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Topic review (newest first)

gAr
2013-09-08 01:22:04

Maxima:

Code:

sum(pdf_hypergeometric(i,3,5,4),i,0,2);
bobbym
2013-09-07 21:56:05

For the other M:

Code:

Probability(RandomVariable(Hypergeometric(8, 3, 4)) < 3)
ElainaVW
2013-09-07 20:10:28

Hello;

Code:

Probability[x < 3, 
 x \[Distributed] HypergeometricDistribution[4, 3, 8]]
bobbym
2013-09-06 17:01:59

Hi;

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



Thank you spotting that everyone.

Fruityloop
2013-09-06 14:21:59

I think the answer is 13/14.
(5C4*3C0 + 5C3*3C1 + 5C2*3C2)/8C4 = 13/14.

anonimnystefy
2013-09-06 11:14:20

Okay. See you later.

bobbym
2013-09-06 11:10:56

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.

anonimnystefy
2013-09-06 10:58:36

Here's a simulation that confirms 13/14:

Code:

(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
anonimnystefy
2013-09-06 10:47:39

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

bobbym
2013-09-06 10:38:43

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.

anonimnystefy
2013-09-06 10:19:35

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

Code:

(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)
Anynomous
2013-09-06 09:45:44

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

bobbym
2013-01-23 15:45:31

Hi;

cooljackiec
2013-01-23 14:48:49

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

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