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Topic: No of Independent Event.. (Read 856 times) |
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k3rn3l
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Posts: 6
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No of Independent Event..
« on: Feb 14th, 2009, 10:21am » |
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Given n ( Sample Space = S) and two events A (subset of S ) and B ( subset of S)..How many possible combo of A & B will be independent event..??
n can be considered to be a dice with n faces wth equal probability of each face..
Like for example n = 3;
Total number of independent event is 13
Here it is :
Code:(1,2,3) ( 1 )
(1,2,3) ( 2 )
(1,2,3) ( 3 )
(1,2,3) ( 1,2 )
(1,2,3) ( 1,3 )
(1,2,3) ( 2,3 )
total = 6;
Reversing the set : again 6 so total = 12;
and one combo is (1,2,3) and (1,2,3)
therefore total = 13;
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Quote: independent event A and B if:
P ( A intersection B) = P(A) * P(B);
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I know the value will increase rapidly with n ...so output will be d value after taking mod of 100000000.
Thnks in advance...i tried it solving for 2 hrs but i m not able..!!
Some ans i may give...for ur checking:
n = 6 ----> 845
7 ----> 253
8 ----> 7509
9 ---->16141
Code:
and i have another question T(n) = 1/sqrt(n)..
what will be T(n)..??
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| « Last Edit: Feb 14th, 2009, 12:09pm by k3rn3l » |
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towr
wu::riddles Moderator
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Some people are average, some are just mean.
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Posts: 13730
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Re: No of Independent Event..
« Reply #1 on: Feb 14th, 2009, 1:26pm » |
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Here's a bit of javascript that solves the first problem in O(n3)-ish time.
There's definite room for improvement, though (e.g. you can solve for c) . But I don't plan on working on this any more tonight.
Code:<script>
function fac(n)
{
if (n < 2) return 1;
return n * fac(n-1);
}
for(s=1; s < 10; s++)
{
t=0;
for(a=0; a <=s; a++)
for(b=0; b <=s; b++)
for(c=1; a+b+c <= s ; c++)
if (c*s == (a+c)*(b+c))
{
t += fac(s)/fac(a)/fac(b)/fac(c)/fac(s-a-b-c)
}
document.write("s: " + s + " " + t + "<br>");
}
document.close();
</script>
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It's interesting how the results of prime numbers jump out. (Although not surprising, once you understand how this algorithm works)
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| « Last Edit: Feb 14th, 2009, 1:32pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Eigenray
wu::riddles Moderator
Uberpuzzler
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Re: No of Independent Event..
« Reply #2 on: Feb 14th, 2009, 6:54pm » |
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Using the same formula, a Mathematica one-liner:
Code:| Sum[Total[s!/(c!(#-c)!(s c/#-c)!(s - # - s c/# + c)!) &/@Select[Divisors[s c], c <= # <= s &]], {c,1,s}] |
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| « Last Edit: Feb 14th, 2009, 6:55pm by Eigenray » |
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