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Math Help - probability

  1. #1
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    probability

    4 persons will be employed for a job but 6 persons applies to the job. What is the probability that Steve will be employed?

    My solution is C(1,1)STEVE IS SELECTED
    C(5,3) 3 WORKERS WILL BE CHOSEN
    C(1,1)C(5,3)/C(6,4)
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  2. #2
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    Re: probability

    Quote Originally Posted by kastamonu View Post
    4 persons will be employed for a job but 6 persons applies to the job. What is the probability that Steve will be employed?
    My solution is C(1,1)STEVE IS SELECTED, C(5,3) 3 WORKERS WILL BE CHOSEN C(1,1)C(5,3)/C(6,4)

    It would look better to write as \frac{\binom{5}{3}}{\binom{6}{4}}
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  3. #3
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    Re: probability

    Kastamonu

    since Steve must be selected ...the favourite outcomes will be the combinations of the remaining 5 applicants by 3 . check it...
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  4. #4
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    Re: probability

    Many Thanks Plato
    Many Thanks Mindanman I will check it
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  5. #5
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    Re: probability

    My be I missed something in the question.

    All applicants have the same chance to get the job, so the probability that Steve (or any one else) get the job is 1/6.
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  6. #6
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    Re: probability

    Quote Originally Posted by Kmath View Post
    All applicants have the same chance to get the job, so the probability that Steve (or any one else) get the job is 1/6.

    What does that have to do with the question? Nothing.
    From a set of six, how many subsets of four are there?
    Of those sets of four, how many have Steve as a member?

    This is a simple counting question.
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  7. #7
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    Re: probability

    Quote Originally Posted by Plato View Post
    What does that have to do with the question? Nothing.
    From a set of six, how many subsets of four are there?
    Of those sets of four, how many have Steve as a member?

    This is a simple counting question.
    So stupid fault! Ok, got it, thanks.
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  8. #8
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    Re: probability

    I know I'm probably wrong but I always use to think of problems like this as:

    4 out of 6 people will be selected or 4/6 which can be reduced to 2/3 which can be written as the percentage 66.6%, of the people will be selected, therefor he has a 66% percent chance. :s is this even close to something correct or was I completely wrong all along?

    --- edit ---
    Quote Originally Posted by Plato View Post
    From a set of six, how many subsets of four are there?
    Of those sets of four, how many have Steve as a member?
    From this I confirmed it's correct. There are 15 subsets of four in a set of six. Ten of the 15 subsets have steve. That works out to 10/15, that is steve will be selected 10 out of 15 times or ten times in 15 difference universes. or 66.6 % of the time.
    Last edited by russ123; February 24th 2013 at 02:06 AM.
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