Probability

- anonymous

Probability

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- chestercat

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- anonymous

The 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?

- anonymous

the answer is 5.

- anonymous

No answer is 32

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## More answers

- anonymous

But I don't know how

- amistre64

there are no paranthesised values

- anonymous

0.43 vs 0.57
0.28vs 0.72
0.11vs 0.89
0.33vs o.67
0.18vs 0.82

- amistre64

assuming 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?

- anonymous

But the answer is 32

- amistre64

it 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

- anonymous

What is the probability that at least one underdog wins?

- anonymous

Underdog is the team who is less likely to win

- amistre64

i 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

you 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.

- anonymous

The favourite is the team with the higher probability of winning

- anonymous

Underdog is the team who is less likely to win

- anonymous

What is the probability that at least one underdog wins?

- anonymous

That's what all the question asking about

- amistre64

thats doesnt really help me out in determining anything past the first round of winners and losers

- amistre64

in any case, this question makes no sense to me so im not going to be of any use in solving it .... good luck

- anonymous

I have included the relevant numbers for each team

- amistre64

yes you have, and they still dont help me sort anything out ....

- anonymous

Tomo help me

- anonymous

C(2,1)*C(2,1)*C(2,1)*C(2,1)*C(2,1) = 2^5 = 32

- anonymous

What is the probability that at least one underdog wins?

- anonymous

1-P(no underdog wins)

- anonymous

What are the non underdog wins?

- amistre64

it almost sounds like a decision tree model
http://www.bus.utk.edu/stat/datamining/Decision%20Trees%20for%20Predictive%20Modeling%20(Neville).pdf

- amistre64

i see the 32 from 5 cases in there

- anonymous

Game 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)]
= 1-P(germany wins)*P(switzerland wins) *P(canada wins) * P(sweden wins) * P(scotland wins)
= 1-(.57*.72*.89*.67*.82)
= 0.7993283536

- kropot72

Each game has 2 possible outcomes. The sample space is 2^5 = 32

- anonymous

I see thank you very much tomo

- anonymous

your welcome

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