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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 160

I have a bag with 5 pennies and 3 nickels. I draw coins out one at a time at random. What is the probability that I haven't removed all 3 nickels after 4 draws?

I tried 1/2 and 11/14. They are wrong

I see you have graph paper.

You must be plotting something

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,747

Hi;

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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**Anynomous****Guest**

This solution is not right, i tried it. Can you show how you got it, or something

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,846

I think he might be right. I am getting 13/14.

`(Select[Permutations[{p, p, p, p, p, n, n, n}], Count[Take[#, 4], n] < 3 &] // Length)/(Permutations[{p, p, p, p, p, n, n, n}] // Length)`

*Last edited by anonimnystefy (2013-09-05 12:20:32)*

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,747

Hi all;

That is all the ways to draw 4 coins from 5 and 3. There are 15 total ways and four winners. So I will have to go with 11 / 15 jist as before. Also, the OP already verified this.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,846

But, the probabilities of those 4-draws are not all the same.

Here lies the reader who will never open this book. He is forever dead.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,846

Here's a simulation that confirms 13/14:

```
(Table[If[Count[
Delete[Delete[
Delete[Delete[{p, p, p, p, p, n, n, n}, RandomInteger[{1, 8}]],
RandomInteger[{1, 7}]], RandomInteger[{1, 6}]],
RandomInteger[{1, 5}]], n] > 0, 1, 0], {100000}]//Total)/100000//N
```

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,747

Hooohaa as the great Al Pacino would say. You have provided some good evidence, I will rethink the whole thing as soon as I come back.

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,846

Okay. See you later.

Here lies the reader who will never open this book. He is forever dead.

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**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 120

I think the answer is 13/14.

(5C4*3C0 + 5C3*3C1 + 5C2*3C2)/8C4 = 13/14.

The eclipses from Algol (an eclipsing binary star) come further apart in time when the Earth is moving away from Algol and closer together in time when the Earth is moving towards Algol, thereby proving that the speed of light is variable and that Einstein's Special Theory of Relativity is wrong.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,747

Hi;

Yes, that is the correct answer. Verified by the using the hypergeometric distribution.

Thank you spotting that everyone.

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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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 270

Hello;

```
Probability[x < 3,
x \[Distributed] HypergeometricDistribution[4, 3, 8]]
```

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,747

For the other M:

`Probability(RandomVariable(Hypergeometric(8, 3, 4)) < 3)`

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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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,462

Maxima:

`sum(pdf_hypergeometric(i,3,5,4),i,0,2);`

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

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