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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

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: 90,886

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 90,886

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.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

I hope not...

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

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: 90,886

Yes an amazing accomplishment.

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

It would give her something to aspire to.

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,886

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

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,886

Hi;

First part looks okay.

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

Hi;

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

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,886

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.

**In mathematics, you don't understand things. You just get used to them.**

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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!

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,886

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

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,886

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.

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,886

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?

**In mathematics, you don't understand things. You just get used to them.**

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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...