STEP I, II, III 2002 Solutions

2002 STEP II question 14

\text{Area of region }R_n \text{ is } \pi \left[n^2-(n-1)^2 \right]= \pi(2n-1)

\text{Probability that observer selects region }R_n \text{ is } k\pi(2n-1) \text{ where } k\pi N^2=1 \Rightarrow k=\dfrac{1}{\pi N^2}

\text{so probability that he selects }R_n \text{ is } \dfrac{2n}{N^2}

\text{overall average number of accidents is thus }\displaystyle \sum_{n=1}^N \left( \dfrac{2n-1}{N^2} \right) \left( 2- \dfrac {n}{N} \right)= \displaystyle \sum_{n=1}^N \left( \dfrac {2n-1}{N^2} \right) \left( \dfrac{2N-n}{N} \right)

= \dfrac{1}{N^3} \displaystyle \sum_{n=1}^N (4nN-2N-2n^2+n)=\dfrac{4}{N^2} \displaystyle \sum_{n=1}^Nn-\dfrac{2N^2}{N^3}-\dfrac{2}{N^3} \displaystyle \sum_{n=1}^Nn^2+\dfrac{1}{N^3} \displaystyle \sum_{n=1}^Nn

= \dfrac{4}{N^2} \dfrac{N(N+1)}{2}- \dfrac{2}{N}- \dfrac{2}{N^3} \dfrac{N(N+1)(2N+1)}{6}+ \dfrac{1}{N^3} \dfrac{N(N+1)}{2}= \dfrac{2(N+1)}{N} - \dfrac{2}{N} - \dfrac{(N+1)(2N+1)}{3N^2}+ \dfrac{N+1}{2N^2}

=2- \dfrac{(N+1)(2N+1)}{3N^2}+ \dfrac{N+1}{2N^2}=2- \dfrac{4N^2+3N-1}{6N^2}=2- \dfrac{(N+1)(4N-1)}{6N^2}=2-\dfrac{1}{6} \left(1+\dfrac{1}{N} \right) \left(4-\dfrac{1}{N} \right)

Pr (X=k)= \displaystyle \sum_{n=1}^N \dfrac{2n-1}{N^2} \times \dfrac{ \text{e}^{-2+n/N}(2-n/N)^k}{k!}= \dfrac{ \text{e}^{-2}N^{-k-2}}{k!} \displaystyle \sum_{n=1}^N(2n-1)(1N-n)^k \text{e}^{n/N}

Unparseable latex formula:

\text{with }N=3, Pr(X=2)= \dfrac{\text{e}^{-2}}{2.3^4} \displaystyle \sum_{n=1}^3(2n-1)(6-n)^2\text{e}^{n/3}= \dfrac{1}{162\text{e}^2}\left[25\text{e}^{1/3}+48\text{e}^{2/3}+45\text{e} \right]}

Pr(R_2 \text{ selected and }X=2)= \dfrac{3}{N^2} \times \dfrac{\text{e}^{-4/3} (4/3)^2}{2!} c \text{ hence}

Pr(R_2 \text{ selected }|X=2)= \dfrac{\frac{1}{3} \times \frac{16}{18}\text{e}^{-4/3}}{\frac{1}{162\text{e}^2}[25\text{e}^{1/3}+48\text{e}^{2/3}+45\text{e}]}= \dfrac{48}{48+45 \text{e}^{1/3}+25 \text{e}^{-1/3}}

Reply 142

2002 STEP III question 9

\text{The problem is equivalent to finding the centre of mass of the solid on the righrt of the diagram}

Unparseable latex formula:

\text{with respect to axes OA and OB where the volume is }k^3 \text{, i.e. }a^2b+ \dfrac{1}{2}a^3 \tan \tehta= ka^3 \Rightarrow b= \dfrac{a(2k-\tan \theta)}{2}

\text{providing }2k> \tan \theta \text{ i.e. }k> \dfrac{1}{2} \tan \theta

\text{splitting into a rectangular block and a triangular prism we have distance of c of m from OA given by }

Unparseable latex formula:

ka^3 \bar{x}=a^2b \times \dfrac{a}{2}+\ \dfrac{1}{2}a^3 \tan \thewta \times \left(b+\dfrac{a \tan \theta}{3} \right)= \dfrac{1}{2}a^3b+ \dfrac{1}{6}a^4 \tan \theta= \dfrac{1}{4}a^4(2k- \tan \theta)+ \dfrac{1}{6}a^4 \tan \theta

\Rightarrow \dfrac{\bar{x}}{a}=\dfrac{1}{2}- \dfrac{ \tan \theta}{12k}\text{ and similarly distance from OB is given by}

ka^3 \bar{y}=a^2b \times \dfrac{1}{2}b+ \dfrac{1}{2}a^3 \tan \theta \times \left(b+ \dfrac{a \tan \theta}{3} \right)= \dfrac{1}{2}a^2b^2+ \dfrac{1}{2}a^3b \tan \theta+ \dfrac{a^4 \tan^2 \theta}{6}

Unparseable latex formula:

\text { so } \drac{\bar{y}}{a}= \dfrac{k}{2}+ \dfrac{\tan \theta}{24k} \text{ as required}

\text{If }k< \dfrac{1}{2} \tan \theta \text{ then the volume of fluid is a triangular prism, (see second diagram)}

\text{Let base now be a rectangle }a \times b \text{ then } \dfrac{1}{2}ab^2 \tan \theta=ka^3 \text{ i.e. }b^2= \dfrac{2ka^2}{\tan \theta}

\text{We now have } \dfrac{ \bar{x}}{a}= \dfrac{b}{3a}= \dfrac{\sqrt{2k}}{3\sqrt{ \tan \theta}} \text{ and } \dfrac{ \bar{y}}{a}= \dfrac{b \tan \theta}{3}a= \dfrac{\sqrt{2k\tan\theta}}{3}

\text{In this latter case container will topple if } \dfrac{ \bar{x}}{y}< \tan \theta \Rightarrow \tan \theta > \dfrac{ \sqrt{2k}}{3 \sqrt{ \tan \theta}}\times \dfrac{3}{ \sqrt{2k\tan \theta}}= \dfrac{1}{ \tan \theta}

\text{i.e. if }\tan^2 \theta<1 \Rightarrow \theta>45 ^\circ

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Reply 143

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2002 STERP III question 10

\text{By conservation of energy, if angular velocity is } \dot{\theta} \text{ and the second particle}

\text{does not slip then we have, loss of PE }=mga\sin \theta-mga(1- \cos \theta)

\text{Gain of KE is }ma^2 \dot{\theta}^2 \text{ hence } a \dot{\theta}^2=g(\sin \theta+ \cos \theta-1) \Rightarrow \dot{\theta}^2= \dfrac{g}{a}(\sin \theta+\cos \theta-1)

\text{Differentiating w.r.t. time we have }2 \dot{\theta} \ddot{\theta}=\dfrac{g}{a}( \cos \theta- \sin \theta) \dot{\theta} \text{ i.e. } \ddot{\theta}= \dfrac{g}{2a} (\cos \theta- \sin \theta)

Unparseable latex formula:

\text{For second particle not to slip we must have }F-mg \sin \theta=ma \ddot{\theta}= \dfrac{mg(\cos \thetra- \sin \theta)}{2}

\text{also }R-mg \cos \theta=m \dot{\theta}^2a \Rightarrow R=mg(\sin \theta+2 \cos \theta-1) \text{ and } F= \dfrac{mg(\cos \theta+ \sin\theta)}{2}

\text{ so when } \theta=60^\circ,R=mg \dfrac{\sqrt3}{2} \text{ and }F=\dfrac{1}{2}mg \left( \dfrac{1+\sqrt3}{2} \right)

\text{Particle is about to slip if }F> \mu R \Rightarrow \dfrac{1+\sqrt3}{2}>\mu \sqrt3 \Rightarrow \mu< \dfrac{1+\sqrt3}{2\sqrt3}= \dfrac{3+\sqrt3}{6}

\text{so since this is the given value of }\mu, \text{ it follows that particle slips when cylinder has rotated through }60^\circ

2002 STEP III question 11

\text{By conservation of momentum and equation of restitution we have }u_1 \cos \theta=v_1 \cos(\theta-30^\circ)

\text{ and } eu_1 \sin \theta=v_1 \sin(\theta-30^\circ)

\text{so } \dfrac{u_1}{v_1}= \dfrac{ \cos(\theta-30^\circ)}{\cos\theta} \text{ and } e\tan \theta= \tan(\theta-30^\circ)

\text{Clearly the same equations apply at }N_1 \text{ so angle of incidence at each impact is the same}

\text{Let distances between successive impacts be }d_1,d_2,d_3 \text{ etc and the times to treavel these distances be }

t_1,t_2,t_3 \text{ etc then } \dfrac{t_1}{t_2}= \dfrac{d_1}{v_1} \div \dfrac{d_2}{v_2}= \dfrac{d_1v_2}{d_2v_1} \text{ and by the sine rule }\dfrac{d_1}{d_2}= \dfrac{\sin \theta}{\sin(\theta-30^\circ)}

\text{ so } \dfrac{t_1}{t_2}= \dfrac{\sin \theta}{ \sin(\theta-30^\circ)} \times \dfrac{cos \theta}{ \cos(\theta-30^\circ)} \text{ i.e. }t_2= \dfrac{\sin(\theta-30^\circ)\cos(\theta-30^\circ)}{\sin\theta \cos\theta}t_1

\text {so the times form a G.P. with common ratio } \dfrac{\sin(\theta-30^\circ)\cos(\theta-30^\circ)}{\sin\theta \cos\theta}

\text{Hence total time is sum of a G.P. with common ratio }\dfrac{\sin(2\theta-60^\circ)}{\sin{2\theta}}

\text {so total time is finite if }\dfrac{\sin(2\theta-60^\circ)}{\sin{2\theta}}<1 \text{ and clearly we must have }\theta>30^\circ

\text{ Now }30^\circ<\theta<60^\circ \Rightarrow60^\circ<2\theta<120^\circ \text{ and }0^\circ<2\theta-60^\circ<60^\circ \Rightarrow \sin(2\theta-60^\circ)< \dfrac{\sqrt3}{2}

Unparseable latex formula:

\text{ and }\sin{2\theta}> \dfrac{\sqrt3}{2} \text{ so } \dfrac{\sn(2\theta-60^\circ)}{\sin{2\theta}}<1 \text{ as required}

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Reply 145

2002 STEP III question 12

\text{Let }A,B,C \text{ be the events } {H},{TH} \text { and } {HT} \text{ respectively}

\text{if }X \text{ is the number of tosses required then E}(X)=\text{E}(X|A).Pr(A)+\text{E}(X|B).Pr(B) +\text{E}(X|C).Pr(C)=E \text{ say}

\text{after }A \text{ or }B, \text{ the number of tosses required is still }E \text { and }Pr(A)=p,Pr(B)=p(1-p),Pr(C)=(1-p)^2

\text{hence }E=p(E+1)+p(1-p)(E+2)+2(1-p)^2 \text{ i.e. }E=pE+p+pE-p^2E+2p-2p^2+2-4p+2p^2

\Rightarrow E(1-2p+p^2)=2-p \text{ hence }E= \dfrac{2-p}{1-2p+p^2}= \dfrac{2-p}{(1-p)^2}

\text{Similarly, if }W\text{ is the expected winnings we have }W=p(W+1)+p(1-p)(W+1)+(1-p)^2.0

Unparseable latex formula:

\text{ i.e. } W=pW+p+pW-p^2W+p-p^2n \RightarrowW= \dfrac{2p-p^2}{1-2p+p^2}= \dfrac{2p-p^2}{(1-p)^2}

\text{In an analogous manner, if the expected number of throws to obtain }r\text{ successive tails is } R

\text{then }R=p(R+1)+p(1-p)(R+2)+p(1-p)^2(R+3)+\dots+r(1-p)^r

=p(R+1)+pq(R+2)+pq^2(R+3)+\dots+rq^r \text { by putting }1-p=q

=pR(1+q+q^2+\dots+q^{r-1})+p+2pq+3pq^2+\dots+rpq^{r-1}+rq^r

= \dfrac{pR(1-q^r)}{1-q}+p(1+2q+3q^2+\dots+rq^{r-1})+rq^r= \dfrac{pR(1-q^r)}{1-q}+p\left[ \displaystyle \sum_{t=1}^rtq^{t-1}\right]+rq^r

=\dfrac{pR(1-q^r)}{1-q}+p\left[\dfrac{d}{dq} \displaystyle \sum_{t=1}^rq^t\right]+rq^r = \dfrac{pR(1-q^r)}{1-q}+p\dfrac{d}{dq}\left(\dfrac{q(1-q^r)}{1-q}\right)+rq^r

=R-Rq^r+\dfrac{p(1-q)[1-(r+1)^r]+pq(1-q^r)}{(1-q)^2}+rq^r \Rightarrow R= \dfrac{(1-q)[1-(r+1)q^r]+q(1-q^r)+rq^r(1-q)}{q^r(1-q)}

= \dfrac{1-q-(r+1)q^r+q(r+1)q^r+q-q^{r+1}+rq^r-rq^{r+1}}{pq^r}=\dfrac{1-q^r}{pq^r} \text{ as required}

(edited 11 years ago)

Reply 146

2002 STEP III question 13

Unparseable latex formula:

\text{If }X_1,X_2 \text{ are independent random variables with }X_1 \sim \text{exp}(\lambda_1) \text{ and }X_2 \sim \text{exp}(\lambda_2) \terxt{ then }

Pr(X_1 \geq X_2)=1-\displaystyle \int_0^x \lambda_1\text{e}^{-\lambda_1t}dt=1-\left[-\text{e}^{-\lambda_1t}\right]_0^x=1+\text{e}^{-\lambda_1x}-1=\text{e}^{-\lambda_1x}

\text{so }Pr(X_1 \text{ or }X_2<x)=1-\text{e}^{-\lambda_1x}\text{e}^{-\lambda_2x}=1-\text{e}^{-(\lambda_1+\lambda_2)x}

\text{which is thus the cumulative probability function for }X

\text{hence, density function for }X \text{ is }\dfrac{d}{dx}(1-\text{e}^{-(\lambda_1+\lambda_2)x})=( \lambda_1+\lambda_2)\text{e}^{-(\lambda_1+\lambda_2)x}

\text{i.e. }X \text{ has an exponential distribution with parameter }(\lambda_1+\lambda_2)

Unparseable latex formula:

\text{if }X\sim \text{e}(\lambda) \text{ then E}(X)=\displaystyle \int_0^{\infty}x\lambda\text{e}^{-\lambda x}dx=\left[-x\text{e}^{-\lambda x}\right]_0^\infty +\displaystyle \int_0^{\infty}\text{e}^{-\lambda x}dx=\left[-\dfrac{1}{\lambda}\text{e}^{-\lambda x}\right]_0^{\infre=ty}=\dfrac{1}{\lambda}

\text{so if }X_1,X_2 \text{ are the waitingt times for A and B respectively we have }\lambda_1=\dfrac{1}{15} \text{ and }\lambda_2=\dfrac{1}{30}

\text{The time to the next bus will be }X\sim\text{exp}\left(\dfrac{1}{15}+\dfrac{1}{30}\right)=\text{exp}\left(\dfrac{1}{10}\right)

\text{so if }m \text{ is the median waiting time we require }m\text{ such that }

\displaystyle \int_0^m \dfrac{1}{10}\text{e}^{-0.1t}dt=\dfrac{1}{2} \Rightarrow \left[-\text{e}^{-0.1t}\right]_0^m=0.5\Rightarrow1-\text{e}^{-0.1m}=0.5

\text{so e}^{-0.1m}=0.5 \Rightarrow0.1m=\ln2 \Rightarrow m=10\ln2=6.93\approx7 \text{ minutes}

Reply 147

2002 STEPIII question 14

I think this will complete the 2002 solutions

\text{Var}(X+Y)=\text{E}(X+Y)^2-(\text{E}(X+Y))^2=\text{E}(X+Y)^2-(\text{E}(X))^2-2\text{E}(X)\text{E}(Y)-(\text{E}(Y))^2

Unparseable latex formula:

=\text{E}(X^2)+2\text{E}(XY)+ \text{E}(Y^2)-2\text{E}(X)\text{E}(Y)-(\ytext{E}(X))^2-(\text{E}(Y))^2

Unparseable latex formula:

=\text{E}(X^2)-(\text{E}(X))^2+2\text{E}(XY)-2\text{E}(X)\text{E}(Y)+ \text{}E}(Y^2)-(\text{E}(Y))^2

=\text{Var}(X)+2\text{Cov}(X,Y)+\text{Var}(Y)

(i) W+L+D=M \text{ a constant, hence }W+L+D \text{ has variance }0

(ii) \text{In each game the probability of a win or a loss is } \dfrac{2}{3} \text{ independently of any other game}

\text {hence }W+L \text{ has the binomial distribution }B\left(m,\dfrac{2}{3}\right)

W \sim B\left(m,\dfrac{2}{3}\right) \text{ so Var}(W)=\dfrac{2}{9}m \text{ and similarly Var}(L)= \dfrac{2}{9}m

\text{hence, Cov}(W,L)= \dfrac{1}{2}\left(\dfrac{2}{9}m-\dfrac{2}{9}m-\dfrac{2}{9}m\right)=-\dfrac{1}{9}m \text{ so } \dfrac{ \text{Cov}(W,L)}{ \sqrt{\text{Var}(W) \text{Var}(L)}}= \dfrac{-\frac{1}{9}m}{\sqrt{\frac{4}{81}m^2}}=-\dfrac{1}{2}

Reply 148

Original post by toasted-lion

Question 5, I'll pick up where Dadeymi left off:

(b-1)\left(ab^2 + (a-1)b +(a-1)\right)=0

Now,

a_1 \not= a \implies k \not= \frac{1}{1-a} \implies b \not= 1

So

\left(ab^2 + (a-1)b +(a-1)\right)=0

\implies \left( b + \frac{a-1}{2a} \right)^2 = \frac{1-a}{a} - \frac{a^2 - 2a +1}{4a^2} = \frac{-5a^2 + 6a - 1}{4a^2}

\implies b = \frac{1-a}{2a} \pm \frac{\sqrt{-5a^2 +6a - 1}}{2a}

\implies k = \frac{1}{2a} \pm \frac{\sqrt{-5a^2 +6a - 1}}{2a(1-a)}

I think it should be

\implies \left( b + \frac{a-1}{2a} \right)^2 = \frac{1-a}{a} + \frac{a^2 - 2a +1}{4a^2} = \frac{-3a^2 + 2a + 1}{4a^2}

\implies b = \frac{1-a}{2a} \pm \frac{\sqrt{-3a^2 +2a + 1}}{2a}

\implies k = \frac{1}{2a} \pm \frac{\sqrt{-3a^2 +2a + 1}}{2a(1-a)}

Reply 149

Blutooth

16

Original post by klausw

I think it should be

\implies \left( b + \frac{a-1}{2a} \right)^2 = \frac{1-a}{a} + \frac{a^2 - 2a +1}{4a^2} = \frac{-3a^2 + 2a + 1}{4a^2}

\implies b = \frac{1-a}{2a} \pm \frac{\sqrt{-3a^2 +2a + 1}}{2a}

\implies k = \frac{1}{2a} \pm \frac{\sqrt{-3a^2 +2a + 1}}{2a(1-a)}

Nice one Klaus from the hill near a mill :smile:

.

Thx!!

Original post by Glutamic Acid

II/7: (Scary scary vectors.)

??????????

Thank you for your solution！

However I think there is a mistake in the final part:

Firstly, after the line
The second equation gives

\alpha = \dfrac{2}{2 - 3 \lambda}

.
When you substitute into the third it gives not

\beta = \dfrac{2}{2 - 3 \lambda}

but

\beta = \dfrac{-2}{2 - 3 \lambda}

and if you substitute these values of alpha and beta into the first equation you will find the

\lambda

terms don't cancel out. Finally you will get a root

\lambda=2/3

But this is not correct as

\lambda=2/3

makes

\alpha = \dfrac{2}{2 - 3 \lambda}

meaningless and also if you substitute it into the second equation you get 0=2/3.

The correct "zero value" of

\lambda

should arise when you substitute

\alpha = \dfrac{2}{2 - 3 \lambda}

into the third equation. Here you must have cancelled a

\lambda

from both side and this is exactly the "zero value" of

\lambda

required by last part of the question.

I don't know if I have made any stupid mistakes here and I hope someone can check my work. And if I am right, I need to say that I still don't know why the root

\lambda=2/3

arises here.

(edited 12 years ago)

Reply 152

8inchestall

Step 2 question 2

the one where you sub

w = z - z^{-1}

followed most of the soln on here but for the second part of the question, after dividing the whole thing by

z^{4}

i ended up with the same equation

2w^{4}-3w^{3}-4w^2+3w+2

and then i just guessed factors which worked to factorise it into

(w-1)(w-2)(2w+1)(w+1)=0

and put each of the values of w into

z^2-wz-1=0

but i got 8 solutions (each value of w giving a +/- value from quadratic formula.. where have i gone wrong?

Reply 153

desijut

10

Original post by 8inchestall

Step 2 question 2

the one where you sub

w = z - z^{-1}

followed most of the soln on here but for the second part of the question, after dividing the whole thing by

z^{4}

i ended up with the same equation

2w^{4}-3w^{3}-4w^2+3w+2

and then i just guessed factors which worked to factorise it into

(w-1)(w-2)(2w+1)(w+1)=0

and put each of the values of w into

z^2-wz-1=0

but i got 8 solutions (each value of w giving a +/- value from quadratic formula.. where have i gone wrong?

Just did this question. Instead of guessing factors I used

a = w - w^{-1}

And got

2a^{2}-3a =0

and then

a(2a-3)=0

and then got to the same solutions of w as you.

And yes i got 8 solutions, should you not get 8 if the polynomial is of order 8? I need someone to confirm this.

Answers i got:

Spoiler

(edited 11 years ago)

Reply 154

8inchestall

Original post by desijut

Just did this question. Instead of guessing factors I used

a = w - w^{-1}

And got

2a^{2}-3a =0

and then

a(2a-3)=0

and then got to the same solutions of w as you.

And yes i got 8 solutions, should you not get 8 if the polynomial is of order 8? I need someone to confirm this.

Answers i got:

Spoiler

i got the same solutions as you except for this one:

Spoiler

Reply 155

desijut

10

Original post by 8inchestall

i got the same solutions as you except for this one:

Spoiler

Yes, my bad, copied it down wrong, i got that too

Reply 156

Original post by SimonM

STEP III, Question 1

Spoiler

Just a typo

\displaystyle \pi \left [ - \frac{(\ln x)^2}{x} - \frac{2 \ln x}{x} - \frac{2}{x} \right ]_1^a =

\displaystyle \pi \left ( 2 - \frac{(\ln a)^2}{a} - \frac{2 \ln a}{a} - \frac{2}{a} \right )

As

a \to \infty

, the volume tends to

2 \pi

(edited 11 years ago)

Reply 157

Original post by Dadeyemi

Some more;

Did these quite a while ago I'm afraid some may be partial solutions.

Agree with most of the III Q4 solution here but for the final bit the only solutions I get is (0, 0), (-1, 0), and(0, 1), any one any comments?

Reply 158

Intriguing Alias

16

Original post by klausw

Agree with most of the III Q4 solution here but for the final bit the only solutions I get is (0, 0), (-1, 0), and(0, 1), any one any comments?

They are correct (assuming you've written it as y^3 - x^3 = y^2 + x^2 and your solutions are of the format (x,y)). But surely you can see that (1, -1) (and vice versa) is also a solution. Even if you don't see where it comes from you can see that it works.

You don't really need much confirmation from other people. Sub it in, if it works then it's a solution.

Reply 159