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Math Help - Probability of 1 chosen in a group of 20 out of 500

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    Probability of 1 chosen in a group of 20 out of 500

    Suppose you have 500 elements and select a group of 20 out of them. What is the probability of any element being in the group?

    I've come up with an answer two different ways, but both of them disagree with the book's answer. The book says that it's 1/500.

    However, I would think that I'd need to calculate \frac{\binom{499}{19}}{\binom{500}{20}}. The numerator, I think, is the number of ways that a particular element could be in some group of 20. The denominator is the number of ways of selecting groups of 20 from 500. This comes out to 1/25.

    Alternately, we could select some random element in the 500, and then find the probability of successively selecting elements to form a group of 20 and not select that element each time. That'd be \frac{499}{500}\cdot \frac{498}{499}\cdot \cdot \cdot \frac{480}{481} = 24/25. Taking 1-24/25 gives the same answer again.

    Did I find an error in the book?
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    Re: Probability of 1 chosen in a group of 20 out of 500

    Suppose you have 500 elements and select a group of 20 out of them. What is the probability of any element being in the group?
    Is this the exact wording for the question? The book's answer seems to reflect a question built this way: "Suppose you have 500 different elements and you select a group of 20, what is the probability of selecting any element from this group?"
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    Re: Probability of 1 chosen in a group of 20 out of 500

    The question from the book asks "What is the probability that each experimental unit [any one of the 500, I take it] is selected for inclusion in the sample [the group of 20]?" Is it instead asking, for any group of 20 elements, what is the probability that this 20 would be the group selected?
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    Re: Probability of 1 chosen in a group of 20 out of 500



    Is it this random number problem?
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    Re: Probability of 1 chosen in a group of 20 out of 500

    Hello, ragnar!

    Suppose you have 500 distinct elements and select a group of 20 out of them.
    What is the probability of a particular element being in the group?


    I've come up with an answer two different ways,
    . . but both of them disagree with the book's answer.
    The book says that it's 1/500. .??

    However, I would think that I'd need to calculate \frac{\binom{499}{19}}{\binom{500}{20}} . Yes!
    The numerator is the number of ways that the element can be in some group of 20.
    The denominator is the number of ways of selecting groups of 20 from 500.
    This comes out to 1/25. . Right!

    Alternately, we could select some random element in the 500, and then
    find the probability of successively selecting elements to form a group of 20
    and not select that element each time.

    That would be: . \frac{499}{500}\cdot \frac{498}{499}\cdot \cdot \cdot \frac{480}{481} = \frac{24}{25}.

    Taking . 1-\frac{24}{25} gives the same answer again. . Outstanding!

    Did I find an error in the book? . You certainly did!

    ?? .The book's answer is way WAY off!
    It does not consider the size of the sample group.

    Suppose we take a sample of 499 elements.
    The sample will almost certainly contain that particular element.

    As the sample size decreases, so does the probability.
    When the sample size is one, the probability is \tfrac{1}{500}


    Thank you for your excellent explanations.
    So many say, "I can't get the book's answer. .What am I doing wrong?"
    neglecting to give us their answer and the book's answer.

    I usually respond:
    . . It's hard to see your work from here,
    . . but I think you played the 7\heartsuit instead of the Q\spadesuit
    . . and transposed to the key of B\flat major instead of C\sharp minor.

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    Re: Probability of 1 chosen in a group of 20 out of 500

    Quote Originally Posted by Soroban View Post
    Hello, ragnar!

    I usually respond:
    . . It's hard to see your work from here,
    . . but I think you played the 7\heartsuit instead of the Q\spadesuit
    . . and transposed to the key of B\flat major instead of C\sharp minor.

    [/size]
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    Re: Probability of 1 chosen in a group of 20 out of 500

    Thank you very much for the high praise—after feeling stupid about not getting the book's answer, that's welcome relief.

    Quote Originally Posted by Stro View Post


    Is it this random number problem?
    Yes, and now I'm wondering how the #%^* you found that, but that's exactly the problem. Am I misunderstanding the question?
    Last edited by ragnar; June 15th 2011 at 01:44 PM.
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    Re: Probability of 1 chosen in a group of 20 out of 500

    And since you've been able to reproduce that table, if it's no great task, can anyone explain what's going on with those "random" numbers? Why do we have two random numbers corresponding to each experimental unit, and why do they look so nonrandom? Is that just an assignment so that we can tell a random number generator to generate any 3-digit number, rather than telling it to generate a random number between 1-500? And presumably, from the generator, we choose the experimental unit that corresponds to the assigned number, skipping any double-selections of the same unit?
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