams that have no brother or one brother
on them, from a total of 300 possible teams.
Prob (team has at least two brothers on it) = 284/300
Prob at least 18 of 36 games have two brothers = [1-binomcdf(36,284/300,18)]
= approximately 1.
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Cheers,
Stan H.
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