That is, how to calculate it given the size of N(that is, logN) and assuming that logN is much greater than M. Its an approximation. There is no exact formula. 


I do know that according to the prime number theorem the probability for N being prime is 1/ln(N) for large positive integers. 


Thanks.



The probability that a random number N has at least M factors
The count of numbers up to $x$ having at most $k$ *prime* divisors (counted with mutiplicity) is 
asymtotically equal to (see for example [Almost prime](http://en.wikipedia.org/wiki/Almost_prime) on Wikipedia)
$$ \frac{x}{\log x} \frac{(\log \log x)^{k-1}}{(k-1)!} .$$


Thus, since the number of divisors is definitely greater than the number of prime divisors, the probabilty (in the informal sense used in the question) that some $N$ has at least $M\ge 3$ divisors is (asymtotically) at least 
$$ 1 - \frac{1}{\log N} \frac{(\log \log N)^{M-3}}{(M-3)!} .$$


It might well be possible to improve this as there are 'few' numbers for which the number of primes divisors (with mutiplicty) is 'close to' the number of divisors. 


A quick guess, possibly wrong, would be that one might be able to improve this towards something like
$$ 1 - c \frac{(\log \log N)^{\log M}}{\log N}, $$ by looking at distinct prime divisors.


And this would be about optimal, which can be seen via noting that if the number of prime divisors (with multiplicty) is $k$ this bounds the number of divisors by $2^k$, and using the above argument in the other direction. 


