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Probability of Choosing 2 Coprime Numbers.md

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+#Math #NT #Probability
+
+# Problem
+
+Calculate:
+
+$$
+P(x, y \in \mathbb{N}: gcd(x, y) = 1)
+$$
+
+# Solution
+
+Each number has a $\frac 1 p$ chance to be divisible by prime $p$, so the probability that two numbers do not share prime factor $p$ is
+
+$$
+1 - p^{-2}
+$$
+
+Therefore, the probability two numbers are coprime is:
+
+$$
+\prod _{p \in \mathbb{P}} 1 - p^{-2}
+$$
+
+Since $1 - x = (\frac 1 {1 - x})^{-1} = (\sum _{n = 0}^{\infty} x^n)^{-1}$, we can express the above as: 
+
+$$
+(\prod _{p \in \mathbb{P}} \sum _{n = 0}^{\infty} p^{-2n})^{-1}
+$$
+
+We can choose any $n$ for $p^{2n}$ for each prime $p$, so by the Unique Factorization Theorem (any natural number can be prime factored one and only one way), we get:
+
+$$
+(\sum _{n = 0} n^{-2})^{-1} \\
+= (\frac {\pi^2} 6)^{-1} \\
+= \frac 6 {\pi^2}
+$$