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**cooljackiec****Member**- Registered: 2012-12-13
- Posts: 185

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: 107,182

Hi;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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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: Harlan's World
- Registered: 2011-05-23
- Posts: 16,018

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)*

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The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,182

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,018

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

The knowledge of some things as a function of age is a delta function.

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,018

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

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,182

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.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,018

Okay. See you later.

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

The knowledge of some things as a function of age is a delta function.

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**Fruityloop****Member**- Registered: 2009-05-18
- Posts: 135

I think the answer is 13/14.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,182

Hi;

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

Thank you spotting that everyone.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 571

Hello;

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,182

For the other M:

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

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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

Maxima:

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

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