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**zetafunc.****Guest**

What do you mean?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

I have lost the link to the STEP questions and I always come up to the TSR room on a google search. I am not getting 1 / 2 for iii.

**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 am getting 1/2 for that question...

Try mathshelper . co . uk, then go to "Oxbridge Page".

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Hi;

I am getting 1 / 2 using multivariable hypergeometric distribution too but I thought that they wanted the probability of first a 3 and then 2 blank and another number.

That probability is 4 / 167

**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 don't know what a multivariable hypergeometric distribution is...

How are you getting 4/167?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

That is a good question. I am counting them all up by computer.

There are only 12 possibilities: Easy to do on paper.

The exact count of all permutations of 1,2,3,4,5,0,0,0,0,0,0 of length 4 is 501.

The 501 can be verified with an exponential gf:

Coefficient of x^4 is 167 / 8 when you times by 24 you get 501.

12 / 501 = 4 / 167

**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 have been staring at your post for 20 minutes but I can't understand why your answer is different -- unless your answer of 501 is wrong or something... I'm getting 1/2 with my method.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

The 501 is correct by ecxponential gf and by computer count. The 12 is correct because it is easy to do by hand and is just a mississippi problem.

I may have over analyzed the question. He places the 3 in front and calls that 1 / 1 probability and then goes on with the question. I calculated the probability of the 3 in front and then the question. They are obviously not the same!

Short break.

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**

You mean 1 / 11 probability?

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Hi;

Yes, looks like I got bizarre. Any way the multivariable hypergeometric is really nothing more than a discrete distribution that is designed to answer urn problems that have more than 2 different colors without replacement. Amounts to just a formula. It gets 1 / 2.

For iv) This is a conditional probability and I get 1 / 2.

v) 1 / 2 straight hypergeometric question.

vi) Tricky wording. 10 / 21

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**

Yes, those are the answers I got. How do you set up a hypergeometric distribution here? Would it save me time in a problem like this?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

No, the major difference between the kind of math I use and the kind you need to learn is big.

The kind of math I use is geared toward things that CAS and tables do easily. That is why I use them. You are not allowed those things so you have to learn to do without them for now.

School math and industrial math are two very different things.

But I will still show you what it is, I will need some time to get it together.

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**

Okay, thanks. Unfortunately, for the remainder of this year, and probably the next 3-4 years, I must be a slave to the exam.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Yes, the school system and industry do not appear to want to synchronize. School trains you one way and jobs train you for another.

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**

Well, they have vocational courses, BTECs, apprenticeships, etc. over here which are replacements for GCSEs and A-levels, which are notoriously easier but also have a rather more applied perspective on things; a typical maths vocational course might be "Use of Mathematics" or a heavily finance-based maths course. Unfortunately, Michael Gove, the moron that is the education secretary in England, is scrapping those and making things backward again. Thanks to our Conservative government, we could easily be about to witness Thatcher 2.0.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Over here they come out of the best universities, notoriously unable to solve a real world problem. They have to be totally retrained on the job.

(iii) Exactly two numbered discs are taken, given that the disc \

numbered "3" was taken first.

This illustrates a hypergeometric distribution well;

Since we used a 3 we have 6 blank and 4 numbered.

winners = 4, (the 4 numbered balls remaining)

total= 4 + 6 ( total number of balls remaining )

trials = 3 ( you are picking 3 more balls )

num = 1 ( you need 1 more numbered ball )

I think I spoke too soon earlier. The hypergeometric might be difficult to remember but the multivariable version is easier to remember! Want to see 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**

Yes please -- I was just thinking that it might be difficult to remember, I was trying to find similar problems to use it on today. How would you use the multivariable one here?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Best to illustrate with a clear example:

Suppose there are 5 blue, 10 yellow, and 15 red marbles in an urn. You reach in and randomly select nine marbles without replacement. What is the probability that you pick 2 blue, 3 yellow and 4 red?

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**

Oh, I see. So for the STEP question, we have 5 numbered discs and 6 blank discs, so the probability of taking 4 numbered discs (without replacement) is

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Yes, that is correct. The multi one is easy to remember because of the obvious pattern.

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 looks useful to remember. I will write it in my notes... hopefully I'll get a chance to use it this exam season.

Only 4 days to go, almost 3.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

Lots of problems can be thought of as urn problems and the hypergeometric is the distribution to use when there is no replacement.

You will do fine on your exams do not worry.

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 knew there must be a quicker way than just writing out all the probabilities... I always thought those 'without replacement' questions were extremely tedious because of that. Too bad they never teach us anything like this in school.

STEP is in 7 weeks -- looks unlikely I'll finish all the questions now...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,654

It is the difference between school math and math ( I hate to say in the real world because it gets people very angry ).

When they present a problem in school it is to illustrate a principle. The numbers always work out to nice neat answers so that the concept is learned without getting balled up in numerics. They want you to do combinatorics in the slow old way because they feel that drums in the concepts. It is almost never the best way of doing the problem.

Real world problems are very different. The numerics is at least as important as the method, probably more so. Their is no principle to be learned just the need to get an answer as quickly and as accurately as possible. You are allowed to use tables, CAS, books, internet, guessing or tarot cards, just as long as you always have a solution faster than the competitors. When you work for the military, and 75% of all researchers do whether they know it or not you can guess how important skunking the competitors is!

GF's, recurrences, series, lattices, markov chains, iteration and programming now are top dog in the methods used.

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 agree, it has always been a criticism of A-levels too -- Michael Gove isn't really solving the problem though. Terence Tao wrote a problem-solving book with the same point in the foreword.