STEP I, II, III 1999 solutions

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Reply 80

Aurel-Aqua

STEP I, Question 5

(Please someone check this, I'm not very confident, because I did last year)

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Reply 81

Aurel-Aqua

STEP I 1999 Question 7

Could somebody verify the method used in the third part? It really would not hold up as any kind of a proof in my eyes. I did this one last year, too.

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Reply 82

DFranklin

Aurel-Aqua

STEP I 1999 Question 7

Could somebody verify the method used in the third part? It really would not hold up as any kind of a proof in my eyes. I did this one last year, too.I think you need to show the initial conditions are satisfied, but it's otherwise OK (if a little indirect).

The following isn't really any different to what you wrote, but it feels a little bit better, somehow:

My version

I don't think you'd lose more than 2 marks for what you did though (and that would basically be because of not checking the initial conditions).

Reply 83

Aurel-Aqua

STEP I 1999 Question 10

(Someone please verify that my assumption is correct).

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Reply 84

Miccoli

Aurel-Aqua

STEP I, Question 5

(Please someone check this, I'm not very confident, because I did last year)

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I think that's right, although I used cos instead of sin.

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Reply 85

DFranklin

I don't see how you can assume the string to be the same as two separate equal length ones, when we're given there are two lengths a and b.

Reply 86

Aurel-Aqua

DFranklin

I don't see how you can assume the string to be the same as two separate equal length ones, when we're given there are two lengths a and b.

Oh, my mistake... I think I misread my working from long time ago.

Reply 87

Aurel-Aqua

STEP I 1999 Question 14

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Reply 88

SimonM

STEP I, Question 2

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Reply 89

Aurel-Aqua

STEP II, Question 10
Please check it :smile:

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Reply 90

SimonM

STEP II, Question 13

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Reply 91

Aurel-Aqua

STEP II, Question 13
Don't even quote me on this question, there is absolutely no guidance within it.

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Reply 92

SimonM

Heh. Different method, different answers. Not great :s-smilie:

Reply 93

DFranklin

Simon: Surely you really want to have p(R < r) = 1 in the case where r=1?

Edit: I don't think you're taking into account that your uniform variable is on [0,1/2], not [0,1], (and so has density 2 instead of density 1).

Reply 94

SimonM

DFranklin

Simon: Surely you really want to have p(R < r) = 1 in the case where r=1?

This is true.. Better check my logic again

Ah... spotted the problem. Not too bad to fix

Edit: Yeah, that was what I ascertained. Stupid mistake

Reply 95

SimonM

STEP II, Question 14

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Reply 96

squidfuji

squeezebox

Ok, so heres my working:

I get the integral down to;

[br] \frac{3a^{2}}{8} \displaystyle\int^{2\pi}_{0} \sin^{2}(2t) \, dt

which becomes:

\frac{3a^{2}}{16} \displaystyle\int^{2\pi}_{0} 1-cos(4t) \, dt

which is equal to

\frac{3a^{2}\pi}{8}

I think i got the same answer as yours :biggrin:

Reply 97

Agrippa

For paper III Q2. PDF attached.

STEP III 1999 Q2.pdf274.3KB

Reply 98

steppity-step

STEP III Question 13

I am not certain that this method is correct, so it would be very much appreciated if somebody were to follow my working and check I have not made any wrong assumptions.

Firstly,

\int_0^1 f(x) dx = 1 \rightarrow A = 2

\int_0^1 x.f(x) dx = \left[ 2 x^3/3 \right]_0^1 = 2/3

ie expected proportion of cake.

Currents are randomly placed so the expected number of currents = 4 * 2/3 = 8/3

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

I have spent a whole day thinking about pdfs and probabilities and venn diagrams and integrals and have got about 100 wrong answers to this question, let alone a lot of messing about with excel formulas to convert rand() into the continuous random variable you want, so now my brain hurts! But, I think I finally understand how to do this question. If anybody has any comments, that would be great, and also if anybody could point me in the direction of similar questions to practice this, that would be good too, as I'm hoping that probability will be one of my specialist subjects in STEP. Now to attempt number 14...

Reply 99

maltodextrin

speedy_s

I was wondering if someone could help me with the last part of STEP II Q5

basically what i tried to do:

cos a = \frac{35-12\sqrt{120}}{169}

and we have to show a < 3pi/4

so

cos a > -\frac{1}{\sqrt{2}}

putting the two together I got

-\frac{1}{\sqrt{2}} < \frac{35-12\sqrt{120}}{169}

which leads to

35\sqrt{2} - 12\sqrt{240} +169 > 0

which is obviously true since

12\sqrt{240}

is just less than 192. Taking

35\sqrt{2}

as just more than 35 the number is > 0. Hence a is <3pi/4.

But, I am not sure if this is a valid argument to secure all the marks.

I am pretty sure there is a better method and if some one can shed some light on it, it would be great!

I used the fact that 120 < 121 thus 120^1/2 < 11, to show that

(12(120^1/2) - 35)/169 > 97/169

Then proved by contradiction that 97/169 < (1/2)^1/2 and with the aid of a sketch it follows that a < 3pi/4. You end up having to calculate 169^2 but it avoids making approximations. It put me off that asked you to proove this 'carefully', makes me feel like i haven't been thorough enough.