\chapter{组合概率}
\section{事件与概率}
略。

\section{独立重复试验}
略。

\section{大数定律}
抛一枚硬币$n$次，正面朝上的次数记为$X$，有$X \sim B \left(n, \dfrac{1}{2}\right)$。对$k \in [0, n]$，有$P(X = k) = \dfrac{\binom{n}{k}}{2^n}$。

\begin{theorem}[Bernolli大数定律]\label{Bernolli's law of large numbers}
    $\forall \varDelta > 0$，当$n \to \infty$有 \[P\left(\left|\dfrac{X}{n} - \dfrac{1}{2}\right| < \varepsilon\right) \to 1 \eqper\]
\end{theorem}

\begin{theorem}\label{corollary of Bernolli's law of large numbers}
    若$X \sim B \left(2n, \dfrac{1}{2}\right)$，则$\forall t \in [0, n]$，有
    \[P(\vert x - n \vert > t) \leq e^{-\frac{t^2}{n+t}} \eqper\]
\end{theorem}

我们先证明定理\ref{corollary of Bernolli's law of large numbers}。
\begin{proof}
    注意到
    \begin{equation*}
        \begin{aligned}
            P(\vert X - n \vert > t) & = P(X < n - t \text{或} X > n + t)\\
            & = P(X = 0) + P(X = 1) + \dots + P(X = n - t - 1)\\
            & \ \ \ + P(X = n + t + 1) + \dots + P(X = 2n)\\
            & = 2[P(X = 0) + \dots + P(X = n - t - 1)]\\
            & = \frac{2\left[\binom{2n}{0} + \dots + \binom{2n}{n-t-1}\right]}{2^{2n}}\\
        \end{aligned}
    \end{equation*}
            应用引理\ref{lemma for Bernolli's law of large numbers}：
    \begin{align*}
        P(\vert X - n \vert > t) & < \binom{2n}{n-t} \bigg/ \binom{2n}{n}\\
        & \leq e^{-\frac{t^2}{n+t}}\eqper \qedhere
    \end{align*}
\end{proof}

我们可利用定理\ref{corollary of Bernolli's law of large numbers}证明定理\ref{Bernolli's law of large numbers}。
\begin{proof}
    我们需要证明
    \[n \to \infty \text{时} P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| < \varepsilon \right) \to 1\]
    只需证
    \[n \to \infty \text{时} P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| > \varepsilon \right) \to 0\]
    注意到
    \begin{equation*}
            P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| > \varepsilon \right) = P\left(\left|X - n\right| > 2n\varepsilon \right)
    \end{equation*}
    应用\ref{corollary of Bernolli's law of large numbers}，这里的$2n\varepsilon$即为定理中的$t$，从而
    \begin{equation*}
        P\left(\left|X - n\right| > 2n\varepsilon \right) \leq e^{- \frac{4n^2\varepsilon^2}{n+2n\varepsilon}}
    \end{equation*}
    当$n \to \infty$时，$e^{- \frac{4n^2\varepsilon^2}{n+2n\varepsilon}} \to 0$，因此$P\left(\left|X - n\right| > 2n\varepsilon \right) \to 0$。
\end{proof}