diff --git a/docs/math/analysis/fourier.md b/docs/math/analysis/fourier.md
new file mode 100644
index 0000000..36b6174
--- /dev/null
+++ b/docs/math/analysis/fourier.md
@@ -0,0 +1,23 @@
+# Fourier 分析
+
+## Parseval
+
+### $\zeta(2)$
+
+考虑对 $f(x) = x$ 做 Fourier:
+
+$$
+c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} x \cdot \mathrm{e}^{-i n x} \mathrm{d} x = \frac{(-1)^n}{n} i \quad \forall\ n\neq 0
+$$
+
+显然有$c_0 = 0$, 因此根据 Parseval:
+
+$$
+\sum_{n=\infty}^{\infty} |c_n|^2 = \frac{1}{2\pi} \int_{-\pi}^{\pi} x^2 \mathrm{d} x = \frac{\pi^2}{3}
+$$
+
+因此
+
+$$
+\zeta(2) = \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
+$$
diff --git a/docs/math/probability/probability.md b/docs/math/probability/probability.md
index a6ddc98..bdac8ea 100644
--- a/docs/math/probability/probability.md
+++ b/docs/math/probability/probability.md
@@ -1 +1,61 @@
 # 概率论
+
+## Questions
+
+### 2024-09-06 两个数互素的概率
+
+> Also see https://math.stackexchange.com/questions/64498/probability-that-two-random-numbers-are-coprime-is-frac6-pi2
+>
+> 因为很有趣所以对照翻译一遍。
+
+按照这样的思路考虑两个数互素 (_coprime_) 的概率：
+
+-   两个数都是 $2$ 的倍数的概率：$\frac{1}{4}$.
+-   两个数都是 $3$ 的倍数的概率：$\frac{1}{9}$.
+-   $\cdots$
+-   两个数都是 $p$ 的倍数的概率：$\frac{1}{p^2}$.
+
+所以两个数互素的概率是：
+
+$$
+\prod_{p \in \text{prime}} \left(1-\frac{1}{p^2}\right)
+$$
+
+接下来的处理方式是：
+
+$$
+\prod_{p \in \text{prime}} \left(1-\frac{1}{p^2}\right) = \left(\prod_{p \in \text{prime}}\frac{1}{1-p^{-2}} \right)^{-1}
+$$
+
+回顾
+
+$$
+\begin{aligned}
+\frac{1}{1-x} &= 1 + x + x^2 + \cdots \\
+\frac{1}{1-p^{-2}} &= 1 + \frac{1}{p^2} + \frac{1}{p^4} + \cdots \\
+\end{aligned}
+$$
+
+因此有：
+
+$$
+\prod_{p \in \text{prime}}\frac{1}{1-p^{-2}} = (1 + \frac{1}{2^2} + \frac{1}{2^4} + \cdots)\times (1 + \frac{1}{3^2} + \frac{1}{3^4} + \cdots) \times \cdots
+$$
+
+注意到这实际上「组合」出了所有正整数，因为这样的映射是一一对应的：
+
+$$
+x = p_1^{i_1} \cdot p_2^{i_2} \cdots p_n^{i_n}
+$$
+
+其中 $p_i$ 是素数，$i_i$ 是非负整数。因此有：
+
+$$
+\prod_{p \in \text{prime}}\frac{1}{1-p^{-2}} = \sum_{n=1}^{\infty} \frac{1}{n^2} = \zeta(2) = \frac {\pi^2}{6}
+$$
+
+$\zeta(2)$ 的计算可以 [参考](/math/analysis/fourier/#zeta2), 原问题的答案是：
+
+$$
+\left(\prod_{p \in \text{prime}}\frac{1}{1-p^{-2}} \right)^{-1} = \zeta(2)^{-1} = \frac{6}{\pi^2}
+$$
diff --git a/mkdocs.yml b/mkdocs.yml
index b4676e7..171f16a 100644
--- a/mkdocs.yml
+++ b/mkdocs.yml
@@ -34,6 +34,7 @@ nav:
       - math/index.md
       - Analysis:
           - math/analysis/calculus.md
+          - math/analysis/fourier.md
           - math/analysis/functional-analysis.md
       - Probability:
           - math/probability/probability.md