Re: Probability of picking a positive rational number at random

On Mar 13, 5:20 pm, quasi <qu...@xxxxxxxx> wrote:
On Thu, 13 Mar 2008 15:03:57 -0700 (PDT), S_Pa...@xxxxxxxxxxx wrote:
I am thinking that if i picked a random integer n>=1 i could use this
distribution:
The probability of the exponent of its 2 prime factor=0 is 1/2
The probability of the exponent of its 2 prime factor=1 is 1/4
...
The probability of the exponent of its 3 prime factor=0 is 2/3
The probability of the exponent of its 3 prime factor=1 is 2/9
The probability of the exponent of its 3 prime factor=2 is 2/27
...
The probability of the exponent of its n'th prime factor=k is (Pn-1)/
Pn^k
Each prime factor would be independent of the others.

It's not a valid distribution since, for every natural number n, the
probability that the result of your proposed experiment yields the
value n is 0.

quasi

I was thinking it was a valid distribution because if i summed the
probabilities for each possible whole number the sum would be 1.

The probability of picking a 1 would be:
p1=(1/2 * 2/3 * 4/5 * 6/7 ...)
The probability of picking a 2 = p(1)/2
The probability of picking a 3 = p(1)/3
The probability of picking a 4 = p(1)/4
...
Thus the sum of all the probalities is (p(1) * harmonic series)
But p(1)= (1 / harmonic series) according to the Euler Product
.

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