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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Please give me a serious answer.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

I thought I did. The book stiffens up after the first two chapters. I can only get small pieces of it. There is some stuff in later chapters.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

And the 3-day streak is broken -- nothing from adriana today. Perhaps she is just busy.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Or she could be out there finding doctors who want to take blood from her.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I hope not...

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

bobbym wrote:

I thought I did. The book stiffens up after the first two chapters. I can only get small pieces of it. There is some stuff in later chapters.

Hm, the third chapter will really bee a tough nut to crack.

Hi zf.

You guys managed to go over 10000 posts in a thread! That is more than most members on the forum have.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Yes an amazing accomplishment.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

10,000 posts... H would be proud.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

It would give her something to aspire to.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

She could be reading this thread, that's a scary thought.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

I would not worry about it. They love attention, good or bad really doesn't matter.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Who knows what she is doing now...

I am stuck on a probability problem -- from STEP III 1999 Q13 (unfortunately there aren't official solutions for this problem, only an unofficial one which I can't follow):

"The cakes in our canteen each contain exactly four currants, each currant being randomly

placed in the cake. I take a proportion X of a cake where X is a random variable with density

function

f(x) = Ax

for 0 ≤ x ≤ 1, where A is a constant.

(i) What is the expected number of currants in my portion?

(ii) If I find all four currants in my portion, what is the probability that I took more than

half the cake?"

First part seems pretty straightforward.

So f(x) = 2x.

The 4 currents are randomly distributed, so the expected number of currents in my portion is 4*(2/3) = 8/3.

I can find the probability of taking more than half the cake (integrating the PDF between 1/2 and 1 gives 3/4), but not sure what to do about the 4 currants in my portion. It's a conditional probability so I'd also need to know how to find the probability of having 4 currants AND more than half the portion, and I don't know how to do that...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

First part looks okay.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Thanks -- what can I do about the second part?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

That is what I am working on right now but not doing so well.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

That's how I feel about most questions in STEP III...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

That is how I feel about math in general, I am checking some texts for a similar problem because I do not have one solved in my notes.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

Sorry to take so long. I have this guys answer

http://www.thestudentroom.co.uk/showthr … st18366432

It seems to correspond to a simulation I ran. I just do not understand his first step!

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Thanks -- that's the one I've been struggling to understand too. I don't understand the x^4 thing...

Unfortunately it is the only solution I can find on the internet.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

They say that solution is correct later on but I do not get the X^4 either.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

He has said P(4 currants in a portion) = E(X^4)... not sure what to make of that.

Does that imply P(1 currant in a portion) = E(X)?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

Hi;

I do not know the answer to that one. I can not see why he can multiply like that or whether that relation you are stating holds.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

He is multiplying because

I'm still not getting that g(X) = X^4 though...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,476

This is what I meant:

Now taking a proportion X of the cake gives probability X^4 that all 4 currents will be in that portion.

Does that integral you posted apply here?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

I still can't understand why he did that. Everybody afterwards just says they agree with it, even though the original poster says he's not certain it's correct.

Then again I can't think of what other options there are...