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What distribution do I use for these? watch

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    In part a do i use the binomial and in part b what do I use?
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    a) is binomial, yes.

    For b) write down the probability of success after one try, then two, etc. Don't actually work then out, but leave them in the raw form. The distribution should be obvious.
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    Is it geometric?
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    yes, part A is binomial and part B is geometric
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    no
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    The first distribution has fixed n trials, which is Binomial as you stated. The second distribution, however, is very similar but fixes x as opposed to n and leaves n as the variable you're looking for.

    It's related to the Binomial. (It's basically an extension of it.) What do you think it is?

    The answer is in the spoiler, but try and think about it before actually looking.

    Spoiler:
    Show
    Negative Binomial
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    (Original post by Whelanmk)
    Is it geometric?
    That's my thinking.
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    This is what I did:

    P(X > 4) = 1 - 0.1 - (0.1 \times 0.9) - (0.1 \times 0.9^2) - (0.1 \times 0.9^3) - (0.1 \times 0.9^4)

    Is this right for part b?
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    \displaystyle P(X=x) = p \cdot (1-p)^{x-1} for \displaystyle x = 1, 2, 3, ...

    \displaystyle P(X>4) = 1 - P(X<5) = 1 - (P(X=1) + P(X=2) + P(X=3) + P(X=4) )
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    (Original post by .ACS.)
    \displaystyle P(X=x) = p \cdot (1-p)^{x-1} for \displaystyle x = 1, 2, 3, ...

    \displaystyle P(X>4) = 1 - P(X<5) = 1 - (P(X=1) + P(X=2) + P(X=3) + P(X=4) )
    Is that essentially the same as what I did before with P(X=0) etc.
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    (Original post by Whelanmk)
    Is that essentially the same as what I did before with P(X=0) etc.
    It's exactly what you did, except your last term is \displaystyle P(X=5), so basically you've one term too many. Lose your very last term, and it's fine.

    So it should be:

    \displaystyle 1 - 0.1 - (0.1 \times 0.9) - (0.1 \times 0.9^2) - (0.1 \times 0.9^3)
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    The wording of (b)(ii) seems misleading to me. Publishing more than 4 before having a success; e.g. on the next one, is it you can publish 5 and then have the success on the 6th, or you publish five and there is success in the 5th volume.

    On balance, I'd go with .ACS.'s interpretation; for what my opinion is worth....

    Edit: Spelling :sigh:
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    Considering part B concerns the number of books that need to be published for ONE success, it is geometric
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    (Original post by Tony2DaMax)
    Considering part B concerns the number of books that need to be published for ONE success, it is geometric
    Who said it isn't? If you want to be really clever though, you could say it's negative binomial where the x parameter is 1, which is exactly the same as geometric.
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    (Original post by .ACS.)
    Who said it isn't? If you want to be really clever though, you could say it's negative binomial where the x parameter is 1, which is exactly the same as geometric.
    Which is all just multiple Bernoulli trials. Yeah, we get it, you know the difference between a Negative Binomial and a Geometric distribution. However, the most PRECISE definition of part B is that it is a Geometric distribution.
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    (Original post by Tony2DaMax)
    Which is all just multiple Bernoulli trials. Yeah, we get it, you know the difference between a Negative Binomial and a Geometric distribution. However, the most PRECISE definition of part B is that it is a Geometric distribution.
    I would disagree on it being the most precise definition, though. (Unless of course he hasn't covered the negative binomial.)

    Say you've a hypothetical distribution, would it be more precise to say it's a \displaystyle Gamma(\frac{1}{2},2) or a \displaystyle \chi^2(1)?
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    (Original post by .ACS.)
    I would disagree on it being the most precise definition, though. (Unless of course he hasn't covered the negative binomial.)

    Say you've a hypothetical distribution, would it be more precise to say it's a \displaystyle Gamma(\frac{1}{2},2) or a \displaystyle \chi^2(1)?
    Perhaps precise is the wrong word. How about "simple"? The OP doesn't seem too confident so perhaps whatever avoids confusion is best.
 
 
 
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