Calculate the probability that found two numbers only to have common divisor 1 in positive whole number(1...n)

Question

Please help if you can.
Show and explain detailed working so that I would understand.
Thank you very much.

Answer 1

The probability that any number is divisible by a particular prime p is 1 / p. Hence the probability that two numbers are both divisible by that prime is 1 / p^2, and the probability that at least one of them is not is (1 − 1 / p^2). Thus the probability that two numbers are coprime is given by the product of that probability, taken over all primes :-

P = (1 - 1/4)*(1 - 1/9)*(1 - 1/25)*(1 - . . .

By very advanced mathematics it can be shown that this product taken to infinity is related to the value of the Riemann Zeta function for argument 2, and precisely that P = 6 / pi^2 or approximately 0.607927...

Answer 2

Finding the distribution of prime numbers in the integers is an area of active research.

As n → ∞, the proportion of prime numbers tends towards 1/ ln n

so i would say that as n → ∞ and you pick two integer values at random via the integer uniform distribution then you would have an approximate probability of (1 / ln n) ^ 2 that both numbers are prime.

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an interesting side note is that Dennis P. Walsh wrote a short article for the Mathematical Association of America called "A Curious Way to Test for Primes" (published in Mathematics Magazine Vol 80, No. 4, Oct 2007, page 302) where he shows that for each positive integer n > 1 we can define a function g_n (x) such that:

g_n (x) = ∑ exp(x ^ k / k)

where the limits on the sum are k = 1 to n -1.

A positive integer n is prime if and only if the nth derivative of g_n evaluated at 0 equals 1.