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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: 84,467

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 have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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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: 84,467

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 have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**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: 84,467

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 have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**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: 84,467

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.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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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: 84,467

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 84,467

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 ...

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

And...?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,467

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

It sounds like a nigh impossible task...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,467

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 84,467

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?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 15,354

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,467

Hi;

I would say -5 is closer to it.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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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: 84,467

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

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