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Math Help - Probability of books on a shelf

  1. #1
    MHF Contributor Unknown008's Avatar
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    Probability of books on a shelf

    Ok, this one is giving me some problems:

    Qu. On a bookshelf there are works by three authors, three volumes by Gilmore, three
    volumes by Lawson and one volume by Patterson. If the books are placed at
    random on the shelf, the probability that the volumes of the same author are
    together is

    A. \frac{3}{70}

    B. \frac{1}{140}

    C. \frac{3}{140}

    D. \frac{1}{70}

    E. \frac{1}{35}

    I could find the probability that three books from Gilmore are together, the separate probability that the books from Lawson being together, but I don't know how to find the intersection between those two sets of probabilities...

    P(3 Gilmore together) = \frac{3!5!}{7!} = \frac{1}{7}

    P(3 Lawson together) = \frac{1}{7}

    P(3 Lawson together and 3 Gilmore together) = \frac{3!3!3!}{7!} = \frac{3}{70}

    Total probability = \frac{1}{7} + \frac{1}{7} + \frac{3}{70} - P(intersections)

    How can I solve this?
    Last edited by Unknown008; August 4th 2010 at 10:48 AM. Reason: forgot the latex tabs
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  2. #2
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    Look at \dfrac{(3!)^3}{7!}. WHY?
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  3. #3
    MHF Contributor Unknown008's Avatar
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    Um... I thought of it like this:

    (G G G) (L L L) P

    G permute among themselves, so 3!
    L permute among themselves, so 3!
    Then, there are three major items, so permuting again gives 3!
    Finally, total different arrangements = 7!

    So, I get ((3!)^3)/7!

    Or... did I miss something?

    EDIT: Ok, I'll see that tomorrow... bedtime here...
    Last edited by Unknown008; August 4th 2010 at 12:05 PM.
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  4. #4
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    That is correct.
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  5. #5
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    Quote Originally Posted by Unknown008 View Post
    Ok, this one is giving me some problems:

    Qu. On a bookshelf there are works by three authors, three volumes by Gilmore, three
    volumes by Lawson and one volume by Patterson. If the books are placed at
    random on the shelf, the probability that the volumes of the same author are
    together
    is

    A. \frac{3}{70}

    B. \frac{1}{140}

    C. \frac{3}{140}

    D. \frac{1}{70}

    E. \frac{1}{35}

    I could find the probability that three books from Gilmore are together, the separate probability that the books from Lawson being together, but I don't know how to find the intersection between those two sets of probabilities...

    P(3 Gilmore together) = \frac{3!5!}{7!} = \frac{1}{7}

    P(3 Lawson together) = \frac{1}{7}

    P(3 Lawson together and 3 Gilmore together) = \frac{3!3!3!}{7!} = \frac{3}{70}

    Total probability = \frac{1}{7} + \frac{1}{7} + \frac{3}{70} - P(intersections)

    How can I solve this?
    The way I'm reading this is....

    the volumes of any author who is the author of more than one book are together,
    in other words, no author has volumes scattered on the shelf.
    Hence, should not only one case be analysed??

    no?
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  6. #6
    MHF Contributor Unknown008's Avatar
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    Ok, I admit that I don't quite know what the question is really asking for.

    If the volumes of each author needs to be grouped together, then the answer is only 3/70. So, this is the answer then?

    It's so hard to understand what is required to find by the question...

    Thanks!
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  7. #7
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    Quote Originally Posted by Unknown008 View Post
    Ok, I admit that I don't quite know what the question is really asking for.

    If the volumes of each author needs to be grouped together, then the answer is only 3/70. So, this is the answer then?

    It's so hard to understand what is required to find by the question...

    Thanks!
    I would think so...

    As if all of Gilmore's books are together,
    but Patterson's book is between Lawson's books,
    then it could be stated

    "the volumes of the same author are not together".
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Ok, thanks. I'll be waiting for the solution and if I don't forget, I'll post it here .
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