Solution to STEP III Question 8 is not quite right. Since a_k-3 is used in the proof, you have to show that it is true for n = 3 and then true for all n >= 3

Reply 163

nerak99

13

A more complete solution to 2000 STEP 11 Q1 for future searchers.
By inspection the first value a is N+1. and b is N(N+1).
The rest I have put in spoiler tags

Spoiler

(edited 7 years ago)

Reply 164

Wu Yizhou2007

1

Beautiful solution. I think it's more of an "even if some approx or assumptions are different they are still gonna end up around 2/3" situation?
For the second part that's exactly what I thought of and I think the main idea is they want you to think lol there really isn't a right or wrong answer. And being aware of the randomness is important as well so I think this answer is completely justified

Original post by Glutamic Acid

III/12:

P(3 or fewer | Holding a winning ticket) = P(3 or fewer AND holding a winning ticket)/P(holding a winning ticket).

3 or fewer and you win: you win and two others win; you win and one other wins; you win and no others win. You winning and other people winning is independent, so P(3 or fewer AND holding a winning ticket)/P(holding a winning ticket) = P(2 others or fewer win).

The others can be modelled as a binomial distribution, X ~ B(2n-1, 1/n)
P(X <= 2) = P(X = 2) + P(X = 1) + P(X = 0)
=

\dbinom{2n-1}{2}\dfrac{1}{n^2} \left(1 - \dfrac{1}{n}\right)^{2n-3} + \dbinom{2n-1}{1}\dfrac{1}{n} \left(1 - \dfrac{1}{n}\right)^{2n-2} + \left(1 - \dfrac{1}{n}\right)^{2n-1}

=

\left(1 - \dfrac{1}{n}\right)^{2n-3}\left[\dfrac{(2n-1)(2n-2)}{2!n^2} + \dfrac{(2n-1)}{n}\left(1 - \dfrac{1}{n}\right) + \left(1 - \dfrac{1}{n}\right)^2 \right]

= \left(1 - \dfrac{1}{n}\right)^{2n-3}\left[5 - \dfrac{8}{n} + \dfrac{3}{n^2}\right] = \left(1 - \dfrac{1}{n}\right)^{2n}(5 - \dfrac{8}{n} + \dfrac{3}{n^2}) \div \left(1 - \dfrac{1}{n}\right)^3 = \left(1 - \dfrac{1}{n}\right)^{2n} \dfrac{5n^2 - 3n}{n^2 - 2n + 1} = \left(1 - \dfrac{1}{n}\right)^{2n} \dfrac{5 - 3/n}{1 - 2/n + 1/n^2}

As n gets large,

\left(1 - \dfrac{1}{n}\right)^{2n} \approx e^{-2} \text{ and } \dfrac{5 - 3/n}{1 - 2/n + 1/n^2} \approx 5/1 = 5

So the probability is approximately

5e^{-2}

.

5e^{-2} \approx 5(1 - 2 + \frac{4}{2} - \frac{8}{6} + \frac{16}{24} - \frac{32}{120} + \frac{64}{720} - \frac{128}{5040}) = \frac{41}{63} \approx \frac{2}{3}

.

The last part is slightly dodgy, plus the arithmetic was rather tedious. I've been light on the algebra at several parts; none of it's particularly difficult.

- Human selections of numbers won't be uniform, and some will be more popular than others. So given that I have a winning ticket, it's likelier that more others will have a winning ticket so this probability will decrease. (Not sure if this is what they want or even correct.)

Consider the other people playing: X ~ B(2N - 1, 1/N) E(X) = "np" = 1/N*(2N-1) = 2 - N^-1. So in total, given that you have a winning ticket, the expectation is 3 - N^-1.

Reply 165

Wu Yizhou2007

1

sure, so long as you got the correct answer in this way. i did it like that as well

Original post by Renewhj

Guys, for the first part of question 2 in paper III, can you do it this way?

Reply 166

CoffeeRob

1

Step 1 2000- Question 2
So for the second bit you can convert the (x^4-x^-2) into (x^2-1) * (x^2+1+x^-2) and the (x-x^-1) into (x^-1) * (x^2-1), meaning that the original expression can be refactorised from (x^4-x^-2)^5 * (x-x^-1)^6 into (x^2-1)^11 * (x^2+1+x^-2)^5 * x^-6 and then into (x^2-1)^11 * (x^4+x^2+1)^5 * x^-16. From this we can realise that the coefficients are the same as in the original expression but of order 16 more, so the coefficient of x^12 is the same as the coefficient of x^-4 from the original expression and the coefficient of x^38 is the same as the coefficient of x^22 from the original expression. Therefore from part a) the x^-4 (x^12) coefficient is -15 and the x^22 (x^38) coefficient is 15 as desired.

(edited 5 years ago)

Reply 167

Yourface1738

1

\displaystyle (2a-N)(2b-N) = N^2 for \displaystyle \frac{2}{N}

Reply 168

UndeadBalazar

1

Step 2 2000 Question 1:
In case anyone cares 11 years on, 2/N is identical to 1/N except you factorise it to (2a-N)(2b-N) = N^2

(edited 3 years ago)

Reply 169

Sxynix

Original post by DFranklin

In that case, you need to show that k exists and is > 0 (since you take the log of it). But it was sloppy reading on my part as well, I admit.

So now I know what k is supposed to be, let's look again:

Two problems here: firstly 201.029996... isn't really a number, (or if it is, I don't know what number it's supposed to be) and yet you're treating it like it is. That's not just me being picky - all they've given you is an approximation to log 2, and for full marks you would need to make sure you only treat it as an approximation and not an exact value.

But the killer if k is big (as it is here), it's almost certain that k and k+1 start with the same digit, so k* = (k+1)*. And yet you argue that log k* < 0.029996 and log (k+1)* > 0.029996. I don't see any reason to think this will be true.

log 2^1000 = 1000 log 2. So log 2^1000 > 301, so 2^1000 > 10^301. And log 2^1000 < 301.1 < 301 + log 2, so 2^1000 < 2 * 10^301.

So 10^301 < 2^1000< 2*10^301 and so the first digit is 1.

I understand why the first part of the inequality is true ( the statement that 2^1000 > 10^301 ) but I'm not quite sure on the statement that 2*10^301 > 2^1000; could you elaborate on that a little more please?

Reply 170

DFranklin

18

Original post by Sxynix

I understand why the first part of the inequality is true ( the statement that 2^1000 > 10^301 ) but I'm not quite sure on the statement that 2*10^301 > 2^1000; could you elaborate on that a little more please?

Taking all logs to base 10:

log 2 = 0.301029996 to 9 d.p., so log 2 < 0.30103, so log 2^1000 < 301.03
But we also have that log 2 > 0.3, and so log(2*10^301) = 301 + log 2 > 301.3

So log 2^1000 < 301.03 < 301.3 < log (2 * 10^301).

So log 2^1000 < log (2 * 10^301)

So 2 ^ 1000 < 2 * 10^301.

Reply 171

Ethan Li

2

Original post by Oh I Really Don't Care

II / 8

\displaystyle - \int \frac{dy}{(y+3)^\frac{1}{2}} = -2 \int -2xe^{-x^2} dx \implies -2(y+3)^{\frac{1}{2}} = -2e^{-x^2} + C \implies y = e^{-2x^2} + 2ke^{-x^2} + (k^2 - 2)

After subbing (0,6) ;

\displaystyle (k+4)(k-2) = 0

\displaystyle \therefore y = e^{-2x^2} - 8e^{-x^2} + 13

and case 2

\displaystyle y = e^{-2x^2} + 4e^{-x^2} + 1

There is a vlaue of k such that as x tends to

\displaystyle \infty

y tends to

\displaystyle 1

(ii) wip. Got

\displaystyle y^2e^{-3x^2}

but the answer requires

\displaystyle ye^{-3x^2}

for the second part my answer is 1/6 , but i didn't use the hint given , is there anyone get the same answer?? Cambridge pro. are too lazy, they dont even write the solutions leaving poor students behind

Reply 172

Ethan Li

2

Original post by Renewhj

Guys, for the first part of question 2 in paper III, can you do it this way?

sure! i believe it's the simplest way to do it by using the hint.

Reply 173

RuairidhW

2

If we select k=2 we can rearrange for ((y+3)^2)/e^6x^2 = 2/12 + c/e^6x^2 , then we can square root both sides, (I apologise for the lack of latex) this can give us ye^-3x^2 and it will tend to sqrt(1/6) or -sqrt(6) my only issue is that I think that’s only true for y≠-3, I’m not sure but in that case I think that we may just get a vertical line at some x value.

Original post by Oh I Really Don't Care

II / 8

\displaystyle - \int \frac{dy}{(y+3)^\frac{1}{2}} = -2 \int -2xe^{-x^2} dx \implies -2(y+3)^{\frac{1}{2}} = -2e^{-x^2} + C \implies y = e^{-2x^2} + 2ke^{-x^2} + (k^2 - 2)

After subbing (0,6) ;

\displaystyle (k+4)(k-2) = 0

\displaystyle \therefore y = e^{-2x^2} - 8e^{-x^2} + 13

and case 2

\displaystyle y = e^{-2x^2} + 4e^{-x^2} + 1

There is a vlaue of k such that as x tends to

\displaystyle \infty

y tends to

\displaystyle 1

(ii) wip. Got

\displaystyle y^2e^{-3x^2}

but the answer requires

\displaystyle ye^{-3x^2}

STEP I, II, III 2000 solutions

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