MHB Problem of Calculating Probabilities

[PeterJ1]

I hope this question is in the right place.

I'm trying to calculate probabilities and struggling. Hopefully someone can help.

Suppose I want to calculate the probability of N being a product of a prime below sqrt N.

N will have a 1:2 chance of being a product of 2.

If N is not a product of 2 then it will have a 1:6 chance of being a product of 3.

If N is not a product of 2,3 then it will have a 1:15 chance of being a product of 5.

And so on...

How would I arrive at combined probability for the primes up to sqrt N?

It could be a different calc. I could sieve out the products of each primes at each step and then calculate the probabilities for only the numbers that remain. So, for the divisor 2 the odds are 1:2. For 3 the odds are 1/3 but half the numbers have already been eliminated so really it is 1/6. Then I could use 1/3p as a simple estimate for each successive potential divisor, but would have no idea how to sum them.

Either way I cannot see an easy way to do it.

(I know the the PNT gives the probability of a number being prime, by the way, but this is not where I'm going.)

Thanks for any help with this. No doubt the solution looks easy for you guys.

Another way to do it seems to be use a table for the sum of the reciprocals of primes up to sqrt N, but I want to be able to deduct from this primes that are irrelevant. So, if I multiply 2,3,5, 7 and deduct 1 to give N, then I know these primes can be ignored. If I deduct the odds of these four primes being divisors then I can deduct this from the sum of the reciprocals up to N and get some sort of result. But if there are 100 possible divisors to deduct then I'm back where I started, trying to work out the sum of probabilities for these 100 possible divisors.

If I've asked a **** fool question then my apologies.

 

[HallsofIvy]

Re: Problem of Calculating Probabilitiers

I hope this question is in the right place.

I'm trying to calculate probabilities and struggling. Hopefully someone can help.

Suppose I want to calculate the probability of N being a product of a prime below sqrt N.

Do you mean "having a prime factor below sqrt(N)"?

N will have a 1:2 chance of being a product of 2.

Do you mean "being a product of two primes" or do you mean "having 2 as prime factor"? In either case, how do you arrive at "a 1:2 chance"?

If N is not a product of 2 then it will have a 1:6 chance of being a product of 3.

If N is not a product of 2,3 then it will have a 1:15 chance of being a product of 5.

And so on...

How would I arrive at combined probability for the primes up to sqrt N?

It could be a different calc. I could sieve out the products of each primes at each step and then calculate the probabilities for only the numbers that remain. So, for the divisor 2 the odds are 1:2. For 3 the odds are 1/3 but half the numbers have already been eliminated so really it is 1/6. Then I could use 1/3p as a simple estimate for each successive potential divisor, but would have no idea how to sum them.

Either way I cannot see an easy way to do it.

(I know the the PNT gives the probability of a number being prime, by the way, but this is not where I'm going.)

Thanks for any help with this. No doubt the solution looks easy for you guys.

Another way to do it seems to be use a table for the sum of the reciprocals of primes up to sqrt N, but I want to be able to deduct from this primes that are irrelevant. So, if I multiply 2,3,5, 7 and deduct 1 to give N, then I know these primes can be ignored. If I deduct the odds of these four primes being divisors then I can deduct this from the sum of the reciprocals up to N and get some sort of result. But if there are 100 possible divisors to deduct then I'm back where I started, trying to work out the sum of probabilities for these 100 possible divisors.

If I've asked a **** fool question then my apologies.

 

[PeterJ1]

Re: Problem of Calculating Probabilitiers

Do you mean "having a prime factor below sqrt(N)"?

Yep.

Do you mean "being a product of two primes" or do you mean "having 2 as prime factor"?

Having 2 as a factor.

In either case, how do you arrive at "a 1:2 chance"?

Er, one in every two numbers is divisible by 2.