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Math Help - "At Least" Combination Problem

  1. #1
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    "At Least" Combination Problem

    I've been banging away at this... and I think I'm close, but would really appreciate someone looking over my shoulder... here goes:

    - 13 students took part in a matching exercise
    - the exercise asked them to match 4 questions to four answers
    - 6 of the 13 students got them all right

    What is the probability of achieving a result at least this good?

    Hereís where Iím at so far:
    - the probability of one student getting them all correct is 4! (1 in 24Ö 0.042)
    - the probability of 6 students matching them all is 0.042 to the power of 6 (191 million to 1)
    - the number of combinations of 6 out of 13 is 13! / 6! * 7! = 1,716
    - so, the probability of getting 6 out of 13 correct is 191 million / 1,716 = 111,000 to 1

    Right so far?

    Since I want to know the probability of getting at least 6/13 do I have to go further? Do I have to calculate the probability of going 7/13, 8/13Ö and add all those to this one? Is there an easier way?

    Many Thanks.
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  2. #2
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    Quote Originally Posted by skeptiko View Post
    I've been banging away at this... and I think I'm close, but would really appreciate someone looking over my shoulder... here goes:

    - 13 students took part in a matching exercise
    - the exercise asked them to match 4 questions to four answers
    - 6 of the 13 students got them all right

    What is the probability of achieving a result at least this good?

    Here’s where I’m at so far:
    - the probability of one student getting them all correct is 4! (1 in 24… 0.042)
    - the probability of 6 students matching them all is 0.042 to the power of 6 (191 million to 1) Mr F says: This is not quite right. There's a factor of {\color{red}^{13}C_6} ..... (The reason why is implied in my main post)

    - the number of combinations of 6 out of 13 is 13! / 6! * 7! = 1,716
    - so, the probability of getting 6 out of 13 correct is 191 million / 1,716 = 111,000 to 1

    Right so far?

    Since I want to know the probability of getting at least 6/13 do I have to go further? Do I have to calculate the probability of going 7/13, 8/13… and add all those to this one? Is there an easier way?

    Many Thanks.
    The number of ways of arranging the answers to the questions is 4! = 24. Only one of these arrangements gives a correct match to all 4 questions.

    So the probability of getting a match in a single trial is 1/24.

    Let X be the random variable number of students who answer all questions correctly.

    X ~ Binomial(n = 13, p = 1/24).

    You have to calculate \Pr(X \geq 6). The easy way to do this is to use technology. Using my TI-89 I get 0.0000069655.

    Conclusion: It's highly improbable that 6 or more students would get a perfect match by chance.
    Last edited by mr fantastic; December 30th 2008 at 03:44 PM.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    The number of ways of arranging the answers to the questions is 4! = 24. Only one of these arrangements gives a correct match to all 4 questions.

    So the probability of getting a match in a single trial is 1/24.

    Let X be the random variable number of students who answer all questions correctly.

    X ~ Binomial(n = 13, p = 1/24).

    You have to calculate \Pr(X \geq 6). The easy way to do this is to use technology. Using my TI-89 I get 0.0000069655.

    Conclusion: It's highly improbable that 6 or more students would get a perfect match by chance.
    Thx a bunch Mr. F...
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