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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: 86,366

Hi;

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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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: 15,522

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.

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: 86,366

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.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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.

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: 86,366

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.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

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: 86,366

Hi;

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

Thank you spotting that everyone.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 291

Hello;

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

For the other M:

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

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**gAr****Member**- Registered: 2011-01-09
- Posts: 3,479

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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