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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 178

A bag has marbles of 4 colors: red, white, blue, and green. Assume that if we take four marbles out at random (without replacement), each of the following is equally likely:

(1) one marble of each color is chosen,

(2) one white, one blue, and two reds are chosen,

(3) one blue and three reds are chosen,

(4) all four are red.

What is the smallest possible number of marbles in the bag?

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Hi;

Sorry, I could not meet the time, I was sleeping.

The smallest I can come up with is a bag containing 11 reds, 3 whites, 2 blues and 5 greens.

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

**If it ain't broke, fix it until it is.**

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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 178

Its okay, I got 21

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Sorry, that is what I got too, I meant 3 whites not 4!

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

**If it ain't broke, fix it until it is.**

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**Tpetrie01****Guest**

Sorry, I know this post is old - but can you give me a hint as to how you figured this problem out, bobbym? I've got a similar problem and i want to know which direction i should be thinking in.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Hi;

You would use the multivariate hypergeometric distribution. After that you would try it until you found the right number. This would require some trial and error best done by computer.

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

**If it ain't broke, fix it until it is.**

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**Tpetrie01****Guest**

Thanks! This helps a lot.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

You know how to use the multivariate hypergeometric?

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

**If it ain't broke, fix it until it is.**

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**deoxysxxxx****Member**- Registered: 2014-12-21
- Posts: 8

Is there an easy way to solve it without using some complex formula thing?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Hi;

I do not see how you can avoid either more complicated math or some trial and error. This is a fairly tough problem to me. Where is it being asked?

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

**If it ain't broke, fix it until it is.**

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**deoxysxxxx****Member**- Registered: 2014-12-21
- Posts: 8

An online class called Art Of Problem Solving. Have you heard of it?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

I am a member of the AOPS. I have never seen any solution to this posted over there.

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

**If it ain't broke, fix it until it is.**

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**deoxysxxxx****Member**- Registered: 2014-12-21
- Posts: 8

That's weird. It's problem number 5, week 12 for Introduction to Counting and Probability

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

What solution do they post for it?

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

**If it ain't broke, fix it until it is.**

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**deoxysxxxx****Member**- Registered: 2014-12-21
- Posts: 8

I need to wait for it to be posted xD

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,052

Hi;

The problem appears here (#15.23), with the solution below it.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**deoxysxxxx****Member**- Registered: 2014-12-21
- Posts: 8

I actually solved it this morning. Wow, my solution was the same as that one!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

It is much easier by computer.

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

**If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,052

Hi Bobby,

I ran 200,000-iteration simulation in Excel approx 5 times for each of the A, B and C scenarios, with the following results:

```
rwbg | wbrr | brrr | rrrr
-----------------------------------
A. For r = 11, w = 3, b = 2, g = 5: 19% | 21% | 25% | 35%
B. For r = 10, w = 3, b = 2, g = 5: 22% | 22% | 25% | 31%
C. For r = 9, w = 3, b = 2, g = 5: 26% | 24% | 24% | 26%
```

If I'm reading that right, C gives the most even spread across the four marble groups, and if so, then my results differ from everyone else's. But I could easily have muffed something because I haven't learnt probabilities...which is one reason for using E instead of M.

The percentage figures relate to the number of times each marble group was successfully chosen in the random selections.

*Last edited by phrontister (2014-12-22 21:48:04)*

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Hi;

If I'm reading that right, C gives the best result,

Why?

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

**If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,052

...each of the following is equally likely

I thought that meant that the four groups should all be chosen approx the same number of times in a simulation.

"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Hi;

It means that if you had the some composition of balls in that urn and you picked 4, each of these choices

(1) one marble of each color is chosen,

(2) one white, one blue, and two reds are chosen,

(3) one blue and three reds are chosen,

(4) all four are red.

would be equally likely.

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

**If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,052

Yes, that's what I understood it to mean, but I thought that a simulation would work on this problem.

I tried various colour combinations and numbers of marbles, but the A, B and C scenarios gave the best results.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 95,085

Let's take a look at A:

For r = 11, w = 3, b = 2, g = 5:

Your chart shows that rwbg, wbrr, brrr, rrrr all have different percentages. You must find the number of r's, w's, b's and g's that when you draw rwbg, wbrr, brrr, rrrr from that urn the percentages are the same.

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

**If it ain't broke, fix it until it is.**

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**phrontister****Real Member**- From: The Land of Tomorrow
- Registered: 2009-07-12
- Posts: 4,052

My simulation method could never give the same answers for the four groups.

Maybe my 5 times simulation @ 200000 iterations isn't enough. I keep getting similar results for each simulation group, so my code might be out.

Is an accurate simulation possible?

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