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**zetafunc.****Guest**

Okay, see you later.

**zetafunc.****Guest**

Also I think I know the solution to the cake problem, it is a lot easier than I thought. I'll post the solution tomorrow.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Hi;

Okay thanks, I have been unable to find any similar problem.

**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**

Nothing from adriana yet, 4 days ago now.

Here's the problem again just as a reminder:

---

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?

---

(i)

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.

(ii) If I take a proportion x of cake, the probability that a currant fell into it is clearly also x. The currant is either in the slice, or not in the slice. So we have a binomial distribution for the number of currants C ~ B(n, p), with p = x and n = 4. So P(C = 4) = x[sup]4[/sup]. The probability distribution for a proportion X AND having 4 currants, is 2x*x[sup]4[/sup] = 2x[sup]5[/sup].

Then we just use P(A n B) = P(A|B)P(B), for which I am getting 63/64.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Uh oh! I like the reasoning for the 2x^5. I think that solves that. But why not use the integral idea that toasted fellow did?

**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.**

**Online**

**zetafunc.****Guest**

That's what I did -- it's a little strange because here we are kind of using individual values in a continuous distribution. You obtain P(4 currants in portion AND more than half a portion) by integrating 2x[sup]5[/sup] between 1/2 and 1. P(4 currants in portion) is obtained just by integrating 2x[sup]5[/sup] between 0 and 1, since the size of the portion doesn't matter for that probability. Putting that together yields the same answer toasted-lion got.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Sure wish we had a few more of these type to try out your idea on..

**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.**

**Online**

**zetafunc.****Guest**

I might come across some more in the 916 more STEP questions I have to do. It is going to be a long summer...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

I do not know how to phrase the problem to search for more. In the meantime would you mind writing up your full solution for it.

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

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**zetafunc.****Guest**

My solution is basically the same as toasted-lion's on The Student Room, the only difference is I explained where the x^4 bit came from...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Okay, then I will use that one with your explanation. I think it is done.

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

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**zetafunc.****Guest**

Okay, I have ticked it off my spreadsheet now.

No comment from adriana about the question. I think I will leave her to figure it out. (That being said, she is scared stiff of STEP III.)

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Good, good, you will be there to maybe comfort her. Tell her it is not possible to explain them with an email or chat. It must be done with paper and pencil and ...

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

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**zetafunc.****Guest**

And...?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Hmmm, can only be done in person is the point.

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

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**zetafunc.****Guest**

It sounds like a nigh impossible task...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Well if you have explained that too her what other choice does she have but to plan something more intimate?

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

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**zetafunc.****Guest**

She could say that she will just not do the question, as there are 1166 others and she is not trying to do all the questions at any rate...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Then you are alleviated from having to explain it to her.

The point is to arrange a meeting not to discuss math by email, is it not?

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

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**zetafunc.****Guest**

I don't know, at the moment I'm not sure what I want to do. Part of me just wants to permanently ignore her, completely forget about her. I don't think she is worth the pursuit. There is now 0 benefit from knowing her.

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,822

Hi zetafunc.

You are studying at the Cambridge university, right?

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Hi;

I would say -5 is closer to it.

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

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**zetafunc.****Guest**

Nope, I was rejected by Cambridge -- for a reason my school found very controversial. I am going to UCL instead, and then maybe go there for my Masters degree instead.

Knowing her does mean she won't be put out of my mind; but I agree, nothing good really comes from talking to her anymore. I wonder if she's told her BF about me at all. Some of the stuff she used to say to me, you'd think that I was her BF! Oh well, I failed this time, but I will get another opportunity at some point, probably.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,350

Probably? I have known mere bumpkins, guys that were barely human that had dozens of opportunities.

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

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**zetafunc.****Guest**

You still read/hear about people who never had any opportunities though, into their middle-ages.