Odds that a random number is prime

A random 321,000 (decimal) digit number is chosen. What is the probability that it's prime if it...

a.) Is an odd number?
b.) Doesn't have any factors below one billion (10^9)?
c.) Doesn't have any factors below one trillion (10^12)?

For part A, I got 1 in 370,000 by taking ln(10^321000) and dividing it by 2. Is this right? Also, could you help me out with parts B and C?

[wblipp]

http://en.wikipedia.org/wiki/Mertens%27_theorems

The third theorem is usually rearranged to get an estimate.

[R.D. Silverman]

Originally Posted by Number theory

A random 321,000 (decimal) digit number is chosen. What is the probability that it's prime if it...

a.) Is an odd number?
b.) Doesn't have any factors below one billion (10^9)?
c.) Doesn't have any factors below one trillion (10^12)?

For part A, I got 1 in 370,000 by taking ln(10^321000) and dividing it by 2. Is this right? Also, could you help me out with parts B and C?

(a) 2* [pi(10^321000) - pi(10^320999)]/(10^321000 - 10^320999)

(b) approx. exp(gamma)/[9 * log(10)]

(c) approx. exp(gamma)/[12*log(10)]

Where pi is the prime counting function and gamma is Euler's constant.
Look up Mertens' Theorem.

Your answer for (a) is approximately correct.

--- In primenumbers@yahoogroups.com, James Wanless <james@...> wrote:

Hello All,
I am interested in trying to derive (an estimate of) an expression for
the likelihood [expressed as a probability between 0 and 1] that a given
generic number (eg random) N is prime, given that no factors of it have
yet been found of a size up to some lesser integer, M.
I found this expression, which makes use of Merten's Theorem:
http://home.earthlink.net/~elevensmo....html#Residual
However, on first seeing Merten's Theorem I (mistakenly?) assumed that
this _in itself_ [ie in entirety] was the answer I needed above, rather
than the somewhat more complex answer in the link. I also (mistakenly?)
had guestimated to myself (on very first thoughts) that my required
expression might actually be independent of N [so long as within certain
limits, eg M<=sqrt(N)].
Note that this is pretty much the case with pure Merten's.
Surely, therefore, what I am saying is that the expression in the link
is too simplistic [or rather actually over-complex], because, for
instance, it does not take into account the fact that not all primes in
its population under consideration are equally likely - in fact the
larger ones will be much rarer, probability-wise.
Am I going nuts here, or can someone else see what I'm getting at?
[obviously - for my nefarious purposes :-) - ie gaining some evidence
that MM127 is prime :-) - I'm looking for a probability answer as high
as possible, but I do genuinely believe atm that the currently accepted
estimate is too low...]
Please help!
James

Firstly, thanks for your input regarding this, folks...
I too have been thinking about it a bit more, and this, I think is my
query/problem/objection:
********************************************************************
Suppose you give me a supposedly 'random' integer N.
Suppose I then discover (empirically) that it is not divisible by any
number (or indeed prime number) less than say 1000000 [ie M=1000000]
Am I not entitled to be surprised - _regardless_ of the size of number N
you originally gave me?
********************************************************************
ie - just a 'pure' Mertens result - rather than the lowering adjustment
to Mertens...
(IMHO) Probability is _so_ tricky, so I'm not claiming infallibility
here, by any means - but I'd really appreciate it if someone could clear
this up for me!


I _think_ I might finally have an answer to this (after much consideration...)??
[Mertens' theorem is a red-herring to this situation]
*********************************************
pr(N prime | no factors up to M)
= ln (M^2) / ln N
*********************************************
J

...which would mean that the 'surprise factor' to discover that N has no factors up to M _would_ be independent of N, and just:
******************************
ln (M^2)
******************************
Now how exactly one interprets that 'surprise factor'...?
J

Thanks.