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Math Help - soccer problem

  1. #1
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    dfsdfdf
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    soccer problem

    Dear Sir,
    I need help some help with the below problems and would appreciate very much if can offer some help.
    thanks
    Kingman

    1) In a competition, 2 teams (A and B) will play each other in the best of 3 games. That is, the final team to win 2 games will be the winner. In the first games, both teams have equal chance of winning. In subsequent games, the probability of team A winning team B given that team A won in the previous game is p and the probability of team A winning team B given that team A lost in the previous game is 1/3.

    a) Illustrate the information with an appropriate tree diagram.
    b) Find the value of p such that the team A has equal chances of winning and losing the competition.

    2) A soccer team of 30 players has 18 Americans and 12 foreigners. Of the
    Americans, 10 are midfielders and 8 are defenders. Of the foreigners, 5 are
    goalkeepers and 7 are strikers. Four players are chosen at random.
    a) If exactly one striker is chosen, find the probability that 3 American are
    chosen.
    b) Find the probability that there is exactly one goalkeeper or exactly one defender (or both)
    Last edited by mr fantastic; April 26th 2010 at 04:32 PM. Reason: Changed font to larger size to make the post easier to read.
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  2. #2
    Junior Member atalay's Avatar
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    Answer 2)

    a) P(X=3) = \binom{18}{3} \binom{11}{0} / \binom{29}{3}

    b) exactly one goalkeeper

    P(goalkeeper=1) = \binom{5}{1} \binom{18}{2} \binom{6}{0} / \binom{29}{3}

    exactly one defender

    P(d=1) = \binom{8}{1} \binom{15}{2} \binom{6}{0} / \binom{29}{3}

    both


    P(both=1) = \binom{8}{1} \binom{5}{1} \binom{10}{1} \binom{6}{0} / \binom{29}{3}

    (if exactly one striker chosen)
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  3. #3
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    Dear atalay,
    Thanks very much for the solution but I wonder why it does not tally with the answer given to question 2 as
    a).461
    b).671
    Thanks very much
    Kingman
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  4. #4
    Junior Member atalay's Avatar
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    Yes, i had trouble fault.

    a) P(X=3) = \binom{18}{3} \binom{7}{1} / \binom{25}{4}

    b) which one calculated?
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  5. #5
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    I have only looked at problem 2a so far.

    You have 18 Americans, 12 Foreigners.

    1 of those Foreigners is chosen (striker). You now randomly select 3 more people. What are the chances that all three are American?

    So now you have 18 Americans, 11 Foreigners.

    the chance of first being an American is 18/(18+11) = 18/29
    the chance of the second being an American is 17/(17+11) = 17/28
    the chance of the third being an American is 16/(16+11) = 16/27

    The chances of those happening consecutively?

    (18/29)*(17/28)*(16/27) = (4896/21924) = 22.31%

    Proof? I just got done making a simulation that would choose 3 players from 29 total. 18 of them are americans, 12 foreigners. Once one is chosen, it cant be chosen again.

    Here were the results of the first test with 100,000 trials.



    which is 22.188%

    and here's the code.

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  6. #6
    Junior Member atalay's Avatar
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    Quote Originally Posted by shenanigans87 View Post
    I have only looked at problem 2a so far.

    You have 18 Americans, 12 Foreigners.

    1 of those Foreigners is chosen (striker). You now randomly select 3 more people. What are the chances that all three are American?

    So now you have 18 Americans, 11 Foreigners.

    the chance of first being an American is 18/(18+11) = 18/29
    the chance of the second being an American is 17/(17+11) = 17/28
    the chance of the third being an American is 16/(16+11) = 16/27

    The chances of those happening consecutively?

    (18/29)*(17/28)*(16/27) = (4896/21924) = 22.31%

    Proof? I just got done making a simulation that would choose 3 players from 29 total. 18 of them are americans, 12 foreigners. Once one is chosen, it cant be chosen again.

    Here were the results of the first test with 100,000 trials.



    which is 22.188%

    and here's the code.

    You should be thinking your answer again.
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  7. #7
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    You are right, let me change it a bit.

    I was counting the rest of the strikers in the probability, I forgot about the part which states that exactly 1 striker is chosen, not more.

    So

    \frac{18}{29}*\frac{17}{28}*\frac{16}{27}=23.3317%

    becomes

    \frac{18}{23}*\frac{17}{22}*\frac{16}{21}=46.0757%
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  8. #8
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    sample space of soccer problem

    Dear shenanigans87
    Thanks very much for providing great insight into the
    soccer problem but I wonder whether you can enumerate
    the sample space in the form of tree diagram or
    ordered n-tuplets in spreadsheet thus enabling one to see how
    the events look like.
    Thanks
    Kingman
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  9. #9
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    Quote Originally Posted by kingman View Post
    Dear shenanigans87
    Thanks very much for providing great insight into the
    soccer problem but I wonder whether you can enumerate
    the sample space in the form of tree diagram or
    ordered n-tuplets in spreadsheet thus enabling one to see how
    the events look like.
    Thanks
    Kingman
    I don't understand what you are asking for...?
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  10. #10
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    dfsdfdf
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    Dear Sir,
    Can you please explain how you get the answer
    " %" by drawing a tree diagram showing the probability of all the outcomes and using Venn diagram
    to prove it.
    Thanks
    Kingman
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