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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: 104,193

Hi;

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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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: 16,015

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

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: 104,193

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.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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: 16,015

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: 104,193

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.** **A number by itself is useful, but it is far more useful to know how accurate or certain that number is.**

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

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: 134

I think the answer is 13/14.

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

**The eclipses from Algol 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: 104,193

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.**

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 510

Hello;

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,193

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.**

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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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