Please Post By 10:30 Eastern Time!!!

cooljackiec · 2013-01-31 12:40:55

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?

bobbym · 2013-01-31 20:56:58

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.

cooljackiec · 2013-02-01 11:48:49

Its okay, I got 21

bobbym · 2013-02-01 18:45:50

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

Tpetrie01 · 2014-02-06 14:11:48

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 · 2014-02-06 14:20:35

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.

Tpetrie01 · 2014-02-06 14:39:40

Thanks! This helps a lot.

bobbym · 2014-02-06 14:42:14

You know how to use the multivariate hypergeometric?

deoxysxxxx · 2014-12-21 10:22:26

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

bobbym · 2014-12-21 12:04:26

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?

deoxysxxxx · 2014-12-21 12:08:22

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

bobbym · 2014-12-21 12:14:41

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

deoxysxxxx · 2014-12-22 03:46:19

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

bobbym · 2014-12-22 07:54:06

What solution do they post for it?

deoxysxxxx · 2014-12-22 10:02:58

I need to wait for it to be posted xD

phrontister · 2014-12-22 11:43:52

Hi;

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

deoxysxxxx · 2014-12-22 12:09:15

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

bobbym · 2014-12-22 12:56:27

It is much easier by computer.

phrontister · 2014-12-22 21:37:28

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)

bobbym · 2014-12-22 21:45:40

Hi;

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

Why?

phrontister · 2014-12-22 21:51:13

...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.

bobbym · 2014-12-22 22:01:09

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.

phrontister · 2014-12-22 22:06:17

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.

bobbym · 2014-12-22 22:14:27

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.

phrontister · 2014-12-22 22:25:33

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?

Math Is Fun Forum

#1 2013-01-31 12:40:55

Please Post By 10:30 Eastern Time!!!

#2 2013-01-31 20:56:58

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#3 2013-02-01 11:48:49

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#4 2013-02-01 18:45:50

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#5 2014-02-06 14:11:48

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#6 2014-02-06 14:20:35

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#7 2014-02-06 14:39:40

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#8 2014-02-06 14:42:14

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#9 2014-12-21 10:22:26

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