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yashar806
 one year ago
Best ResponseYou've already chosen the best response.0The following five games are scheduled to be played at the World curling championship one morning. The values in parentheses are the probabilities of each team winning their respective game. Game 1: Finland vs Germany Game 2:USA vs Switzerland Game3:Japan vs Canada Game4:Denmark vs Sweden Game 5:France vs Scotland The outcome of interest is the winners of the five games.how many out comes are contained in the sample space?

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0But I don't know how

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0there are no paranthesised values

yashar806
 one year ago
Best ResponseYou've already chosen the best response.00.43 vs 0.57 0.28vs 0.72 0.11vs 0.89 0.33vs o.67 0.18vs 0.82

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0assuming each game can only have one winner ... and he outcome of interest is the winners of the five games. a.b > a c.d > c e.f > e g.h > g i .j > i which leads one to believe that the set of outcomes has a cardinality of 5 is there something else to the setup that we might be overlooking?

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0But the answer is 32

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0it might be asking, of the 10 teams, how many ways are there to choose 5 winners? (10P5)/5! 10.9.8.7.6  5.4.3.2 2.9.7.2 ; .... is greater than 32 so its not that

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0What is the probability that at least one underdog wins?

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0Underdog is the team who is less likely to win

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i dont know the rules of how winners are paired up with losers ... so i cant even begin to make an educated run at this

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0you have 5 teams in the running, 4 teams get paired up with a 1 team in the wings .... to do what? there is either assumed knowledge that is not present, or missing information that i cant deduce.

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0The favourite is the team with the higher probability of winning

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0Underdog is the team who is less likely to win

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0What is the probability that at least one underdog wins?

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0That's what all the question asking about

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0thats doesnt really help me out in determining anything past the first round of winners and losers

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0in any case, this question makes no sense to me so im not going to be of any use in solving it .... good luck

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0I have included the relevant numbers for each team

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0yes you have, and they still dont help me sort anything out ....

tomo
 one year ago
Best ResponseYou've already chosen the best response.2C(2,1)*C(2,1)*C(2,1)*C(2,1)*C(2,1) = 2^5 = 32

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0What is the probability that at least one underdog wins?

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0What are the non underdog wins?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0it almost sounds like a decision tree model http://www.bus.utk.edu/stat/datamining/Decision%20Trees%20for%20Predictive%20Modeling%20(Neville).pdf

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i see the 32 from 5 cases in there

tomo
 one year ago
Best ResponseYou've already chosen the best response.2Game 1: Finland vs Germany Game 2:USA vs Switzerland Game3:Japan vs Canada Game4:Denmark vs Sweden Game 5:France vs Scotland 0.43 vs 0.57 0.28 vs 0.72 0.11 vs 0.89 0.33 vs 0.67 0.18 vs 0.82 since they are independent P(A AND B) = P(A)*P(B) you want 1[P(germany wins AND switzerland wins AND canda wins AND sweden wins AND scotland wins)] = 1P(germany wins)*P(switzerland wins) *P(canada wins) * P(sweden wins) * P(scotland wins) = 1(.57*.72*.89*.67*.82) = 0.7993283536

kropot72
 one year ago
Best ResponseYou've already chosen the best response.0Each game has 2 possible outcomes. The sample space is 2^5 = 32

yashar806
 one year ago
Best ResponseYou've already chosen the best response.0I see thank you very much tomo
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