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STEP I, II, III 1999 solutions

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    (Original post by SimonM)
    What is X? The expected value of one role?

    What if there were no rules, you could stop when you like? How would your argument scale to only stopping when you role a 6?

    Ok, the argument using variances would go like this:

    The variance for "stopping >3" is:
    \displaystyle \frac{1^2+4^2+5^2+6^2}{4} - 16 = 3 + \frac{1}{2}

    and "stopping >4" is:
    \displaystyle \frac{1^2+5^2+

6^2}{3} - 16 = 4+\frac{2}{3}

    With stopping at >4 there is a bigger spread, so depending on what properties you want your system to have, you might pick 3 more than 4 or vice-versa.
    Ah, makes a lot more sense now. Thanks for that, when I finally have the ability to rep on this thing I'll be sure to send you some.
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    (Original post by SimonM)
    STEP I, Question 1

    By symmetry, their average value will be \boxed{5 \times 10^5}
    What symmetry? I think I want to die, I can never beat STEP.
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    (Original post by Physics Enemy)
    What symmetry? I think I want to die, I can never beat STEP.
    Well, I was probably doing it to show off, there are much more "standard" ways to approach that question. However, consider the numbers 0-10

    0,1,2,3,4,5,6,7,8,9,10

    Striking out multiples of 2 and 5 leaves things evenly balanced on both sides: Symmetry
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    (Original post by SimonM)
    Well, I was probably doing it to show off, there are much more "standard" ways to approach that question. However, consider the numbers 0-10

    0,1,2,3,4,5,6,7,8,9,10

    Striking out multiples of 2 and 5 leaves things evenly balanced on both sides: Symmetry
    Yes I was gonna add and subtract sums of various arithmetic series.
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    (Original post by SimonM)
    STEP I, Question 6

    Spoiler:
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    Let y = bx + a, \x \in [-10,10]

    If b=0, y is constant, so \max y = \min y = a
    If b>0, y is strictly increasing so \min y = a-10b, \max y = a+10b
    If b<0, y is strictly decreasing so \min y = a+10b, \max y = a-10b

    Let y = cx^2 + bx+a
    If c=0, treat it as above

    \frac{dy}{dx} = 2cx+b

    This is the case where b&gt;0 from earlier.

    If b-20c \ge 0 then it is strictly increasing and \min y = 100c-10b+a, \max y = 100c+10b+a

    If b+20c&lt;0 then it is strictly decreasing \min y = 100c+10b+a, \max y = 100c-10b+a

    Otherwise 2cx+b changes sign in our interval. This means that

    \min y = a - \frac{b^2}{4c}

    The maximum will occur on an end point. Therefore it depends solely on the sign of b. Therefore if b>0

    \max y = 100c+10b+c otherwise 100c-10b+c
    LOL @ this Q, what a gift. I wish my exam on Mon has Q's like these. This isn't even A-Level standard surely.
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    (Original post by SimonM)
    STEP I, Question 8

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    \displaystyle \int_0^1 \frac{t}{n(n-t)} \, dt = \frac{1}{n} \int_0^1 \frac{t}{n-t} \, dt = \frac{1}{n} \int_0^1 \left ( -1 +\frac{n}{n-t} \right ) \, dt =
    \displaystyle \frac{1}{n} \left [ -t - n \ln ( n-t) \right ]_0^1 = \frac{1}{n} \left ( n \ln \left ( \frac{n}{n-1}\right) - 1 \right) = \boxed{\ln \left ( \frac{n}{n-1}\right) - \frac{1}{n}}

    However

    \displaystyle 0 \leq \frac{t}{n(n-t)} = \frac{1}{n-t} - \frac{1}{n} \leq \frac{1}{n-1} - \frac{1}{n}

    Therefore

    \displaystyle 0 \leq \int_0^1 \frac{t}{n(n-t)} \, dt  \leq \frac{1}{n-1} - \frac{1}{n}
    or
    \displaystyle \boxed{ 0 \leq \ln \left ( \frac{n}{n-1}\right) - \frac{1}{n}  \leq \frac{1}{n-1} - \frac{1}{n}} (as required)

    \displaystyle \sum_{n=2}^N 0 \leq \sum_{n=2}^N \left ( \ln \left ( \frac{n}{n-1}\right) - \frac{1}{n} \right ) \leq \sum_{n=2}^N \frac{1}{n-1} - \frac{1}{n}
    \displaystyle 0 \leq \sum_{n=2}^N \left ( \ln n -\ln (n-1) \right ) - \sum_{n=2}^N \frac{1}{n} \leq 1 - \frac{1}{n} \leq 1
    \displaystyle \boxed{0 \leq \ln N - \sum_{n=2}^N \frac{1}{n} \leq 1}

    \displaystyle \sum_{n=1}^{N} \frac{1}{n} \leq 1 + \sum_{n=2}^{10^{30}} \frac{1}{n} \leq 1+\ln 10^{30} \leq 1+\ln 2^{100} = 1+100 \ln 2

    Crudely approximating e &gt; 1+1 +\frac{1}{2} = \frac{5}{2}

    Therefore \ln 2 &lt; 1

    So \boxed{ \sum_{n=1}^{N} \frac{1}{n} &lt; 101}
    Haha yes the last part is basically a fiddle, though I noticed they didn't use the word prove, just 'show' ...

    I said N < 10^30 => lnN < ln(10^30) = 30ln10
    But using our inequality before: ln10 <= (1 + 1/2 + 1/3 + ... + 1/10) < 3
    So ln10 < 3 thus lnN < 90

    I then said the summation approximates well to lnN for large N, thus is only slightly more than lnN. In addition, we used ln10 < 3 in our deduction of lnN < 90 thus providing even more margin of error (albeit uneeded).

    Thus Summation < 90 + (0.05*90) < 101 (5% more than adequate margin for error)

    Dunno if that would do, they may say it's not rigorous enough, decent stab though, would expect something for my efforts. :p:
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    How do you know that 1+1/2+1/3+...+1/10 < 3? (It is, but it's close enough that I would want some fairly careful justification).

    The rest is blatantly unjustified but I assume you already knew that.

    To be honest, I wouldn't give you much for any of the waffle.
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    (Original post by DFranklin)
    How do you know that 1+1/2+1/3+...+1/10 < 3? (It is, but it's close enough that I would want some fairly careful justification).
    First 6 terms give it away.

    (Original post by DFranklin)
    The rest is blatantly unjustified but I assume you already knew that.
    I haven't proved it approximates well for large N, true. But based on that, I don't think it's unjustified. If you know it approximates well, how could the error be that large?

    (Original post by DFranklin)
    To be honest, I wouldn't give you much for any of the waffle.
    Apart from not proving the approximation, I don't think it's too bad. I think you're being a little harsh, though I would say that naturally!

    TBH I don't think either me or Simon knew how to do the Q properly, god knows what that thing about 2^10 was at the start.
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    (Original post by Physics Enemy)
    First 6 terms give it away.
    Not to me they don't. I honestly can't see any examiner accepting that as an explanation.

    I haven't proved it approximates well for large N, true. But based on that, I don't think it's unjustified. If you know it approximates well, how could the error be that large?
    You can't expect to get marks based on "It would be nice if this was true, so I'll say that it is true, and then I can answer the question".

    Apart from not proving the approximation, I don't think it's too bad. I think you're being a little harsh, though I would say that naturally!
    It's possible I'm being harsh - but if so, it's basically because you'd be getting "pity marks". That is: "this is all nonsense, but he's only an A-level student, we can't really expect any better".

    TBH I don't think either me or Simon knew how to do the Q properly, god knows what that thing about 2^10 was at the start.
    I don't see anything wrong with what Simon did, other than maybe a tiny bit more explanation about why ln 2 < 1.
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    (Original post by DFranklin)
    ...
    if he just said e^3>10 => 3>ln10 =>91>1+30ln10, then surely that would do, everyone knows e^3>10, but 125/8 does it quite easily.
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    (Original post by DFranklin)
    I don't see anything wrong with what Simon did, other than maybe a tiny bit more explanation about why ln 2 < 1.
    e>5/2>2 => ln e > ln 2 => 1 > ln 2
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    (Original post by Physics Enemy)
    TBH I don't think either me or Simon knew how to do the Q properly, god knows what that thing about 2^10 was at the start.
    The 2^10 thing was so you can use that 2^10 > 10^3
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    (Original post by SimonM)
    e>5/2>2 => ln e > ln 2 => 1 > ln 2
    Sure. What you'd written before was fine as an exam answer - I would not in a million years expect to see an examiner complain about it. But it was perhaps a little abrupt for people stuck and trying to understand how to do the question. :dontknow:
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    (Original post by DFranklin)
    Sure. What you'd written before was fine as an exam answer - I would not in a million years expect to see an examiner complain about it. But it was perhaps a little abrupt for people stuck and trying to understand how to do the question. :dontknow:
    I still don't understand how to do the Q lol. No offence to Simon, but when I read his solutions I don't understand what's going on at all. That's probably because he's very bright and answers it so well. That's why he's at Cambridge doing Maths.
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    Basically he says
    10^30=1000^10<1024^10
    Follow from there, (using the result given)
    the inequality is preserved by ln'ing both sides since ln is a stricly increasing function (if your concerned)
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    (Original post by toasted-lion)
    STEP III Question 13

    Now taking a proportion X of the cake gives probability X^4 that all 4 currents will be in that portion.

    So taking a portion X and getting all 4 currents in that portion has pdf:

     f(x) = 2x.x^4 = 2x^5

    Hence,  P(all4inportion) = \int_0^1 2x^5 dx = 1/3

    and,  P(all4inportionbiggerthanhalf) = \int_{1/2}^1 2x^5 dx = 1/3 - 1/192 = 63/192

    So finally  P(portionbiggerthanhalf \mid all4inportion) = (63/192)/(1/3) = 189/192
    The number of currants X in a portion x is actually \sim Po(4x)

    So the integrals are actually \displaystyle \int (2x)[e^{-4x}\frac{(4x)^{4}}{4!}] \, dx
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    (Original post by jasperleeabc)
    The number of currants X in a portion x is actually \sim Po(4x)

    but this gives P(X>4)>0.
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    (Original post by Get me off the £\?%!^@ computer)
    but this gives P(X>4)>0.
    Agreed. The original approach looks fine to me. (Poisson might make sense if the question said the average number of currents was exactly 4).
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    I don't think anything in the question is Poisson distributed. (The integral you have is not something you could reasonably be expected to evaluate, either).
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    Sorry my bad, I know what's actually wrong now.

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