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    Think I'm pretty screwed now... need a 1 for Uni offer but I'm only getting like 65-70. Considering it's a relatively easy paper compared to past papers, I feel like this year's grade boundary will be pretty high😢
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    (Original post by IrrationalRoot)
    I swear they stole Q6 from a Putnam competition lol.
    Nice looking paper though.
    EDIT: Lol @ Q13, STEP examiners having way too much fun again. Almost as bad as the 'crusty bread' question from long ago.
    q6 could easily be a tripos question on riemann integrability haha. what type of rigour dye think they wanted for i) me and zain thought it was bit weird
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    (Original post by physicsmaths)
    q6 could easily be a tripos question on riemann integrability haha. what type of rigour dye think they wanted for i) me and zain thought it was bit weird
    Yeah it's pretty weird. I imagine some sort of explicit statement that the integral is the area under the curve between those x values would be required.
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    STEP I Q12

    Solution
    (a)
    Let X and X_n denote the number of winning tickets purchased by the N participants, and the profit made by the organiser in the event of n winning tickets respectively.

    Clearly, the probability of a given participant choosing a losing number is 1-\frac{1}{N}. Thus, if all participants choose a losing ticket:

     \ \boxed{\mathbb{P} (X=0) = \left(1-\dfrac{1}{N}\right)^N}

    Hence:

    \begin{array}{lcl}

\mathbb{E}[\text{Profit}] &=& \displaystyle\sum_{n=0}^{N} X_n \mathbb{P}(X=n) \\

\\

&=& Nc \mathbb{P} (X=0) + (Nc-J) \mathbb{P}(X \geq 1) \\

\\

&=& Nc \mathbb{P}(X=0) + (Nc-J)(1-\mathbb{P}(X=0)) \\

\\

&=& Nc-J + J \mathbb{P}(X=0) \\

\\

&=& \boxed{Nc - J+ J \left(1-\dfrac{1}{N}\right)^N}

    If 2Nc = J, applying the approximation and the fact that e>2 yields:

     \ \mathbb{E}[\text{Profit}] \approx Nc(-1+2e^{-1}) < Nc (-1+2 \cdot 2^{-1}) = 0

    Hence the organiser should expect to make a loss if J=2Nc, as required.

    (b)
    Given that \gamma is the fraction of the numbers that are popular, there are N\gamma and N(1-\gamma) popular and unpopular numbers respectively. Requiring the probabilities to sum to unity yields the desired relationship between a, b, \gamma:

     \ N\gamma \mathbb{P}(\text{Choosing pop. no.}) + N(1-\gamma) \mathbb{P}(\text{Choosing unpop. no.}) = 1  \iff \boxed{\gamma a + (1-\gamma) b = 1} \ (\star)

    Note:

    \begin{array}{lcl}

\mathbb{P}(X=0) &=& \mathbb{P}(\text{N losers} \ | \ \text{Winner is popular}) \mathbb{P}(\text{Winner is popular}) \\

 && + \ \mathbb{P}(\text{N losers} \ | \  \text{Winner is unpopular}) \mathbb{P}(\text{Winner is unpopular})\\

\\

&=& \left(1-\dfrac{a}{N}\right)^N \gamma+ \left(1-\dfrac{b}{N}\right)^N (1-\gamma)\\

\\

&\approx& \gamma e^{-a} + (1-\gamma) e^{-b}

    Thus the approximate expected profit is given by:

    \begin{array}{lcl}

\mathbb{E}[\text{Profit}] &=& Nc-J + J \mathbb{P}(X=0) \\

\\

&\approx& Nc-J + J[\gamma e^{-a} + (1-\gamma)e^{-b}] \\

\\

&=& J\gamma e^{-a} + J(1-\gamma)e^{-b} + (Nc-J)

    Which is of the desired form with (A,B,C) =(J\gamma, J(1-\gamma), Nc-J).

    Finally, in the special case \gamma = 1/8, a=9b, (\star) easily gives (a,b) = (9/2, 1/2). Setting J=2Nc:

     \ \mathbb{E}[\text{Profit}] \approx Nc\left(\dfrac{1}{4}e^{-9/2} + \dfrac{7}{4} e^{-1/2} - 1 \right)> Nc \left(\dfrac{7}{4}3^{-1/2} - 1\right)

    But note (7/4)^2 = 49/16 > 48/16 = 3 \iff (7/4)3^{-1/2} - 1> 0 and it follows that the organiser should expect to make a profit in this case.



    I'm a little uneasy about my solution to the last part for three reasons - 1) the profit is extremely small, which may indicate a mistake in finding the relationship (\star) as it leaves little slack for handwritten approximation; 2) In STEP, one usually expects there to be a sequence of reasonable bounds to use in order to prove inequalities of this type, whereas I felt almost forced to make a 'brutish' calculation to reach the conclusion here; and 3) I'm too tired to check it carefully myself tonight.

    Thoughts?
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    (Original post by Farhan.Hanif93)
    STEP I Q12

    Solution
    (a)
    Let X and X_n denote the number of winning tickets purchased by the N participants, and the profit made by the organiser in the event of n winning tickets respectively.

    Clearly, the probability of a given participant choosing a losing number is 1-\frac{1}{N}. Thus, if all participants choose a losing ticket:

     \ \boxed{\mathbb{P} (X=0) = \left(1-\dfrac{1}{N}\right)^N}

    Hence:

    \begin{array}{lcl} 

\mathbb{E}[\text{Profit}] &=& \displaystyle\sum_{n=0}^{N} X_n \mathbb{P}(X=n) \\ 

\\ 

&=& Nc \mathbb{P} (X=0) + (Nc-J) \mathbb{P}(X \geq 1) \\ 

\\ 

&=& Nc \mathbb{P}(X=0) + (Nc-J)(1-\mathbb{P}(X=0)) \\ 

\\ 

&=& Nc-J + J \mathbb{P}(X=0) \\ 

\\ 

&=& \boxed{Nc - J+ J \left(1-\dfrac{1}{N}\right)^N}

    If 2Nc = J, applying the approximation and the fact that e>2 yields:

     \ \mathbb{E}[\text{Profit}] \approx Nc(-1+2e^{-1}) < Nc (-1+2 \cdot 2^{-1}) = 0

    Hence the organiser should expect to make a loss if J=2Nc, as required.

    (b)
    Given that \gamma is the fraction of the numbers that are popular, there are N\gamma and N(1-\gamma) popular and unpopular numbers respectively. Requiring the probabilities to sum to unity yields the desired relationship between a, b, \gamma:

     \ N\gamma \mathbb{P}(\text{Choosing pop. no.}) + N(1-\gamma) \mathbb{P}(\text{Choosing unpop. no.}) = 1  \iff \boxed{\gamma a + (1-\gamma) b = 1} \ (\star)

    Note:

    \begin{array}{lcl} 

\mathbb{P}(X=0) &=& \mathbb{P}(\text{N losers} \ | \ \text{Winner is popular}) \mathbb{P}(\text{Winner is popular}) \\ 

 && + \ \mathbb{P}(\text{N losers} \ | \  \text{Winner is unpopular}) \mathbb{P}(\text{Winner is unpopular})\\ 

\\ 

&=& \left(1-\dfrac{a}{N}\right)^N \gamma+ \left(1-\dfrac{b}{N}\right)^N (1-\gamma)\\ 

\\ 

&\approx& \gamma e^{-a} + (1-\gamma) e^{-b}

    Thus the approximate expected profit is given by:

    \begin{array}{lcl} 

\mathbb{E}[\text{Profit}] &=& Nc-J + J \mathbb{P}(X=0) \\ 

\\ 

&\approx& Nc-J + J[\gamma e^{-a} + (1-\gamma)e^{-b}] \\ 

\\ 

&=& J\gamma e^{-a} + J(1-\gamma)e^{-b} + (Nc-J)

    Which is of the desired form with (A,B,C) =(J\gamma, J(1-\gamma), Nc-J).

    Finally, in the special case \gamma = 1/8, a=9b, (\star) easily gives (a,b) = (9/2, 1/2). Setting J=2Nc:

     \ \mathbb{E}[\text{Profit}] \approx Nc\left(\dfrac{1}{4}e^{-9/2} + \dfrac{7}{4} e^{-1/2} - 1 \right)> Nc \left(\dfrac{7}{4}3^{-1/2} - 1\right)

    But note (7/4)^2 = (1.7+0.05)^2 = 2.89 + 0.17 + 0.0025 > 3 \iff 7/4> 3^{1/2} and it follows that the organiser should expect to make a profit in this case.



    I'm a little uneasy about my solution to the last part for three reasons - 1) the profit is extremely small, which may indicate a mistake in finding the relationship (\star) as it leaves little slack for handwritten approximation; 2) In STEP, one usually expects there to be a sequence of reasonable bounds to use in order to prove inequalities of this type, whereas I felt almost forced to make a 'brutish' calculation to reach the conclusion here; and 3) I'm too tired to check it carefully myself tonight.

    Thoughts?
    Pretty much got stuff along these lines (and the working looks pretty sound to my also tired eyes). But I mean I don't see the need for the decimal expansion. I would just write it as a fraction and the bound follows quickly no?
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    (Original post by 13 1 20 8 42)
    Pretty much got stuff along these lines (and the working looks pretty sound to my also tired eyes). But I mean I don't see the need for the decimal expansion. I would just write it as a fraction and the bound follows quickly no?
    Quite right. I'm clearly half asleep.
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    (Original post by physicsmaths)
    q6 could easily be a tripos question on riemann integrability haha.
    That's pushing it a bit... it's out of the STEP I comfort zone in terms of rigour but there's no way that would show up on Tripos.
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    (Original post by Lordtangent)
    Hi ppl
    I got all of Q3 Q4
    All of Q2 except for that (x/y is less than 1 and it still holds) bit
    First two integrals of Q1 (embarrassing)
    (i) half proved(ii) and (iv) of Q10
    Proved (i) and (ii) and obtained the three equations of the parameters of the line in Q6

    Do I have a shot at 1?
    (Original post by Lordtangent)
    Hi ppl
    I got all of Q3 Q4
    All of Q2 except for that (x/y is less than 1 and it still holds) bit
    First two integrals of Q1 (embarrassing)
    (i) half proved(ii) and (iv) of Q10
    Proved (i) and (ii) and obtained the three equations of the parameters of the line in Q6

    Do I have a shot at 1?
    (Original post by Lordtangent)
    Hi ppl
    I got all of Q3 Q4
    All of Q2 except for that (x/y is less than 1 and it still holds) bit
    First two integrals of Q1 (embarrassing)
    (i) half proved(ii) and (iv) of Q10
    Proved (i) and (ii) and obtained the three equations of the parameters of the line in Q6

    Do I have a shot at 1?
    Zacken
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    A tired answer to Q13
    Spoiler:
    Show




    Clearly  s_1 = 0 , since the first slice used cannot possibly be used as the second slice in a sandwich.

    Now, to make toast with the "r"th slice,  2 \leq r \leq n - 1 , he must have either made toast with the previous slice, or used the slice as the second one in a sandwich. The conditional probability upon doing either of these is p in each case, and so in total we have  t_r = t_{r-1} p + s_{r-1} p = (s_{r-1} + t_{r-1})p.

    Next, we think about the "r"th slice again,  2 \leq r \leq n and note that one of the following mutually exclusive events occurs:
    -This is the second slice in a sandwich.
    -The previous slice was the second slice in a sandwich.
    -The previous slice was used to make toast.

    Hence the associated probabilities here sum to 1, giving  s_r + s_{r-1} + t_{r-1} = 1 , from which the required result follows.

    Combining the equations, restricting to  2 \leq r \leq n - 1 , we obtain  t_r = (1 - s_r)p .
    Subbing back into the second equation, for  3 \leq r \leq n we have  s_r = 1 - (s_{r-1} + (1 - s_{r-1})p) = (1-p)(1 - s_{r-1}) = q(1 - s_{r-1}). Note also that  s_2 = 1 - s_1 - t_1 = 1 - 0 - p = 1 - p = q = q(1 - s_1) . So we have infact shown the required identity for  2 \leq r \leq n .

    Next, we use induction. The base case is immediate since  q - q = 0 = s_1 .
    In general, for  2 \leq r \leq n - 1 , the inductive hypothesis gives  \displaystyle s_r = q \left(1 - \frac{q + (-q)^{r-1}}{1 + q} \right) = q\left(\frac{1 - (-q)^{r-1}}{1 + q} \right) = \frac{q + (-q)^r}{1 + q} as required.

    Using  t_r = (1 - s_r)p we obtain  \displaystyle t_r = \left(1 - \frac{q + (-q)^r}{1 + q} \right)(1 - q) = \left(\frac{1 - (-q)^r}{1 + q} \right)(1 - q), 2 \leq r \leq n - 1 . Indeed, it is easily checked that this also holds for r = 1.

    As shown earlier,  s_n = q(1 - s_{n-1}) and the inductive method hence extends to  s_n . Therefore  s_n = \frac{q + (-q)^n}{1 + q} . Note that the following two events are mutually exclusive and exhaustive:

    -He makes toast with the nth slice.
    -He uses the nth slice as the second slice of a sandwich.

    Then  \displaystyle t_n = 1 - s_n = \frac{1 - (-q)^n}{1 + q} .











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    Apropos de not much, isn't question 2 a direct or near copy from an earlier STEP paper? I seem to have seen exactly this before.
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    (Original post by physicsmaths)
    q6 could easily be a tripos question on riemann integrability haha. what type of rigour dye think they wanted for i) me and zain thought it was bit weird
    It looks odd to me too. Doesn't STEP I assume only A level knowledge? If so, then the concept of continuity can't be assumed (surely there are A level students who have not heard the term?) and it's not clear what concept of continuity they are expecting - I based my answer to part i) on the idea that continuity => no jumps, so that a +ve f(b) implies a +ve integral in some neighbourhood of b. I think that's the most you could expect from an A level candidate.
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    (Original post by atsruser)
    It looks odd to me too. Doesn't STEP I assume only A level knowledge? If so, then the concept of continuity can't be assumed (surely there are A level students who have not heard the term?) and it's not clear what concept of continuity they are expecting - I based my answer to part i) on the idea that continuity => no jumps, so that a +ve f(b) implies a +ve integral in some neighbourhood of b. I think that's the most you could expect from an A level candidate.
    I feel like they ought to just give a brief informal definition of continuity. I had some idea of what the term meant when I set STEP but probably would have avoided this question due to being unsure and I imagine a lot of people would have.
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    Do people think the paper was easier/harder than last year's? Just trying to guess grade boundaries haha
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    (Original post by Zacken)
    That's pushing it a bit... it's out of the STEP I comfort zone in terms of rigour but there's no way that would show up on Tripos.
    this could easily be put in tripos ass a riemann integrability question. u know it
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    (Original post by physicsmaths)
    this could easily be put in tripos ass a riemann integrability question. u know it
    Are you mad fam?
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    (Original post by Zacken)
    Are you mad fam?
    are u mad
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    (Original post by atsruser)
    Apropos de not much, isn't question 2 a direct or near copy from an earlier STEP paper? I seem to have seen exactly this before.
    I seemed to think it was familiar as well - but the only thing I could find was
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    (Original post by 13 1 20 8 42)
    A tired answer to Q13 which I feel is wrong. The bounds in the question seemed at times unnecessary, and I seemed to be able to show things which trivialised the last parts. If someone could demonstrate why the argument breaks down that'd be cool, the gist of everything is here.

    Spoiler:
    Show














    Clearly  s_1 = 0 , since the first slice used cannot possibly be used as the second slice in a sandwich.

    Now, to make toast with the "r"th slice,  2 \leq r \leq n - 1 , he must have either made toast with the previous slice, or used the slice as the second one in a sandwich. The conditional probability upon doing either of these is p in each case, and so in total we have  t_r = t_{r-1} p + s_{r-1} p = (s_{r-1} + t_{r-1})p.

    Next, we think about the "r"th slice again,  2 \leq r \leq n and note that one of the following mutually exclusive events occurs:
    -This is the second slice in a sandwich.
    -The previous slice was the second slice in a sandwich.
    -The previous slice was used to make toast.

    Hence the associated probabilities here sum to 1, giving  s_r + s_{r-1} + t_{r-1} = 1 , from which the required result follows.

    Combining the equations, restricting to  2 \leq r \leq n - 1 , we obtain  t_r = (1 - s_r)p .
    Subbing back into the second equation, for  3 \leq r \leq n we have  s_r = 1 - (s_{r-1} + (1 - s_{r-1})p) = (1-p)(1 - s_{r-1}) = q(1 - s_{r-1}). Note also that  s_2 = 1 - s_1 - t_1 = 1 - 0 - p = 1 - p = q = q(1 - s_1) . So we have infact shown the required identity for  2 \leq r \leq n .

    Next, we use induction. The base case is immediate since  q - q = 0 = s_1 .
    In general, for  2 \leq r \leq n - 1 , the inductive hypothesis gives  \displaystyle s_r = q \left(1 - \frac{q + (-q)^{r-1}}{1 + q} \right) = q\left(\frac{1 - (-q)^{r-1}}{1 + q} \right) = \frac{q + (-q)^r}{1 + q} as required.

    Using  t_r = (1 - s_r)p we obtain  \displaystyle t_r = \left(1 - \frac{q + (-q)^r}{1 + q} \right)(1 - q) = \left(\frac{1 - (-q)^r}{1 + q} \right)(1 - q), 2 \leq r \leq n - 1 . Indeed, it is easily checked that this also holds for r = 1.

    As shown earlier,  s_n = q(1 - s_{n-1}) and the inductive method hence extends to  s_n . Therefore  s_n = \frac{q + (-q)^n}{1 + q} . Note that the following two events are mutually exclusive and exhaustive:

    -He makes toast with the nth slice.
    -He uses the nth slice as the second slice of a sandwich.

    Then  \displaystyle t_n = 1 - s_n = \frac{1 - (-q)^n}{1 + q} .





















    Agree with above solution. Here's an 'or otherwise' solution to get the second part directly:

    r^{th} slice makes a sandwich iff (r-1)^{th} slice starts a sandwich. This happens if, at the (r-1)^{th} slice, you choose to make a sandwich (with probability q) AND you have the option to choose (which you have so as long as your are not forced to finish a sandwich you started with the previous slice).

    Hence s_r = q(1-s_{r-1})
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    (Original post by Lordtangent)
    Zacken
    18 +18 + 14 + 10 + 8 + 8 - the last two numbers are somewhat arbitrary, I haven't done those questions myself, so take it with a pinch of salt. That could/should scrape a 1, I reckon.
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    (Original post by Zacken)
    18 +18 + 14 + 10 + 8 + 8 - the last two numbers are somewhat arbitrary, I haven't done those questions myself, so take it with a pinch of salt. That could/should scrape a 1, I reckon.
    Thanks m8 , I just hope the grade boundary wont be too atrocious this year .
 
 
 
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