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STEP maths I, II, III 2007 solutions

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Reply 80
Avatar for Dystopia
Dystopia
10
STEP II, Q11.

Spoiler



Am I to assume that STEP II, Q1 has already been 'claimed'?
Reply 81
Avatar for ukgea
ukgea
6
Dystopia
Am I to assume that STEP II, Q1 has already been 'claimed'?


Yes, by insparato in this post.

But yeah, it's not like II/1 was hard, so you better cough up a solution! :rolleyes:

And Dystopia, your solution to II/8 looks structured and fine to me. :smile:
Reply 82
Avatar for Dystopia
Dystopia
10
STEP II, Q10.

Note: the front page says this question has been started, but that should be STEP I, Q10.

I have uploaded a diagram to make things clearer.

Spoiler

STEPDiagram.gif7.3KB
Reply 83
Avatar for coffeym
coffeym
10
Dystopia
STEP II, Q10.

Note: the front page says this question has been started, but that should be STEP I, Q10.

I have uploaded a diagram to make things clearer.

Spoiler



That drawing is nice, what program do you use?
Reply 84
Avatar for Dystopia
Dystopia
10
coffeym
That drawing is nice, what program do you use?

Thanks. :smile:

Anyway, I used Adobe Photoshop CS3. I recently needed to download the trial version for something else, and thought I'd make the most of it before the trial period ran out.

I'm sure there are other programs more suitable for the job, but for me it was a choice between that and Paint. :p:
As noone seems eager to do any stats I've had a look at step I:

Anyone got a hint for I/13 vi) ?


Also, for I/14, how are we meant to justify that there is just one value? Just by sketching the graphs? As for finding λ\lambda, how do I do that:redface: ?


(yes I'm hopeless at statistics!)
Reply 86
Avatar for Kolya
Kolya
17
Did you do I/12, nota?
Reply 87
Avatar for SsEe
SsEe
13
I've got a solution to II/1. I know insparato has also mentioned it though. Should I post mine?
Yeah, i know what ive got to do, i just havent got the time atm.
Reply 89
Avatar for SsEe
SsEe
13
STEP II, Q1

Spoiler

Lusus Naturae
Did you do I/12, nota?

Nope, haven't done that one - feel free to post up an answer:biggrin:
I had a good try, didn't quite get there though :frown:
Reply 92
Avatar for SsEe
SsEe
13
STEP II, Q13

Spoiler

Reply 93
Avatar for Kolya
Kolya
17
calcium878
I had a good try, didn't quite get there though

I assume you are talking about I/12? I am quite confident over my expression for the probability that the second pick is red, but I can't transform the bloody thing into what I want! (What I want being ap+bqN\frac{ap + bq}{N}) Perhaps it is my method of approaching the question that is at fault; I may have the right expression but it is too unwieldy to manipulate, and, actually, a far simpler solution exists and is for what the examiners are looking.
Yeah, something along those lines ;yes; Just comes out in a big annoying mess :s: Must be missing something simple...
Reply 95
Avatar for ukgea
ukgea
6
Okay I've finally done with it I/12

(i)

P(First sweet is red) = a/N

P(Second sweet is red) =

= P(Second sweet is red AND first sweet is red) + P(Second sweet is red AND first sweet is not red)

= P(Second sweet is red GIVEN first sweet is red) P(first sweet is red) + P(Second sweet is red GIVEN first sweet is not red) P(first sweet is not red)

=a−1N−1aN+aN−1N−aN \displaystyle = \frac{a-1}{N-1}\frac aN + \frac a{N-1} \frac{N-a}N

=aNa−1+N−aN−1\displaystyle = \frac aN \frac{a-1 + N - a}{N-1}

= a/N

= P(first sweet is red)

as required.

(ii) Now, for this part, there's a hard way and an easy way. The hard way is really rather easy, but seriously boring, and there's no way I'm gonna LaTeX it up. (Basically, just consider all the 16 different cases that can arise and their associated probabilities; and draw a tree, for heavens sake, otherwise it's impossible to keep track of things). But the hard way basically boils down to showing that

paa−1N−1+pqab+1N+1+p2(N−a)aN−1+pq(N−a)bN+1+pqba+1N+1+q2bb−1N−1+pq(N−b)aN+1+q2(N−b)bN−1=pa+qbp^a\frac{a-1}{N-1}+pqa\frac{b+1}{N+1}+p^2(N-a)\frac a{N-1} + pq(N-a)\frac b{N+1} + pqb\frac{a+1}{N+1} + q^2b\frac{b-1}{N-1} + pq(N-b)\frac a{N+1} + q^2(N-b) \frac b{N-1} = pa + qb

which, as the interested reader will unavoidably find, is true. (Just group the pq terms together, the p^2 terms together and the q^2 terms together and it comes out fairly easily, note that pa + qb = (pa + qb)(p + q) for the very last step)

Now, that was the hard, or rather very boring method. There is of course an easy method too, the only catch being it's probably impossible to find without first churning through the hard method of solution, unless you have Genuine Statistical Insight. But anyway:

Spoiler

Reply 96
Avatar for SsEe
SsEe
13
STEP III, Q12... Attempt 3. Most likely of many.

Spoiler

SsEe
STEP III, Q12... Attempt 3. Most likely of many.For what it's worth, that's the same answer I got. (It's also essentially the same method, so we might have made the same mistake...)

Now for STEP III, Q14: The question wording is slightly unclear: it could be that the "smallest square" is supposed to have the same orientation as the square dart board, or it could be that it is allowed to have arbitrary orientation. A strict reading of the question would imply the latter, but I don't see any approach to tackling the question in that case. Any thoughts?
Reply 98
Avatar for SsEe
SsEe
13
DFranklin
For what it's worth, that's the same answer I got. (It's also essentially the same method, so we might have made the same mistake...)

Now for STEP III, Q14: The question wording is slightly unclear: it could be that the "smallest square" is supposed to have the same orientation as the square dart board, or it could be that it is allowed to have arbitrary orientation. A strict reading of the question would imply the latter, but I don't see any approach to tackling the question in that case. Any thoughts?


As it says. I've had 3 attempts. First one used an incorrect method (though I had been awake for the previous 24 hours to be fair). Next one was this method but with a slip up. This one I corrected the slip up.
Each time I've done it, I've put n=1 as a sanity check and it's been ok. Got unlucky there!

For Q14, that's the reason I didn't try it. Can't see any way of approaching an arbitrary orientation either. For now I'd read it as having the same orientation just to get something down.
STEP III, Q13:

(i) We can think of pn(j)p_n(j) as the probability that the frog travels less than n meters by jump j−1j-1 and at least n meters by jump j. So p2(2)p_2(2) is probability that the frog travels < 2 m on its first jump and at least 2m by the 2nd jump. So the first jump must have been 1m (with prob p), and then any jump on the 2nd jump is sufficient. So p2(2)=pp_2(2) = p.

(ii) Starting 1−121-\frac{1}{2}m from the pond, any jump will land in the pond, so u1u_1 = 1.

Starting 2−122-\frac{1}{2}m from the pond, the chance of splashing in one jump is q, and the chance of splashing in two jumps is p. So u2=q+2p=2−qu_2 = q+2p = 2-q.

Starting 3−123-\frac{1}{2}m from the pond, getting to the pond in 1 jump is impossible (so p3(1)=0p_3(1) = 0). 3 jumps are required if and only if the first 2 jumps are small (so p3(3)=p2p_3(3) = p^2). So p(2)=1−p3(1)−p3(2)=1−p2p(2) = 1 - p_3(1)-p_3(2)= 1-p^2.

So u3=2(1−p2)+3p2=2+p2=3−2q+q2u_3 = 2(1-p^2)+3p^2 = 2+p^2 = 3-2q+q^2.

(iii)u1=A+B+C=1, u2=−qA+B+2C=2−q, u3=q2A+B+3C=3−2q+q2u_1 = A+B+C = 1,\,u_2 = -qA+B+2C=2-q,\,u_3=q^2A+B+3C=3-2q+q^2.

So u1−u2=(1+q)A−C=q−1u_1-u_2 = (1+q)A-C = q-1, u3−u2=q(1+q)A+C=1−q+q2u_3-u_2 = q(1+q)A+C=1-q+q^2.

Adding these two gives (1+q)2A=q2  ⟹  A=q2(1+q)2(1+q)^2A=q^2 \implies A=\frac{q^2}{(1+q)^2}.

Then C=1−q+(1+q)A=1−q+q21+q=(1−q)(1+q)+q21+q=11+qC = 1-q+(1+q)A = 1-q+\frac{q^2}{1+q} = \frac{(1-q)(1+q)+q^2}{1+q} = \frac{1}{1+q}.

Finally B=1−A−C=(1+q)2−q2−(1+q)(1+q)2=q(1+q)2B = 1-A-C = \frac{(1+q)^2-q^2-(1+q)}{(1+q)^2}=\frac{q}{(1+q)^2}

For large n, the term in C dominates and so un≈n1+q=np+2qu_n \approx \frac{n}{1+q} = \frac{n}{p+2q}.

This is to be expected because the expected distance travelled per jump is p+2qp+2q, so if dnd_n is the distance travelled in n jumps then E(dn)=n(p+2q)E(d_n) = n(p+2q). Moreover, by the central limit theorem, dn≈N(n(p+2q),npq)d_n \approx N(n(p+2q), \sqrt{npq}). This tells us that 99% of d_n lies within n(p+2q)±3npqn(p+2q)\pm 3\sqrt{npq}, or more relevantly, for any ϵ>0\epsilon > 0, P(∣dn−n(p+2q)∣<ϵn)→1asn→∞P(|d_n-n(p+2q)| < \epsilon n) \to 1 as n\to \infty. So after 'n' jumps we are almost certain to have travelled a distance n(p+2q)±ϵnn(p+2q)\pm\epsilon n. It follows that it is almost certain to require np+2q±ϵn\frac{n}{p+2q}\pm \epsilon n to travel a distance 'n', and so ∣E(jn)−np+2q∣<ϵn|E(j_n) - \frac{n}{p+2q}| < \epsilon n. Thus jn≈np+2qj_n \approx \frac{n}{p+2q}.

Comment: I found it quite hard to know what they were expecting for the last bit. Intuitively, the expected distance travelled is n(p+2q), and the CLT (or Chebychev's inequality) says this is a good enough estimate that we can invert it to say the expected number of jumps is to travel a distance n is n/(p+2q). But I don't really see a way of making this argument 'rigourous' that's not a lot more work than the actual question.

On a slightly different note: If we set u0=u−1=0u_0=u_{-1} = 0 then it's fairly obvious that un=pun−1+qun−2+1u_n = p u_{n-1}+q u_{n-2}+1, so un−pun−1−qun−2=1u_n-pu_{n-1}-qu_{n-2} = 1. The aux equation is x2−px−q=0x^2-px-q = 0, or (x+q)(x−1)=0(x+q)(x-1)=0. Set u_n = Cn for the P.I. and find Cn−p(Cn−1)−q(Cn−2)=1Cn-p(Cn-1)-q(Cn-2) = 1, so (p+2q)C=1(p+2q)C = 1, or C=(1+q)−1C=(1+q)^{-1} for a general solution un=A(−q)n+B+n(1+q)−1u_n = A(-q)^n + B +n(1+q)^{-1}. Then u0=0u_0 = 0 gives A+B=0  ⟹  B=−AA+B=0 \implies B=-A, while u−1=0u_{-1} = 0 gives 0=−A/q+B−1/(1+q)0 = -A/q + B -1/(1+q), so −A/q−A=1/(1+q)  ⟹  A=−q/(1+q)2-A/q -A = 1/(1+q) \implies A = -q/(1+q)^2. This is the same solution as that found in the question (only we have derived the formula for unu_n rather than being given it).

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